vedanta Excel In Opt. Mathematics - Book 10
(d) cot2A = cot2A - 1
2cotA
cotA.cotB - 1
Here, cot(A + B) = cotB + cotA
Now, cot2A = cot(A + A) = cotA.cotA -1 = cot2A - 1
cotA + cotA 2cotA
cot2A - 1
? cot2A = 2cotA
II. Trigonometric Ratios of 3A.
(a) sin3A = 3sinA - 4sin3A
Now, sin3A = sin(A + 2A)
= sinA.cos2A + cosA.sin2A
= sinA (1 - 2sin2A) + cosA. 2sinA.cosA
= sinA - 2sin3A + 2sinA.cos2A
= sinA - 2 sin3A + 2sinA(1 - sin2A)
= sinA - 2sin3A + 2sinA - 2sin3A = 3sinA - 4sin3A
? sin3A = 3sinA - 4sin3A
Also, 4sin3A = 3sinA - sin3A
(b) cos3A = 4cos3A - 3cosA
Here, cos3A = cos(A + 2A)
= cosA.cos2A - sinA.sin2A
= cosA. (2cos2A - 1) - sinA.2sinA.cosA
= 2cos3A - cosA -2sin2A.cosA
= 2cos3A - cosA - 2(1 - cos2A) cosA
= 2cos3A - cosA - 2cosA + 2cos3A = 4cos3A - 3cosA.
Also, 4cos3A = 3cosA + cos3A
(c) tan3A = 3tanA - tan3A
1 - 3tan2A
Here, tan3A = tan (A +2A)
= tanA + tan2A = tanA + 2tanA
1-tanA.tan2A 1-tan2A
1- tanA.
tanA-tan3A +2tanA 2tanA
1-tan2A
= 1 - tan2A = 3tanA - tan3A
1 - tan2A -2tan2A 1 - 3tan2A
1-tan2A
? tan3A = 3tanA - tan3A
1- 3tan2A
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vedanta Excel In Opt. Mathematics - Book 10
(d) cot3A = cot3A - 3cotA
3cot2A - 1
Here, cot3A = cot(A + 2A)
= cotA.cot2A - 1 = cotA. cot2A -1 - 1
cot2A + cotA 2cotA
cot2A -1 + cotA
2cotA
= cot3A - cotA - 2cotA . 2cotA
2cotA cot2A - 1+2cot2A
cot3A- 3cotA
= 3cot2A - 1
? cot3A = cot3A - 3cotA
3cot2A - 1
III. Geometrical Interpretation of Trigonometric Ratios of 2A.
Let a revolving line OP make an angle XOP = 2A with an initial line OX.
Now, taking O as the centre and OP as radius of a circle, circle PMN is drawn. MN is the
diameter of the circle. OP, MP, PN are joined. PR is drawn perpendicular to MN. MPN
is semi-circle with a diameter MN, MPN = 90° (angle at semi-circle is right angle).
MNP = 1 NOP = 1 × 2A = A
2 2
(Circumference angle is half central angle standing on the same arc PN)
OPM = OMP = A (? ∆ MOP is an isosceles base angles are equal).
MPR = 90° - A, MPN = 90°, RPN = 90° - MPR = 90° - 90° + A = A
? RPN = A Y
In right angled ∆ PRM, P
sinA = PR , cosA = RM , tanA = PR X' A 2A A X
PM PM MR M RN
In right angled ∆MPN, O
sinA = PN , cosA = PM
MN MN
Now, in right angled ∆ ORP, Y'
(i) sin2A = PR = PR = 2PR = 2PR . PM = 2sinA.cosA
OP MN PM MN
1 MN
2
OR 2OR OR + OR
(ii) cos2A = OP = 2OP = 2OP
= (ON - RN) + (RM - OM)
2OP
OM - RN + RM - OM
= 2OP
= RM - RN
MN
RM RN RM PM RN PN
= MN - MN = PM . MN – PN . MN
= cosA.cosA - sinA.sinA = cos2A - sin2A
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vedanta Excel In Opt. Mathematics - Book 10
(iii) tan2A = PR
OR
PR PR
= 2PR = 2PR = 2. RM = 2. RM
2OR RM-RN
1- RN 1- RN . PR
RM PR MR
2tanA 2tanA
= 1 - tanA.tanA = 1 - tan2A
Worked Out Examples
Example 1. If sinA = 3 , find sin2A, cos2A and tan2A.
Solution: 5
3
Here, sinA = 5
cosA = 1 - sin2A = 1 - 9 = 4
25 5
3
5 3
tanA = sinA = 4 = 4
cosA
5
Now, sin2A = 2sinA. cosA = 2. 53 . 4 = 24
5 25
cos2A = cos2A - sin2A
= 16 - 9 = 16 - 9 = 7
25 25 25 25
24
25
tan2A = sin2A = 7 = 24
cos2A 7
25
Example 2. Prove that, cotT = ± 1+cos2T
Solution: 1-cos2T
We have 2cos2T = 1 + cos2T............ (i)
Example 3.
Solution and 2sin2T = 1 - cos2T ........... (ii)
Dividing (i) by (ii), we get,
2cos2T = 1 + cos2T
2sin2T 1-cos2T
1 + cos2T
or, cot2T = 1-cos2T
? cotT = ± 1+cos2T = RHS Proved.
1-cos2T
4
If sinA = 5 , then, find the values of sin3A, cos3A and tan3A.
Here, sinA = 4
5
16
Now, cosA = 1 - sin2A = 1 - 25 = 3
5
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vedanta Excel In Opt. Mathematics - Book 10
4
5
tanA = sinA = 3 = 4
cosA 3
5
sin3A = 3sinA - 4sin3A
= 3. 4 - 4. 64
5 125
= 12 - 256 = 300 - 256 = 44
5 125 125 125
cos3A= 4cos3A - 3cosA
=4 27 - 3. 3 = 108 - 9 = 108-225 =- 117
125 5 125 5 125 125
3tanA - tan3A
tan3A = 1 - 3tan2A
3. 4 - 64 4– 64 108 - 64 44
3 27 1- 27 7(3 - 16) 117
= = 16 = = –
16 3
1- 3. 9
Example 4. (If = 1 m + 1 (, then, show that :
Solution: sinA 2 m ((
((a) 1 m2 (+1 ((b) = - 1 m3 + 1
cos2A = - 2 (m2 sin3A 2 m3
((
(Here, sinA = 1 m + ((1
2 m
(
(a) cos2A = 1 - 2sin2A
(
(= 1 - 2 . ( 1 m + 1 2
4 m
(= ( 1 1 1
((1 - 2. 4 m2 + 2.m. m + m2
=( 1- 1 m2 - 1 - 1 . 1
2 2 m2
(=
- 1 m2 + 1 = LHS. Proved.
2 m2
(b) sin3A = 3sinA - 4sin3A
(= 3 1 m+ 1 (- 4 .1 m + 1 3
2 m 8 m
[ ( ]3 -
(= 1 m+ 1 m + 1 2
2 m m
[ ]3
(= 1 m+ 1 - m2 - 2.m. 1 - 1
2 m m m2
[ ]-
(= 1 m+ 1 (m2 - 1 + 1
2 m m2
( (=
– 1 m + 1 m2 - m. 1 + 1
2 m m m2
(= –
1 m3 + 1 = RHS Proved.
2 m3
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Example 5. Prove the following:
Solution:
(a) sin2A - sinA = tanA (b) 1- cosA = tan2A
1-cosA + cos2A 1+ cos2A
(d) 1 + sec2A = cotA
(c) tanA + cotA = 2cosec2A tan2A
(a) LHS = sin2A - sinA
1-cosA + cos2A
2sinA.cosA - sinA sinA(2cosA - 1)
= 1 - cosA + 2cos2A - 1 = cosA(2cosA -1)
= sinA = tanA = RHS. proved
cosA
1 - cos2A 2sin2A sin2A
(b) LHS = 1 + cos2A = 2cos2A = cos2A
= tan2A = RHS Proved.
(c) LHS = tanA + cotA = sinA + cosA
cosA sinA
sin2A + cos2A 1 2
= sinA.cosA = 2. 2sinA.cosA = sin2A
= 2cosec2A = RHS Proved.
(d) LHS = 1 +sec2A = 1+ = cos2A + 1 × cos2A
tan2A cos2A cos2A sin2A
sin2A
cos2A
2cos2A cosA
= 2sinA.cosA = sinA = cotA = RHS Proved.
Example 6. Proved that:
Solution:
(a) sin6T + cos6T = 1 (1 + 3cos22T) (b) cos4T = 1 - 8sin2T + 8sin4T
4
(a) LHS = sin6T + cos6T
= (sin2T)3 + (cos2T)3
= (sin2T + cos2T) (sin4T - sin2T.cos2T + cos4T)
= 1.(sin4T + cos4T - sin2T.cos2T)
= (sin2T + cos2T)2 - 2sin2T.cos2T - sin2T.cos2T
= 12 - 3sin2T.cos2T
=1- 3 (2sinT . cosT)2
4
3
=1- 4 sin22T
= 1 (4 - 3sin22T)
4
1
= 4 (1 + 3 - 3sin22T)
= 1 {1 + 3 (1 - sin22T)} = 1 (1 + 3cos22T) = RHS. Proved.
4 4
(b) LHS = cos4T
= cos2(2T)
= 1 -2sin22T
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vedanta Excel In Opt. Mathematics - Book 10
= 1 - 2[2sinT .cosT]2 = 1 – 8sin2T (1 - sin2T)]
= 1 - 8sin2T + 8sin4T = RHS. Proved.
Example 7. PLHroSved=thsaitn110s°in1-10ct°aons-6100c°°os130° =4
Solution:
= 1 - sin60°
Example 8. sin10° cos60°.cos10°
Solution: cos60°.cos10° - sin60°.sin10°
= sin10°.cos60°.cos10°
Example 9.
Solution: = cos (60° + 10°) = 2 . 2cos70° = 4cos70°
2 sin10°.cos10° sin20°
206 sin10° . 1 cos10°
2
4cos70° 4cos70°
= sin(90° -70°) = cos70° = 4 = RHS. Proved.
Proved that (a) cosec2T + cot4T = cotT - cosec4T
(b) 4cosec2T . cot2T =cosec2T - sec2T
(a) LHS = cosec2T + cot4T
= 2ssiicnn11o22sTT2T+++cs2coisnosi4sn4c2TT2o(T2s.4TcT)os2T
= 2sin2T . cos2T
=
= 2cos2T + 2cos22T - 1
2sin2T.cos2T
2cos2T(1 + cos2T) - 1
= 2sin2T.cos2T
= 2cos2T.2cos2T - 1 = 2cos2T – 1
= 2sin2T.cos2T sin2(2T) 2sinT.cosT sin4T
cosT 1
sinT - sin4T = cotT - cosec4T = RHS. Proved.
(b) LHS = 4cosec2T.cot2T
= 4. 1 . cos2T
sin2T sin2T
4cos2T
= 2sinT.cosT.2sinT.cosT
= cos2T - sin2T = cos2T - sin2T
sin2T.cos2T sin2T.cos2T sin2T.cos2T
1
= 1 - cos2T = cosec2T - sec2T = RHS. Proved.
sin2T
Proved that (2cosT + 1) (2cosT - 1) (2cos2T - 1) (2cos4T - 1) = 2cos8T + 1
LHS = (2cosT + 1) (2cosT - 1) (2cos2T - 1) (2cos4T - 1)
= {(2cosT)2 - 12} (2cos2T - 1) (2cos4T -1)
= (4cos2T –1) (2cos2T - 1) (2cos4T - 1)
= {2(2cos2T - 1) + 1} (2cos2T - 1) (2cos4T - 1)
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vedanta Excel In Opt. Mathematics - Book 10
= (2cos2T + 1) (2cos2T - 1) (2cos4T - 1)
= (4cos22T - 1) (2cos4T - 1)
= {2 (2cos22T - 1) + 1} (2cos4T - 1)
= (2cos4T + 1) (2cos4T – 1)
= 4cos24T - 1
= 2(2cos24T - 1) + 1
= 2cos8T + 1 = RHS. Proved.
Example 10 Prove that cos2D + sin2D . cos2E = cos2E + sin2E . cos2D
Solution: LHS = cos2D + sin2D . cos2E
= cos2D + sin2D(1 - 2sin2E)
= cos2D + sin2D - 2sin2E . sin2D
= 1 - 2sin2E . sin2D = cos2E + sin2E - 2sin2E . sin2D
= cos2E + sin2E(1 - 2sin2D)
= cos2E + sin2E . cos2D = RHS Proved.
Example 11. Prove that 2+ 2+ 2+ 2 + 2cos16T = 2cosT
Solution: LHS = 2+ 2+ 2+ 2 + 2cos16T
= 2+ 2+ 2+ 2(1+cos16T)
= 2+ 2+ 2+ 2.2cos28T
= 2+ 2+ 2 + 2cos8T
= 2+ 2+ 2(1+cos8T)
= 2+ 2+ 2 . 2cos24T
= 2+ 2+2cos4T
= 2+ 2(1+cos4T) = 2+ 2.2cos22T)
= 2(1+cos2T) = 2.2cos2T = 2cosT = RHS. Proved.
Example 12. Prove that 1 - 1 cotA = cot2A
tan3A - tanA cot3A -
1 1
Solution: LHS = tan3A - tanA - cot3A - cotA
= 1 - 1
tan3A -
tanA 1 - 1
tan3A tanA
= 1 tanA - tanA.tan3A
tan3A - tanA - tan3A
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vedanta Excel In Opt. Mathematics - Book 10
= 1 + tanA.tan3A = 1= 1
tan3A - tanA tan3A - tanA tan(3A - A)
= 1 = cot2A 1 +tanA.tan3A
tan2A = RHS. Proved.
Example 13. Proved that 4(cos320° + cos340°) = 3(cos20° + cos40°)
Solution: LHS = 4 (cos320 + cos340°)
= 4cos320 + 4cos340°
= 3cos20° + cos(3.20°) + 3cos40° + cos(3.40°)
= 3cos20° + cos60° +3cos40° + cos120°
= 3(cos20° + cos40°) + 1 - 1
2 2
= 3(cos20° + cos40°) = RHS. Proved.
Example 14. Prove that sin3T + sin3T = cotT
Solution: cos3T - cos3T
sin3T + sin3T 3sinT - 4sin3T + sin3T
LHS = cos3T - cos3T = cos3T - 4cos2T + 3cosT
= 3sinT - 3sin3T = 3sinT(1-sin2T) = sinT . cos2T
3cosT - 3cos3T 3cosT(1-cos2T) cosT . sin2T
cosT
= sinT = cotT = RHS. Proved
Example 15. Prove that : tanT + 2tan2T + 4tan4T + 8cot8T = cotT
Solution: LHS = tanT + 2tan2T + 4tan4T + 8cot8T
= tanT + 2tan2T + 4tan4T + 8
tan(2.4T)
8
= tanT + 2tan2T + 4tan4T + 2tan4T
1 - tan24T
(= tanT + 2tan2T + 8(1-tan24T)
2tan4T
4tan4T + (
(= tanT + 2tan2T + 4 tan4T + 1-tan24T(
tan4T(
(
(= tanT + 2tan2T + 4 tan24T + 1-tan24T
tan4T
4
= tanT + 2tan2T + tan4T
= tanT + 2tanT 4
2tan2T
(= tanT + 2 1 - tan22T
1-tan22T
tan2T + tan2T
= tanT + 2 tan22T +1- tan22T
tan2T
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vedanta Excel In Opt. Mathematics - Book 10
= tanT + 2
2tanT
1 - tan2T
= tanT + 2(1 - tan2T)
2tanT
tan2T + 1 - tan2T 1
= tanT = tanT = cotT = RHS. Proved.
Example 16. Prove that
(a) 4sin3A.cos3A + 4cos3A.sin3A = 3sin4A
(b) cos3A.cos3A + sin3A. sin3A = cos32A
Solution: (a) We have, 4sin3A = 3sinA - sin3A
4cos3A = 3cosA + cos3A
Now, LHS = 4sin3A.cos3A + 4cos3A.sin3A
= (3sinA – sin3A). cos3A + (3cosA + cos3A).sin3A
=3sinA.cos3A - sin3A.cos3A + 3sin3A.cos3A + cos3A.sin3A
= 3(sinA.cos3A + sin3A.cosA)
= 3.sin(A + 3A) = 3sin4A = RHS Proved.
(b) LHS = cos3A cos3A + sin3A.sin3A
= 1 (3cosA + cos3A).cos3A + 1 (3sinA - sin3A).sin3A)
4 4
1
= 4 [3cosA.cos3A + cos23A + 3sinA.sin3A - sin23A]
= 1 [3(cosA.cos3A + sinA.sin3A) + (cos23A - sin23A)]
4
1
= 4 [3cos(A - 3A) + cos6A]
= 1 (3cos2A + cos6A)
4
1 1
= 4 [3cos2A + cos(3.2A)] = 4 [3cos2A + 4cos32A – 3cos2A]
= 4cos32A
4
= cos32A =RHS. Proved.
Example 17. Prove that cot(45° + T) + tan(45° - T) = 2cos2T
1 + sin2T
Solution: LHS = cot(45° + T) + tan(45 - T)
= cot45°.cotT - 1 + tan45° - tanT
cot45° + cotT 1 +tan45°.tanT
cotT - 1 1-tanT
= 1 + cotT + 1 + tanT
1 -1 1 - tanT
tanT 1 + tanT
= +
1 + 1
tanT
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vedanta Excel In Opt. Mathematics - Book 10
= 1-tanT + 1-tanT
1 + tanT 1 + tanT
sinT
2(1-tanT) 1- cosT
1+tanT
= = 2. 1 + sinT
cosT
(cosT - sinT) cosT - sinT cosT + sinT
=2 cosT + sinT = 2. (cosT + sinT) × cosT + sinT
= 2. cos2T = 2cos2T = RHS. Proved.
sin2T + cos2T + 2sinT.cosT 1 + sin2T
Example 18. Prove that 2sinT + 2sin3T + 2sin9T = tan27T - tanT
cos3T cos9T cos27T
2sinT 2sin3T 2sin9T
Solution: LHS = cos3T + cos9T + cos27T
= 2sinT.cosT + 2sin3T.cos3T + 2sin9T.cos9T
cos3T.cosT cos9T.cos3T cos27T.cos9T
= sin2T + sin6T + sin18T
cos3T.cosT cos9T cos3T cos27T.cos9T
= sin(3T - T) + sin(9T - 3T) + sin(27T - 9T)
cos3T.cosT cos9T.cos3T cos27T.cos9T
sin3T.ccoossT3T-.ccoossT3T.sinT+ sin9T.cos3T - cos9T.sin3T
= cos9T.cos3T
+sin27T.ccooss99TT.-cocos2s277TT.sin9T
= tan3T - tanT + tan9T - tan3T + tan27T - tan9T
= tan27T - tanT = RHS. Proved.
Example 19. Without using table or calculator. Show that sin18° = 5-1 .
Solution: Let T = 18° 4
Then 5T = 90°
or, 2T + 3T = 90°
or, 3T = 90° - 2T
Taking cosine on both sides, we get
cos3T = cos(90° - 2T)
or, cos3T = sin2T
or, 4cos3T - 3cosT = 2sinT cosT
since sinT = sin18°≠ 0, dividing both sides by cosT.
4cos2T - 3 = 2sinT
or, 4 - 4sin2T - 2sinT - 3 = 0
or, -4sin2T - 2sinT + 1 = 0
or, 4sin2T + 2sinT - 1 = 0
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This equation is quadratic is sinT in the form of ax2 + bx + c = 0, where
a = 4, b = 2, c = -1
x= b± b2 - 4ac
2a
or, sinT = - 2 ± 22 - 4.4(-1) = 2± 4 + 16
2.4 8
= 2± 20 = 2±2 5 = -1 ± 5
8 8 4
since value of sin18° is positive. We take only positive sign.
Hence, sinT = sin18° = 5 -1 . Proved.
4
( ( (Prove that cosSc 2Sc 3Sc 1
Example 20. 7 (, cos 7 , cos 7 = 8
Solution:
( ( (LHS = cos Sc ( (( 2Sc 3Sc
7 ( (( , cos 7 , cos 7
2sin Sc . cos Sc . cos 2Sc . cos 3Sc
= 7 7 7 7
Sc
2sin 7
2.sin 2Sc . cos 2Sc . cos 2Sc
7 7 7
=
2.2sin Sc
7
4Sc 3Sc
sin 7 . cos 7
= 4sin Sc
7
3Sc 3Sc
2sin 7 . cos 7 [ ( ) ]
sin 4Sc Sc = sin 3Sc
= 4. 2sin Sc 7 = sin S - 7 7
7
(=
sin 6Sc sin Sc - Sc sin Sc 1
7 7 7 8
= = = = RHS. Proved.
8sin Sc 8sin Sc 8sin Sc
7 7 7
Exercise 9.1 211
Very Short Questions
1. (a) Define multiple angles with an example.
(b) Express sin2T in terms of sinT, cosT and tanT.
(c) Express cos2T in terms of sinT, cosT and tanT.
(d) Express cos2T in terms of cotT.
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vedanta Excel In Opt. Mathematics - Book 10
2. (a) Express sin3T in terms of sinT (b) Express cos3T in terms of cosT
(c) Express tan3T in terms of tanT. (d) Express tanT in terms of sin2T and cos2T
Short Questions
3. (a) If sinT = 1 , find the value of cos2T.
2
1
(b) If tanT 3 , find the value of tan3T
(c) If tanT = 1 , find the value of tan3T.
2
4
4. (a) If sinT = 5 , find the value of sin2T, cos2T, tan2T.
(b) If sinT = 1 , find the values of sin2T, cos2T, tan2T.
2
3
(c) If tanT = 4 , find the values of sin2T, cos2T and tan2T
(d) If sinT = 3 , find the value of sin3T and cos3T.
2
1 7
5. (a) If sinT = 4 , show that cos2T = 8
(b) If cosT = 6 , show that cos2T = 11
52 25
(c) If tanT = 5 , show that tan2T = 120
12 119
6. By using the formula of cos2T, establish the following :
(a) sinT = ± 1-cos2T (b) cosT = ± 1+cos2T
2 2
(c) tanT = ± 1-cos2$
1+cos2A
(7. 1 1
(a) If sinT = 2 p + P ((, show that :
(
((
(
( ((i) 1 1 (ii) 1 1
cos2T = – 2 p2 + P2 sin3T = – 2 p3 + P3
((b) 1 1 , show that :
If cosT = 2 p + P
((i) 1 1 ((ii) 1 1
cos2T = 2 p2 + P2 cos3T = 2 p3 + P3
8. Prove that :
(a) 1 - cos2T = tanT (b) sin2T = cotT
sin2T 1-cos2T
sin2T 1 - cos2T
(c) tanT = 1 + cos2T (d) 1 + cos2T = tan2T
(e) sin2T = 2cotT 1 (f) 1 - sin2T = 1 - tanT
cot2T + cos2T 1 +tanT
cos2T 1 - tanT cosT cosT
(g) 1 + sin2T = 1 + tanT (h) cosT - sinT - cosT + sinT = tan2T
(i) 1 + sin2A = sinA + cosA (j) cotT - tanT = cos2T
cos2A cosA - sinA cotT + tanT
4tanT cosT - sinT
(k) tan2T + sin2T = 1 - tan4T (l) cosT + sinT = sec2T - tan2T
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sin5T cos5T sin3T + cos3T 1
sinT cosT cosT + sinT 2
(m) - = 4 cos2T ((n) = 1 - sin2T
1 - sin2T Sc
cos2T 4
(o) cos4T - sin4T = cos2T ( (p) = tan - T
(
((q) Sc
cos2T = tan 4 - T
1 +sin2T
9. Prove that
(a) sin2T - cosT = cotT (b) 1 + sin2T - cos2T = tanT
1 - sinT - cos2T 1 + sin2T + cos2T
(c) sinT + sin2T = tanT (d) 1-cos2T + sin2T = tanT
1 + cosT + cos2T 1 + sin2T + cos2T
(e) (sinT + cosT)2 - (sinT - cosT)2 = 2 sin2T
(f) sinT + cosT + cosT - sinT = 2 sec2T
cosT - sinT cosT + sinT
(g) (1 + sin2T + cos2T)2 = 4cos2T(1 + sin2T)
(h) 1 - 1 cotT = cotT
tan2T - tanT cot2T -
10. Prove that following :
((a) cotT -1 2 (((b) 1 + tan2 Sc - T(
cotT+1 4 (
1-sin2T = ( = cosec2T
1+sin2T ( Sc -
((c) sin2 1 - tan2 4 T
Sc
4 - T = 1 (1 - sin2T)
2
b
11. If tanT = a , prove that a cos2T + b sin2T = a.
Long Questions
12. Prove that
(a) cos4T + sin4T = 1 (3 + cos4T) (b) cos6T - sin6T = cos2T (1 - 1 sin22T)
4 4
1 1
(c) sin4T = 8 (3 - 4 cos2T + cos4T) (d) cos8T + sin8T = 1 - sin22T + 8 sin42T.
(e) cos5T =16cos5T – 20cos3T + 5cosT (f) sin5T = 16sin5T - 20sin3T + 5sinT
13. Prove that
(a) 3 + 1 =4 (b) cosec10° - 3 sec10° = 4
sin40° cos40°
14. Prove that
(a) (2cosT + 1) (2cosT - 1) = 2cos2T + 1)
(b) sec8T - 1 = tan8T
sec4T - 1 tan2T
(c) tanT + 2tan2T + 4tan4T + 8cot8T = cotT
(d) sin2D - cos2D.cos2E = sin2E - cos2E.cos2D
(e) 2+ 2+ 2 + 2cos8T = 2cosT
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(f) sin2D - sin2E = tan (D + E)
sinD.cosD - sinE.cosE
15. Prove that
(a) 4(cos310° + sin320°) = 3(cos10° + sin20°)
(b) sin310° + cos320° = 3 (cos20° + sin10°)
4
(c) 4(cos320° + sin350°) = 3(cos20° + sin50°)
( ((d) Sc Sc
tanA + tan ( ( ((3 + $ – tan 3 - A = 3tan3A
(( ( (
16. Prove that ((( (
(a) cot (A + 45°) - tan (A - 45°) =(( 2cos2A
1 + sin2A
(b) tan(A + 45°) - tan (A - 45°) = 2sec2A.
(c) tan(A + 45°) + tan (A - 45°) = 2tan2A.
17. (a) If 2tanD = 3tanE, then prove that,
(i) tan (D - E) = sin2E (ii) tan(D + E) = 5sin2E 1
5 - cos2E 5cos2E -
1 1
(b) If tanT = 7 and tanE = 3 , prove that : cos2T = sin4E
18. Prove that
(a) cosA - 1 + sin2A = tanA (b) 1 - 1 = cot4T
sinA - 1 + sin2A tan3T + tanT cot3T + cotT
(c) cotA - tanA = 1.
cotA - cot3A tan3A - tanA
19. Prove that
( ( ( ((a)
8 1 + sin Sc 1+sin 3Sc 1- sin 5Sc 1-sin 7Sc =1
8 8 8 8
( ( ( ((b)
1-cos Sc 1- cos 3Sc 1- cos 5Sc 1- cos 7Sc = 1
8 8 8 8 8
( ( ( ((c)
sin4 Sc + sin4 3Sc + sin4 5Sc + sin4 7Sc = 3
8 8 8 8 2
20. Prove that cos233° – cos257° = 2
sin210.5° – sin234.5°
3
21. (a) Prove that cos3A + cos3(120° + A) + cos3(240° + A) = 4 cos3A
(b) Prove that sin3A + sin3(60° + A) + sin3(240° + A) = – 3 sin3A
4
Project Work
22. Prepare a report by calculating the values of sin18°, cos18°, sin36°, cos36°, cos54°,
tan18°, and tan54° by using relations of multiple angle ratios in trigonometry. For this
project work, the students of class can be divided in groups as required.
3. (a) 1 (b) f (c) 11 4.(a) 24 , - 7 , - 24
2 2 25 25 7
3 1 24 7 24
(b) 2 , 2 , 3 (c) 25 , 25 , 7 (d) 0, -1
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9.2 Trigonometric Ratios of Sub-multiple Angles
Let A be any angles. Then A , A , A , ....... etc. are called sub-multiple angles of A. In this
sub-unit, we discuss 2 3 4 ratios of sub-multiple angles $2 , $
the trigonometric 3 , etc.
(I) Trigonometric Ratios of Half Angles :
sinA = 2sin A A 2tan A2 = 2cot A
((a) 2 cos 2 = 2
= A A
1 + tan2 2 1 + cot2 2
A
Here, sinA A + 2 (
2 (
A A A A A A
= sin 2 cos 2 + cos 2 . sin 2 = 2sin 2 cos 2 .
⸫ sinA = 2sin A cos A ............ (i)
2 2
A A
sinA = 2sin 2 cos 2
2sin A . cos A
2 2
= A A
sin2 2 + cos2 2
Dividing numerator and denominator by cos2 A
2
A A
2tan 2 2tan 2
= 1sh+owtanth2 aA2t ⸫ sinA = 1+ tan2 A ........... (ii)
can 2
Similarly we
2cot A
2
sinA = A .......... (iii)
1+ cot2 2
Combing (i), (ii) and (iii), we get,
A A 2tan A 2cot A
2 2 2 2
⸫ sinA = 2sin . cos = A = A
1+ tan2 2 1+ cot2 2
A A 1-tan2 A cot2 A - 1
2 2 2 cot2 2
cosA - sin2
((b) = cos2 = 1+ tan2 A = A +
2 21
Here, cosA = cos A + A
2 2
= cos A . cos A - sin A . sin A
2 2 2 2
A A
= cos2 2 - sin2 2
cosA = cos2 A - sin2 A ........... (i)
2 2
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= cos2 A - 1 + cos2 A
2 2
A
= 2cos2 2 - 1
Also, 2cos2 A = 1 + cosA
2
A A
Again, cosA = cos2 2 - sin2 2
= 1 - 2sin2 A
2
A
2.sin2 2 = 1 - cosA
cosA = cos2 A - sin2 A
2 2
A A
cos2 2 - sin2 2
= sin2 A + cos2 A
2 2
Dividing numerator and denominator by cos2 A .
2
A
1-tan2 2
⸫ cosA = 1+ tan2 A ....... (ii)
2 cot2
A - 1
Similarly, we can show cotA = 2 ............ (iii)
A
cot2 2 + 1
Combining (i), (ii) and (iii), we get
A A 1-tan2 A cot2 A -1
2 2 2 2
⸫ cosA = cos2 - sin2 = A = .
1+ tan2 2 A
cot2 2 + 1
2tan2 A
2
tanA =
((c) 1- tan2 A
2 A A
tan 2 + tan 2
A( + A
Here, tanA = tan 2 ( 2 = 1- tan A . tan A
2 2
2tan A
2
⸫ tanA = A
1- tan2 2
((d) A + A
cotA = cot 2 2
cot A . cot A - 1 cot2 A - 1
2 2 2
Here, cotA = A A = A
cot 2 + cot 2 2cot 2
cot2 A - 1
2
⸫ cotA = A
2cot 2
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II. Trigonometric Ratios of A in terms of A .
3
((a) A 2A ( A A
sinA = sin 3 + 3 = 3sin 3 - 4sin3 3
(
(Here, sinA ( 2A A
= sin 3 + 3
= sin 2A . cos A + cos 2A . sin A
3 3 3 3
(= A A A A ( A
2sin 3 . cos 3 cos 3 + 1- 2sin2 3 . sin 3
(= A 1- sin2 A A A
2sin 3 3 + sin 3 - 2sin3 3
= 2sin A - 2sin3 A + sin A - 2sin3 A
3 3 3 3
A A
? sinA = 3sin 3 - 4 sin3 3
= 4cos3 A - 3cos A
((b) cosA 3 3
Here, cosA = cos (2A+ A
( 3 3
= cos 2A (.cos A - sin 2A . sin A
3 ( 3 3 3
(= A A A A A
2cos2 3 -1 . cos 3 - 2sin 3 . cos 3 . sin 3
(= A - cos A - A 1 - cos2 A (
2cos3 3 3 2cos 3 3
= 2cos3 A - cos A - 2cos A + 2cos3 A = 4cos3 A - 3cos A
3 3 3 3 3 3
A A
? cosA = 4cos3 3 - 3cos 3
((c) tanA = tan A 3tan A - tan2 A
3 3 3
3 . = A
1 - 3tan2 3
(Here, tanA 2A A
= tan 3 + 3
2tan A A
3 3
+ 2tan
tan 2A + tan A 1 - tan2 A
3 3 3
= =
1 - tan 2A . 3tan A A
3 3 1 - 2tan 3 . A
3
1 - tan2 A tan
3
A A A A A
2tan 3 + tan 3 - tan3 3 3tan 3 - tan3 3
= 1 - tan2 A - 2tan2 A = 1 - 3. tan2 A
3 3 3
A A
3tan 3 - tan3 3
? tanA = 1 - 3. tan2 A
3
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((d) cotA = cot A ( cot3 A - 3cot A
3 3 3
3 . = A (For practice do yourself.)
3cot2 3 - 1
Worked out Examples
Example 1. If sin T = 3 find the values of.
Solution: 2 5
(a) sinT (b) cosT (c) tanT
Example 2.
Solution: Here, sin T = 3
2 5
cos T = 1 - sin2 T = 1 - 9
2 2 25
= 25-9 = 16 = 4
25 25 5
T 3
T sin 2 5 3
2 4 4
tan = cos T = 5 =
2
3 4 24
(a) sinT = 2 sin T . cos T = 2. 5 . 5 = 25
2 2
T T 16 9 16-9 7
(b) cosT = cos2 2 - sin2 2 = 25 - 25 = 25 = 25
2tan T 2. 3 3 16 3 16 24
2 4 2 16-9 2 7 7
(c) tanT = = 9 = × = × =
1 - tan2 T 1- 16
2
Alternative 24
tanT sinT = 25 = 24
cosT 7 7
25
If cos45° = 1 , show that cos2212 °= 1 . 2+ 2
2 2
Here, cos45° = 1 2
(or, 45° =1
2 2
cos2 (
or, 224cc25oo°ss22is442255p°°osi-=t1iv=e12[( 1 ( cosT = 2cos2 T - 1)
or, 2 quadrant] 2
of cos +
45° 1
2 lies
( (Value in 1st
? cos 45° = 1 + 1 = 2+1 × 2 (
2 22 2 22 2
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= 2+ 2 = 1 . 2+ 2
2 2
? cos 45° = 1 . 2 + 2 Proved.
2 2
Example 3. Prove that
Solution:
(a) tan T = sinT (b) 1 + sinT - cosT = tan T
Example 4. 2 1 + cosT 1 + sinT + cosT 2
Solution: sinT
(a) RHS = 1 + cosT
2sin T . cos T sin T T
2 2 2 2
= = = tan = LHS. Proved.
2cos2 T cos T
2 2
1 + sinT - cosT
1 + sinT + cosT
((b) LHS =
1 + (2sin T . cos T - 1- 2sin2 T
(( 2 2 2
=
( T T T
1 + 2sin 2 . cos 2 + 2cos2 2 -1
(
1 + (2sinT . cos T - 1+ 2sin2 T
(= ( 2 2 2
((2cosT sin T + cos T
2 2 2
2sin T cos T + sin T
(= 2 2 2
= RHS. Proved.
T T T
2cos 2 cos 2 + sin 2
Prove that
((a) tan Sc + A = secA + tanA
4 2
( ((b) cos2 Sc D Sc D D
4 - 4 - sin2 4 - 4 = sin 2
((a) Sc + A
LHS = 4 2
tan Sc + tan A 1+ sin A
4 2 2
= =
Sc $ $
1 - tan 4 . tan 2 1 - tan 2
1 + sin A
2
cos A cos A + sin A cos $
2 2 2 2
= = ×
$ $ A A
1 - sin 2 cos 2 cos 2 + sin 2
cos A
2
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cos $ + sin $ cos $ + sin $
2 2 2 2
= ×
A A A A
cos 2 - sin 2 cos 2 + sin 2
cos2 $ + sin2 $ + 2sin $ .cos $ 1 + sinA
2 2 2 2 cosA
= =
A A
cos2 2 - sin2 2
= 1 + sinA = secA + tanA = RHS. Proved.
cosA cosA
( ((b)
LHS = cos2 Sc - D (- sin2Sc-D
( (= cos2 4 4 (( 4 4
((
Sc - D = cos Sc - D = sin D = RHS Proved.
(Prove that (cosD + cosE)2 + (sinD + sinE)2 = 4cos244 2 2 2
Example 5: D-E
Solution: 2
LHS = (cosD + cosE)2 + (sinD + sinE)2
= cos2D + 2cosD cosE + cos2E + sin2D + 2sinD.cosD + sin2E
= (cos2D + sin2D) + (sin2E + cos2E) + 2(cosD.cosE + sinD.sin2E)
(= 2[1 + cos(D - E)] = 2.2cos2
= 1 + 1 + 2cos (D - E)
(= 4cos2 D-E ((
(( D-E 2 ((
( 2
= RHS. Proved. (
((
Example 6: (Proved that (( ( ( (1+ cos3Sc
1 + cos Sc ((
1+ cos5Sc 1+ cos7Sc = 1
(( 8
Solution: Here,
( ( { ( )}{ ( )}LHS. = (
( ( ( (= 1+ cos Sc 1+ cos3Sc 1 + cos S - 3Sc 1 + cos Sc - Sc
( (= 8 8
( { ( )}=
( (= 1+ cos Sc 1+ cos3Sc 1- cos3Sc 1- cosSc
1- cos2 Sc 1- cos23Sc
1- cos2 Sc 1 - cos2 Sc - Sc
2 8
1- cos2Sc 1- sin2Sc
= sin2 Sc. cos2 Sc
( (= 1
2
2sinSc. cos Sc
( (=1
2
sin2Sc
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( ( (= 1 sin2Sc = 1 1 = 1 = RHS Proved.
(
( (Example 7.
( ( ( (Solution: D 1 1 1 a2 +a12
If cos 2 = 2 a + a , prove that cosD = 2
Here, cos D = 1 a + 1 ,
a
cosD = 2cos2 D - 1
[ ( (]= 2
1 a + 1 2 -1
( (= a
2. 1 a2 + 2.a 1 +a12 -1
( (= 1 a
( (
a2 + 1 +1-1= 1 a2 + 1 = RHS. Proved.
a2 a2
Exercise 9.2
Very Short Questions
1. (a) Define sub-multiple angle with an example.
(b) Express sinT in terms of sin T and cos T .
T T T
(c) Express cosT in terms of cos , sin and tan .
(d) Express tanT in terms of tan T .
T
2. (a) Express sinT in terms of sin .
(b) Express cosT in terms of cos T .
T
(c) Express tanT in terms of tan .
3. (a) If sin T = , find the value of sinT.
T 3
(b) If cos = 2 , find the value of cosT.
(c) If tan T = 3, find the value of tanT.
(d) If = find the value of cosT.
T 4
cos 5 ,
Short Questions
4. (a) If cos30° = 23, find the values sin15°, cos15° and tan15°.
1
(b) If cos45° = 2 , then prove that :
(i) sin2212° = 1 2- 2 (ii) cos2212° = 1 2+ 2 (iii) tan2212° = 3-2 2
2 2
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(c) If cos330° = 3 prove the following :
2
3-1 3-1
(i) cos165° = - 22 (ii) sin165° = 22 (iii) tan165° = - (2 - 3)
5. Prove that:
( ( ( ((a)
1 - tan2 x (b) sinx = 3 sin x - 4sin3 x
2
cosx = x
1 + tan2 2
( ( ( ( ( ( ( (3tan x - tan3 x
cot3 x - 3cot x
1 - 3tan2 x 3cot2 x - 1
( ( ( ((c) tanx =
(d) cotx =
( ( ( (6.
(a) If cos T = 1 p + 1 , prove that cosT = 1 p2 + 1 .
( ( ( ((b) 2 p 2 p2
If cos T = 1 p + 1 , prove that cosT = 1 p3 + 1 .
( ( ( ((c) 2 p 2 p3
If sin T = 1 p+ 1 , prove that cosT = - 1 p2 + 1 .
2 p 2 p2
(7. Prove that:
((b) 1 + sinT =
(a) 1 - sinT = sin T - cos T ((2 sin T +cos T 2
2 2 2 2
(c) 1 + cosT = cot T (d) 1 + secT = cot T
sinT 2 tanT 2
1 + cosT
(e) cosecT - cotT =tan T (f) 1 - cosT = cot2 (2T)
2
A
1 - sinA 1 - tan 2 sin3 T - cos3 T 1
cosA 2 2 2
(g) = A (h) = 1 + sinT
1 + tan 2 sin T - cos T
2 2
1+ secT T 1+ sinT cos T + sin T
secT 2 cosT 2 2
(i) = 2cos2 (j) =
cos T - sin T
(((l) 2 2
((k) (( Sc T 1 - tan2 Sc - T ( T
(4 2 4 4 2
1 - 2sin2 - = sinT ( = sin
Sc T (
1 +tan2 4 – 4
Long Questions
( (8. Prove that:T T 2sinT - sin2T = tan2 T
(a) cos4 2 - sin4 2 = cosT 2sinT + sin2T 2
((b)
(c) 1 sin2T . cosT = tan T (d) 1 +sinT = tan2 Sc + T
+ cos2T 1 + cosT 2 1 - sinT 4 2
cos T 1+sinT T sin T 1+sinT T
2 2 2 2
(e) = tan (f) = cot
T T
sin 2 1+sinT cos 2 1+sinT
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9. Prove that
((a)
tan Sc + T(((( = secT + tanT
((b) 4 2( (((
Sc 1 - sinT
tan 4 – T (=1 + sinT
( ((c) 2 (
sec Sc + T(( . sec Sc – T = 2secT
((d) 4 2(( 4 2
tan Sc – T( = cosT
( ((e) 4 2(( 1 +sinT
2cosT
cot T + Sc ( – tan T – Sc = 1 +sinT
( ((f) 2 4 ( 2 4
tan Sc + T + tan Sc - T = 2secT
4 2 4 2
10. Prove that
((a) (cosD - cosE)2 + (sinD - sinE)2 = 4sin2
((b) (sinD + sinE)2 + (cosD + cosE)2 = 4cos2 D-E
2
D-E
2
11. Prove that
( ( ( ((a)
cos 2Sc . cos 4Sc . cos 8Sc . cos 16Sc = 1
( ( ( ((b) 15 15 15 15 16
1+cos Sc 1+cos 3Sc 1+cos 5Sc 1+cos 7Sc = 1
8 8 8 9 8
Project Work
12. Discuss how to find the value of tan 7 1° . Then show that
2
1°
tan 7 2 = 6– 3+ 2 – 2.
3. (a) 3 (b) 1 (c) - 3 (d) - 44
2 2 125
3 -1 3 + 1
4. (a) 2 2 , 2 2 , 2 – 3
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9.3 Transformation of Trigonometric Formulae
The sum or difference forms of trigonometric ratios can be transformed into the product
forms and the product forms can be transformed into the sum or difference forms.
(a) Transformation of product into sum or difference form formula of compound angles;
we have
sin (A + B) = sinA cosB + cosA. sinB ................ (i)
sin (A - B) = sinA. cosB - cosA.sinB ................. (ii)
cos (A + B) =cosA.cosB - sinA.sinB ................... (iii)
cos (A -B) = cosA. cosB - sinA.sinB ................... (iv)
Adding identities (i) and (ii), we get,
sin (A + B) + sin (A - B) = 2cosA.sinB.
Again, adding identities (iii) and (iv)
cos (A + B) + cos(A - B) = 2cosA.cosB
Subtracting identity (iii) from(iv), we get,
cos(A - B) - cos(A + B) = 2 sinA. sinB
Hence, we have the following formulae:
(I) 2sinA. cosB = sin(A + B) + sin(A - B)
(II) 2cosA. sinB = sin(A + B) - sin(A - B)
(III) 2cosA. cosB = cos(A + B) + cos(A - B)
(IV) 2sinA.sinB = cos(A -B) – cos(A + B)
(b) Transformation of Sum or difference into product
Let, A + B = C ..................... (v)
and A - B = D ...................... (vi)
Adding (v) and (vi), we get,
2A = C + D
or, A= C+D
2
Again subtraction (vi) from (v), we get,
2B = C - D
or, B= C–D
2
Now, express above four formulae (i), (ii), (iii) and (iv) in terms of C and D. We get,
(V) sinC + sinD = 2 sin C + D. cos C-D
2 2
C + D C-D
(VI) sinC - sinD = 2cos 2 . sin 2
(VII) cosC + cosD = 2cos C + D . cos C-D
2 2
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(VIII) cosD - cosC = 2sin C + D. sin C-D .
2 2
C + D. D-C
or, cosC - cosD = 2 sin 2 sin 2
Above formulae (i) to (viii) are the transformation formulae of trigonometric ratios.
Worked out Examples
Example 1. Express the following product forms into sum or difference forms:
Solution:
(a) sin24° cos12° (b) cos42° sin22°
Example 2.
Solution: (c) cos50°.cos35° (d) sin40° sin20°
(a) sin24°.cos12° = 1 (2sin24°cos12°)
2
1
= 2 [sin(24° + 12°) + sin(24° - 12°)]
= 1 [sin36° + sin12°]
2
1
(b) cos42°. sin22° = 2 [2cos42°.sin22°]
= 1 [sin(42° + 22°) - sin(42° - 22°)]
2
1
= 2 [sin64° - sin20°]
(c) cos50° cos35° = 1 [2cos50° cos35°]
2
1
= 2 [cos(50° + 35°) + cos(55° - 35°)]
= 1 [cos85° + cos20°]
2
1
(d) sin40°. sin20° = 2 [2sin40°.sin20°]
1 [cos(40° - 20°)- cos(40° + 20°)]
(= 2
1 1 1
= 2 [cos20° - cos60°] = 2 cos20°- 2 (
Express the following sum or difference into product form:
(a) cos6T + cos4T (b) sin50° - sin40°
(c) sin50° - sin40° (d) cos35° - cos25°
(a) cos6T + cos4T = 2cos 6T + 4T . cos 6T - 4T
2 2
= 2cos5T. cosT
(b) sin50° + sin40° = 2sin 50° + 40° . cos 50° - 40°
2 2
= 2sin45°.cos5°
(c) sin50° - sin40° = 2cos 50° + 40° . sin 50° - 40°
2 2
= 2cos45°. sin5°
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(d) cos35° - cos25° = 2sin 35° + 25° . sin 25° - 35°
2 2
= 2sin30°.sin(-5°)
= –2sin30°.sin5° [ sin(-T) = - sinT@
= –2 . 1 . sin15° = – sin15°
2
Example 3. Prove that
Solutions:
(a) sin75° + sin15° = 3 (b) sin18° + cos18° = 2 cos27°
2
1
(c) sin(45° - A) sin (45° + A) = 2 cos2A
(a) LHS = sin75° + sin15°
= 2sin 75° + 15° . cos 75° - 15°
2 2
= 2sin45°. cos30°
=2. 1 . 3 = 3 = RHS Proved.
2 2 2
(b) LHS = sin 18° + cos18° = sin18° + cos(90° - 72°)
= sin18° + sin72°
( ) ( )= 2sin
18° + 72° . cos 72° - 18°
2 2
1
= 2sin45°. cos27° = 2. 2 . cos27°
= 2 cos27° = RHS. Proved.
(c) LHS = sin (45° - A). sin (45° + A)
= 1 [2sin(45° - A). sin(45° + A)]
2
1
= 2 [cos(45° - A - 45° - A) - cos(45° - A + 45° + A)]
= 1 [cos (-2A)- cos90°] = 1 cos2A = RHS. Proved.
2 2
Example 4. Proved that
Solution:
(a) sin4A + sin2A = tan3A (b) cos8° + sin8° = tan53°
cos4A + cos2A cos8° – sin8°
(a) LHS = sin4A + sin2A = tan3A
cos4A + cos2A
( ) ( )2sin
4A + 2A . cos 4A - 2A = sin3A . cosA
= 2 . cos 2 cos3A. cosA
( ) ( )2cos4A + 2A 4A - 2A
2 2
= tan3A = RHS Proved
(b) LHS = cos8° + sin8°
cos8° – sin8°
cos8° + sin(90° - 82°) cos8° + cos82°
= cos8° - sin(90° - 82°) = cos8° - cos82°
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(( )) (( ))=2cos8 + 82° . cos 8°-82° = sin45°. cos37°
2sin 2 . sin 2 sin45°.sin37°
cos37° 8° + 82° 82° - 8
sin37° 2 2
= = cot 37°
= cot (90° - 53°) = tan53° = RHS. Proved.
Example 5. Proved that sin7T - sin5T - sin3T + sinT = tan2T
Solution: cos7T + cos3T - cos5T - cosT
Example 6. LHS = sin7T - sin5T - sin3T + sinT
Solution: cos7T + cos3T - cos5T - cosT
Example 7. = (sin7T + sinT) - (sin5T + sin3T)
Solution: (cos7T - cosT) + (cos3T - cos5T)
2sin 7T + T . cos 7T – T – 2sin 3T + 5T . cos 5T – 3T
2sin 2 2 2 2
=
7T + T T – 7T 3T + 5T 5T – 3T
2 . sin 2 + 2sin 2 . sin 2
= sin4T . cos3T - sin4T . cosT = sin4T (cos3T - cosT)
sin4T . sin(–3T) + sin4T .sinT sin4T (sinT - sin3T)
2sin 3T + T . sin T - 3T
2 2
sin2T
= 2cos T + 3T . sin T - 3T = cos2T = tan2T = RHS. Proved.
2 2
Prove that sin8T . cosT - sin6T . cos3T = tan2T
cos2T . cosT - sin3T . sin4T
sin8T . cosT - sin6T . cos3T
LHS = cos2T . cosT - sin3T . sin4T
= 2sin8T . cosT - 2sin6T . cos3T
2cos2T . cosT - 2sin3T . sin4T
= sin(8T + T) + sin(8T - T) - sin(6T + 3T) - sin(6T - 3T)
cos(2T + T) + cos(2T - T) - cos(3T - 4T) + cos(3T + 4T)
= sin9T + sin7T - sin9T - sin3T = sin7T - sin3T
cos3T + cosT - cosT + cos7T cos3T + cos7T
= 2cos 7T +3T . sin T - 3T = sin2T
2 2 cos2T
2cos T + 7T . cos 3T - 7T
2 2
= tan2T = RHS. Proved.
Prove that sin20°.sin40°.sin60°.sin80°.Sun120° = 33
32
LHS = sin20°.sin40°.sin60°.sin80°.sin120°
= sin20° sin40° 3 sin80° 3
2 2
3
= 4×2 sin20° [2sin40°. sin80°]
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= 3 . sin20° [cos(40° - 80°) - cos(40° + 80°)]
8
3
= 8 sin20° [cos40° - cos120°]
= 3 sin20° .cos40° - 3 sin20° (– 1 )
8 8 2
3 3
= 8 sin20° .cos40° + 16 sin20°
= 3 [sin(20° + 40°) + sin(20° - 40°)] + 3 sin20°
16 16
3 3
= 16 [sin60° + sin(-20°)] + 16 sin20°
= 3 . 3 - 3 sin20° + 3 sin20° = 33 = RHS. Proved.
16 2 16 16 32
1
Example 8. Prove that cos20°.cos40°. cos60°.cos80° = 16 .
Solution:
LHS = cos20°. cos40°.cos60°.cos80°
Example 9.
Solution: = cos20°.cos40° 1 . cos80°
2
228 1
= 4 cos20° (2cos40°.cos80°)
= 1 cos20° [cos(40° + 80°) + cos(40° - 80°)]
4
1
(= 4
cos20° [cos120° + cos40°]
= 1 cos20° - (+ 1 cos20°.cos40°
4 ( 4
1 1
= – 8 cos20° + 8 (2cos20° . cos40°)
= – 1 cos20° + 1 [cos(20° + 40°) + cos(20° - 40)]
8 8
1 1 1 1 1 1
= – 8 cos20° + 8 cos60° + 8 cos20° = 8 . 2 = 16 = RHS. Proved.
Alternative Method
LHS = cos20°. cos40°.cos60°.cos80°
= 1 (2 sin20°cos20° ) cos40° 1 cos80°
2sin20° 2
1
= 4sin20° sin40°cos40° cos80°
= 1 (2sin40°cos40°).cos80° = 1 (2sin80°cos80°)
8sin20° 16sin20°
1 1 1
= 16sin20° sin160° = 16sin20° . sin20° = 16 = RHS Proved.
Prove that sinT.sin(60° - T) sin(60° + T) = 1 sin3T
4
LHS = sinT.sin(60° - T) sin(60° + T)
= 1 sinT [2sin(60° - T). sin (60° + T)]
2
1
= 2 sinT [cos(60° - T - 60 - T) - cos(60° - T + 60° + T)]
(= 1 [cos
2 sinT [-2T] - cos120°]
= 1 sinT.cos2T - 1 . sinT -
2 2
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= 1 [sin (T + 2T) + sin (T - 2T)] + 1 sinT
4 4
1 1 1 1
= 4 sin3T - 4 sinT + 4 sinT = 4 sin3T = RHS. Proved.
Example 10. Prove that cos Sc . cos 2Sc . cos 4Sc . cos 7Sc = 1
Solution: 15 15 15 15 16
Sc 2Sc 4Sc 7Sc
Example 11. (LHS=cos 15 , cos 15 , cos 15 . cos 15
Solution:
Sc Sc cos 2Sc . cos 4Sc . cos 7Sc
15
2sin .cos ((
((
= 2sin Sc
( 15
= sin 2Sc . cos 2Sc ( . cos 4Sc . cos 7Sc
(= 15 15 ( 15 15
2sin 2Sc .cos (2Sc . cos 4Sc . cos 7Sc
15 (
(
4sin Sc
(= 15 (
1
2sin 4Sc .cos 4Sc . cos 7Sc
8sin Sc 15
(= 15
1
sin 8Sc . cos cos 7Sc
8sin Sc
(= 15
1
sin Sc - 7Sc .cos 7Sc
8sin Sc
(= 15
1
2sin 7Sc cos 7Sc
16sin Sc
15
1
16sin sin 14Sc
(= Sc
15
1
= 16sin Sc . sin Sc - Sc
15
= 1 . sin Sc = 1 = RHS. Proved.
16sin Sc
15
Prove that 2cos Sc . cos91S3c + cos31S3c + cos51S3c = 0
13
Sc cos91S3c 3Sc cos51S3c
2cos 13 13 +
( (LHS= . + cos
= cos Sc + 9Sc + cos Sc – 9Sc + 3Sc + cos 5Sc
13 13 13 13 13 13
10Sc 8Sc 3Sc 5Sc
13 13 13 13
( (= cos + cos + cos + cos
= cos Sc – 3Sc + cos Sc – 5Sc + cos 3Sc + cos 3Sc
13 13 13 13
3Sc 5Sc 3Sc 5Sc
= - cos 13 - cos 13 + cos 13 + cos 13 = 0 = RHS Proved.
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Example 12. (HIfesrei1n, Dsi1n+Dc+o1sDco=1sDsi=1nEs+i1nEco1+sEprove that cot D+E (, = tanD.tanE.
Solution: 1 2
cosE
1 1 1 1
or, sinD - sinE = cosE - cosD
or, sinE - sinD = cosD - cosE
sinD.sinE cosD.cosE
+ E-D E-D
2cos E 2 D . sin 2 2sin D E . sin 2
2
or, =
((or, sinD . sinE cosD . cosE
cos D+E ( sinD . sinE
2 ( cosD.cosE
=
D+E (
sin 2 (
(? (
cot D+E = tanD.tanE Proved.
2
Example 13. Prove that sin2D - sin2E = tan(D + E)
sinD.cosD - sinE.cosE
sin2D - sin2E 2sin2D - 2sin2E
Solution: LHS = sinD.cosD - sinE.cosE = 2sinD.cosD - 2sinE.cosE
1-cos2D - 1 + cos2E cos2E - cos2D
sin2D - sin2E sin2D - sin2E
( (= =
2sin 2E + 2D . sin 2D - 2E (
( (= 2 2 ( sin(D + E)
2cos 2D + 2E . sin 2D - 2E = cos(D + E)
22
= tan (D + E) = RHS. Proved.
Example 14. If sin2x + sin2y = 1 and cos2x + cos2y = 1 , then show that
2 3 2
tan (x + y) = 3 .
Solution: ( (Here, sin2x + sin2y = 1
3
1
2x + 2y 2x - 2y 3
2
or, 2sin (( . cos 2 =
or, (( ............. (i)
2sin (x + y).cos (x - y) = 1
3
1
( (Again, cos2x + cos2y = 2 1
2
2cos 2x + 2y . cos 2x - 2y =
or, 2 + y). cos(x y) =221
- ............. (ii)
2cos(x
Dividing (i) by (ii), we get,
2sin(x + y). cos(x - y) 1
2cos(x + y).cos(x - y) 3
= 1
tan (x + y) = 2 Proved. 2
3
?
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Exercise 9.3
Very Short Questions
1. Express each of the following as sum or difference form:
(a) 2sinA. sinB (b) 2cosA.cosB (c) 2cosA.sinB (d) 2sinA.cosB
(e) 2sin5T. cos3T (f) 2sin3x. sinx (g) 2sin20°.sin10° (h) 2cos40°.cos20°
2. Express each of the following as product form:
(a) sin6T + sin2T (b) sinx - siny (c) cosx - cosy
(d) sinx + siny (e) sin50° + sin40° (f) cos70° - cos40°
(g) cos40° - sin20° (h) sin5x - sin3x (i) cos7x - cos11x
(j) cos5x + cos2x
3. Prove the following:
(a) cos75° + cos15° = 3 (b) sin75° - sin15° = 1
2 2
(d) sin50° + sin70° = 3 cos10°
(c) sin18° + cos18° = 2 cos27°
(e) sin50° + sin10° = sin70° (f) cos52° + cos68° + cos172° = 0
Short Questions
4. Prove the following :
(a) cos5A + cos3A = cotA (b) cos40° - cos60° = tan50°
sin5A - sin3A sin60° - sin40°
cos80° + cos20° cos8° + sin8°
(c) sin80° - sin20° = 3 (d) cos8° - sin8° = tan53°
(e) cos10° - sin10° = cot55° (f) cos(40° + A) + cos(40° - A) = cotA
cos10°+ sin10° sin(40° + A) - sin(40° - A)
Long Questions
5. Prove the following :
(a) sinA.sin2A + sin3A.sin6A = tan5A
sinA.cos2A + sin3A.cos6A
cos2A.cos3A - cos2A.cos7A sin7A + sin3A
(b) sin4A.sin3A - sin2A.sin5A = sinA
(c) sinA + sin3A + sin5A + sin7A = tan4A
cosA + cos3A + cos5A + cos7A
cos7A + cos3A - cos5A -cosA
(d) sin7A - sin3A - sin5A + sinA = cot2A
(e) sin5A - sin7A - sin4A + sin8A = cot6A
cos4A - cos5A - cos8A + cos7A
sin(p+2)T - sinpT
(f) cospT - cos(p + 2) T = cot (p + 1) T
(g) (sin4A + sin2A) (cos4A - cos8A) =1
(sin7A + sin5A) (cosA - cos5A)
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6. Prove the following :
(a) ssccsiiioonnnss2124200000°°°°°.....cssscoiiionnnss4434100000°°°°0....css°siio.inncsno8665s00001°°°°.6.=.sc0ison°isn8=8837000°°18°===136111616
(b)
(c)
(d)
(e)
(f) tan20° tan40° tan80° = 3
7. Prove the following :
(a) sin(45° + T). sin(45° - T) = 1 cos2T
2
1
(b) cos(45° + T). cos(45° - T) = 2 cos2T
(c) cosT. cos(60°
– T) . cos(60 +T) = 1 cos3T
4
1
(d) sinx. sin(60° – x). sin(60° + x) = 4 sin3x
8. Prove that cos(36° - T).cos(36° + T) + cos(54° + T). cos(54° - T) = cos2T
((a)
9. Prove the following :
If 1 – 1 =- 1 – co1sB, prove that cot A+B ( = – tanA.tanB
sinA cosA sinB 2 (
((b)
If 4 + 2 = 2 + 4 , prove that tan D+T = 1
secT secD cosecD secD 2 2
((a)
10. Prove the following : A-B (
2 (
(cosA + cosB)2 + (sinA + sinB)2 = 4cos2
((b)
( ((c) = 4sin2 A -B
2
(d) 3
(cosB - cosA)2 + (sinA - sinB)2 2
sin2
sin2A Sc + A (( Sc A 1 sinA
8 2 - sin2 8 - 2 = 2
+ sin2 (A + 120°) + sin2 (A - 120°) =
(11. (a)
1 1 D+E 1
4 2 2 2
If sinD+ sinE= and cosD+cosE = , then(( prove that : tan =
(
(
((b) 1 1 D+E 3
If cosD + cosE = 3 and sinD + sinE = 4 , prove that : tan 2 = 4
( ((c) If sinx = k siny, prove that : tan
x-y = k-1 tan x+y
2 k +1 2
(d) If sin (A + B) = k sin (A - B), then prove that : (k - 1) tanA = (k + 1) tanB.
12. Prove that sin2x - sin2y = tan(x + y)
sinx.cosx - siny.cosy
1
(13. Prove that cos 2Sc + cos 4Sc + cos 6Sc = - 2
7 7 7
D+E (
14. Prove that x = y cot 2 if xcosD + ysinD = xcosE + ysinE
15. (a) Prove that 1 – cos10° + cos40° – cos50° = tan5°.cot20°
1 + cos10° – cos40° – cos50°
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(b) Prove that 1 – cosT + cosD – cos(D T) = tan T .cot D
1 + cosT – cosD – cos(D + T) 2 2
1. (a) cos(A - B) - cos(A + B) (b) cos(A + B) + cos(A - B)
(c) sin(A + B) - sin(A - B) (d) sin(A + B) + sin(A - B)
(e) sin8T + sin2T (f) cos2x - cos4x (g) cos10° - 3 (h) 1 + cos20°
2 2
x-y x+y
2. (a) 2sin4T . cos2T (b) 2sin 2 . cos 2
(c) 2sin x+y . sin y-x (d) 2sin x+y . cos x-y
2 2 2 2
(e) 2 cos5° (f) -2sin15° . sin55°
(g) 2sin55° . sin15° (h) 2cos4x . sinx
(i) 2sin9x . sin2x (j) 2cos 7x . cos 3x
2 2
9.4 Conditional Trigonometric identities
Identities which are true under given certain conditions are known as conditional identities.
In this section, we deal with some trigonometric identities like A + B + C = Sc,
then A + B = Sc - C, B + C = Sc - A, C + A = Sc - B
We can write,
(i) sin (A + B) = sin (Sc - C) = sinC
sin(B + C) = sin (Sc - A) = sinA
sin(C + A) = sin (Sc - B) = sinB
(ii) cos (A + B) = cos(Sc - C) = - cosC
cos (B +C) = cos (Sc - A) = - cosA
cos(C + A) = cos (Sc - B) = - cosB
(iii) tan(A + B) = tan (Sc - C) = - tanC
tan (B + C) = tan(Sc - A) = - tanA
tan(C + A) = tan (Sc - B) = - tanB
(b) If A + B + C = Sc,
then A + B + C = S2c.
2 2 2
A B C B C A C A B
So, we write 2 + 2 = Sc - 2 , 2 + 2 = Sc - 2 , 2 + 2 = Sc - 2
2 2 2
We write,
((i) (= sin
sin A + B (( (= sin Sc - C = cos C
(sin 2 2 (( 2 2 2
B C A A
2 + 2 Sc - 2 = cos 2
2
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(sin C A (= sin Sc B B
((ii) 2 + 2 (= cos (2-2= cos 2
cos A + B (( Sc - C = sin C
(cos 2 2 (((= cos 2 2 2
B C (( A A
(cos 2 + 2 (= cos(( Sc - 2 = sin 2
(( 2
C A (( B B
((iii) tan2 + 2 (= tan(( Sc - 2 = sin 2 .
2
A B C C
(tan 2 + 2 (= tan Sc - 2 = cot 2 ,
2
B C A A
(tan 2 + 2 (= tan Sc - 2 = cot 2
2
C A B B
2 + 2 Sc - 2 = cot 2
2
Worked Out Examples
Example 1. If A + B + C = Sc, prove that
Solution: (a) tanA + tanB + tanC = tanA . tanB . tanC
(b) tan A2 . tan B + tan B2 . tan C + tan C2 . tan A = 1
2 2 2
(a) tanA + tanB + tanC = tanA. tanB.tanC
Example 2. Here, A + B + C = Sc.
or, A + B = Sc - C
? tan (A + B) = tan (Sc - C)
or, tanA + tanB = – tanC
1 – tanA.tanB
or, tanA + tanB = – tanC + tanA.tanB.tanC
? tanA + tanB + tanC = tanA. tanB.tanC proved.
(b) tan A2 . tan B + tan B2 . tan C + tan C2 . tan A = 1
2 2 2
Here, A + B + C = Sc
or, A + B = Sc – C
2 2 2 2
tan(A2 B2 ) tan(S2c C
? + = – 2 )
tan A + tan B C
2 2 2
(or, A B = cot
1 - tan 2 . tan 2
or,
tan A + tan B = 1 - tanA2 . tan B 1
2 2 2
tan C
2
tanA2 C B C tanA2 B
or, . tan 2 + tan 2 . tan 2 = 1 - . tan 2 .
? tanA2 . tan B + tan B . tan C + tanC2 tanA2 = 1 Proved.
2 2 2
If A + B + C = 180°, then prove the following:
(a) sin2A + sin2B + sin2C = 4sinA.sinB.sinC
(b) cos2A + cos2B + cos2C = - 4cosA.cosB.cosC - 1
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Solution: (c) cos2A + cos2B - cos2C = 1 - 4sinA.sinB. cosC
(a) Here, A + B + C = 180°
sin (A + B) = sin (180° - C) = sinC
and cos(A + B) = cos (180° - C) = - cosC
( (LHS = sin2A + sin2B + sin2C 2A - 2B
2A +2B 2
= 2sin 2 ((. cos + 2sinC.cosC.
((
= 2sin(A + B). cos(A -B) + 2sinC.cosC ((
= 2sinC . cos (A -B) + 2sinC.cosC
= 2sinC[cos(A – B) + cosC]
= 2sinC[cos(A-B)- cos (A + B)]
= 2sinC. 2sinA.sinB
= 4sinA - sinB. sinC = RHS. Proved.
(b) Here, A + B + C = 180°
? cos(A + B) = cos(180° - C) = - cosC
( (LHS = cos2A + cos2B + cos2C 2A - 2B
= 2cos 2A + 2B . cos 2 + 2cos2C - 1
2
= 2cos(A + B). cos(A -B) + 2cos2C - 1
= 2(-cosC). cos (A -B) + 2cos2C - 1
= -2cosC [cos(A-B) - cosC] - 1
=-2cosC. [cos(A - B) + cos(A + B)]- 1
= - 2cosC.[2cosA.cosB]- 1
= -4cosA . cosB . cosC - 1 = RHS. Proved.
(c) Here, A + B + C = 180°
cos(A + B) = cos(180° - C) = -cosC
( (LHS = cos2A + cos2B - cos2C
= 2cos 2A+2B . cos 2A - 2B - 2cos2C + 1
2 2
= 2cos(A + B).cos(A - B) - 2cos2C + 1
= 2cos(A + B).cos(A -B) - 2cos2C + 1
= - 2cosC [cos (A - B) + cosC] + 1
= -2cosC[cos(A - B) - cos(A + B)] + 1
= - 2cosC. 2sinA.sinB + 1
= 1 - 4 sinA. sinB. cosC = RHS. Proved.
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Example 3. If A + B + C = 180°, then prove the following:
Solution:
(a) sinA + sinB + sinC = 4cos A cs.ocinossAA22 B2.. .sciconossB2B2 C2.. ssiinnC2C2
236 (b) cosA + cosB + cosC = 1 + 2
(c) cosA + cosB - cosc = - 1 + 4
4
Here, A + B + C = 180°
( ( ( (Now,
A + B (( C C
2 2 (( 2 2
sin = sin 90° - = cos
and cos A + B = cos 90° - C = sinC2
2 2 2
(a) LHS = sinA + sinB + sinC
( (=
(= A+2- B2sin+C2 s,incoCs
22scions C2A. +2coBs . cos( C
A-B ( 2
(
2((
[ ( ]= 2cosC2
cos A-B + sinC2
2
( ( ]= 2cosC2 [cos
A-B + cos A+B (
2 2
= 2cosC2 2cosA2 B
. . cos 2
= 4 cosA2 . cos B . cos C2 =RHS, Proved.
2
(b) LHS = cosA + cosB + cosC
( (=
(= 2cos A+ B (. cos A-B +1 - 2sin2 C
2 (A-B 2 2sin2 2
C ( C
2sin 2 . cos (( 2 +1- 2
[ ( ]=
2sin C . cos A-B - sin C +1
2 2 2
[ ( ( ]= 2 sinC2
cos A-B - cos A+ B ( +1
2 2
C 2sinA2 B
= 2sin 2 . . sin 2 + 1
= 1 + 4sinA2 . sin B . sin C = RHS. Proved.
2 2
(c) LHS = cosA + cosB - cosC
( (= 2cos
A+ B (. cos A- B - cosC
2 ( 2
(= C (A- B C
2sin 2 .cos (( 2 - 1 + 2sin2 2
[ ( ]= 2 sinC2
cos A-B + sinC2 - 1
2
[ ( ( ]= 2 sinC2
cos A-B + cos A+B ( -1
2 2
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= 2sin C . 2cosA2 . cos B - 1
2 2
= - 1 + 4 sinC2 . cosA2 . cosB2 = RHS. Proved.
Example 4. If A + B + C. = 180°, then prove the following:
Solution:
(a) sin2A + sin2B + sin2C = 2 + 2cosA.cosB.cosC
(b) cos2A + cos2B + cos2C = 1 - 2cosA.cosB.cosC.
(c) sin2A - sin2B + sin2C = 2sinA.cosB.sinC
We have, A + B + C = 180°
or, A + B = 180° - C
? sin (A + B) = sin(180° - C) = sinC
and cos(A +B) = cos(180° - C) = -cosC
Now,
(a) LHS = sin2A + sin2B + sin2C
= 1 [2sin2A + 2sin2B] + 1 - cos2C
2
1
= 2 [1 - cos2A + 1 - cos2B] + 1 - cos2C
1 - 1 (cos2A + cos2B) - cos2C
( (= + 1 2
1 2A+2B 2A - 2B
2 2 2
= 2 - .2 cos ((, cos . - cos2C
((
= 2 - cos(A + B).cos (A - B) - cos2C
= 2 + cosC . cos(A - B) - cos2C
= 2 + cosC [cos(A - B)- cosC]
= 2 + cosC [cos(A - B) + cos (A + B)]
= 2 + 2cosA. cosB. cosC = RHS. Proved.
(b) LHS = cos2A + cos2B + cos2C
= 1 [2cos2A + 2cos2B] + cos2C
2
1
= 2 [1 + cos2A + 1 + cos2B] + cos2C
1 [2 + (cos2A + cos2B)] + cos2C
( (=2
1 2A+2B 2A-2B
=1+ 2 .2.cos 2 . cos 2 + cos2C
= 1 + cos(A +B) . cos(A -B) + cos2C
= 1 + (- cosC). cos (A -B) + cos2C
= 1 - cosC [cos(A -B)- cosC]
= 1 - cosC [cos(A -B) +cos(A +B)]
=1 - cosC . 2cosA . cosB = 1 - 2cosA. cosB. cosC =RHS. Proved.
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(c) LHS = sin2A - sin2B + sin2C
= 1 [2sin2A - 2sin2B] + sin2C = 1 [1–cos2A–1+cos2B] + sin2C
2 2
1
2 (cos2B - cos2A) + sin2C
( (=
1 2B+2A 2A-2B
2 2 2
= . 2 sin (( . sin + sin2C
(
= sin(A +B). sin(A -B) +sin2C
((
= sinC. sin(A -B) + sin2C
((
= sinC [sin (A - B) + sinC]((
= sinC [sin (A -B) + sin (A + B)]((
= sinC. 2sinA. cosB
= 2 sinA cosB.sinC = RHS. Proved.
Example 5. If A + B + C = Sc, then prove that
Solution:
(a) sin2 A + sin2 B - sin2 C = 1 - 2cos A . cos B sin C
238 2 2 2 2 2 2
A B C A B C
(b) cos2 2 + cos2 2 - cos2 2 = 2cos 2 . cos 2 sin 2
Here, A + B + C = Sc
or, A+B = Sc - C
2 2
( (sin Sc C
A+B = sin 2 - 2 = cos C
2 2
( (cos Sc C C
A+B = cos 2 - 2 = sin 2
2
Now,
(a) LHS = sin2 A + sin2 B - sin2 C
2 2 2
] ]=1 A B C
2 2 sin2 2 + 2sin2 2 - sin2 2
= 1 [1 - cosA + 1 - cosB] - sin2 C
2 2
1 C
=1 - 2 [cosA + cosB] - sin2 2
( (= 1
- 1 .2cos A+B . cos A-B - sin2 C
2 2 2 2
] ( ]= C A-B C
1 - sin 2 cos 2 + sin2 2
] ( ( ]=
1 - sin C cos A-B + cos A+B
2 2 2
C A B
= 1 - sin 2 . 2cos 2 . cos 2
= 1 - 2cos A . cos B . sin C = RHS. Proved.
2 2 2
A B C
2 + 2 - cos2 2
] ](b)LHS = cos2 cos2
= 1 2cos2 A + 2cos2 B - cos2 C
2 2 2 2
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= 1 [1 + cosA + 1 + cosB] - 1 + sin2 C
2 2
1 C
= 1+ 2 [cosA + cosB] – 1 + sin2 2
( (=
1 A+B A-B C
2 2 2 2
. 2cos ((. cos + sin2
(= (
sin C . cos A-B ( + sin2 C
2 2 2
((
] ( ]= C
sin 2 . cos A-B + sin C
2 2
] ( ( ]=C
sin 2 . cos A-B + cos A-B
2 2
C A B
= sin 2 . 2cos 2 . cos 2
= 2cos A . cos B . sin C = RHS. Proved.
2 2 2
Example 6. (If A + B + C = Sc, prove thatABC Sc - A (cos Sc - B (. cos Sc - C
Solution: 2 2 2 4 4 4
cos + cos + cos = 4cos (( (
Example 7.
Solution: Here, A + B + C = Sc
or, A + B = Sc - C
B + C = Sc - A
C + A = Sc - B ? cos(C + A) = cos(Sc – B) = –cosB
Now,
LHS = cos A + cos B + cos C
2 2 2
= 2cos A/2 + B/2 . cos A/2 – B/2 + cos C + cosS2c
2 2 2
( ( ( (= 2 cos
A+B A-B C + Sc C - Sc
4 4 4 4
(( . cos + 2cos . cos((
( ( ( (= 2cos
Sc - C A-B Sc + C C - Sc
4 4 4 4
. cos(( + 2cos . cos((
( ] ( ( ]= 2cos
Sc - C A-B Sc + C
4 4 4
(( ( cos
+ cos ( cos(–T) = cos T
( ( (= 2cos
Sc - C A-B+Sc+ C A-B-Sc- C
4 8 8
(((
.2cos . cos
( ( (= 4cos
Sc - C Sc+A+C -B A-(B + C) - Sc
4 8 8
(( cos . cos (
( ] ( (= 4cos
Sc - C Sc+Sc-B -B A-Sc + A - Sc
4 8 8
((( cos . cos
( ( (= 4cos
Sc - C Sc - B A - Sc
4 4 4
( ((. cos . cos
( ( (= 4cos
Sc - A Sc - B Sc - C
4 4 4
.cos( (( cos = RHS. Proved.
If A + B + C = Sc, then prove that
cos (B + C - A) + cos (C + A - B) + cos (A + B - C) = 1 + 4cosA. cosB.cosC.
We have A + B +C = Sc
or, A + B = Sc - C, B + C = Sc - A, C + A = Sc - B
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Now,
LHS = cos (B + C - A) + cos (C + A - B) + cos (A + B - C)
= cos (Sc - A - A) + cos (Sc - B - B) + cos (Sc - C - C)
= cos (S - 2A) + cos (Sc - 2B) + cos(Sc - 2C)
= - cos2A - cos2B - cos2C
( (= - [cos2A + cos2B] - (2cos2C - 1)
2A+2B 2A-2B
2 2
= - 2cos (((. cos - 2cos2C + 1
((
= 1 - 2cos (A + B).cos(A -B)- 2cos2C
= 1 + 2cosC. cos(A -B) - 2cos2C
= 1 + 2cosC [cos(A - B) - cosC]
= 1 + 2cosC [cos(A - B) + cos (A +B)]
= 1 + 2cosC.2cosA. cosB
= 1 + 4 cosA. cosB. cosC = RHS. Proved.
Example 8. If A + B + C = 180°, then prove that
Solution: ( ( (sin(B+2C)+sin(C+2A)+sin(A+2B) = 4sin
B-C .sin C-A .sin A-B
2 2 2
Here, A + B + C = 180°
B + C = 180° - A
C + A = 180° - B , A + B = 180° - C
Now,
LHS = sin (B + 2C) + sin (C + 2A) + sin (A + 2B)
= sin (B + C + C) +sin (C + A + A) + sin (A + B + B)
= sin (Sc - A + C) + sin (Sc - B +A) + sin (Sc - C + B)
= sin{Sc - (A - C)} + sin {Sc -(B - A)} + sin {Sc - (C - B)}
= sin(A -C) +sin(B -A) + sin (C - B)
( (= 2sin
A-C+B-A A-C-B+A
2 2
(( . cos - sin (B - C)
( ( ( (= 2sin
B-C 2A - B - C B-C B-C
2 2 2 2
(( (. cos (
– 2sin . cos (
( ] ( ( ]= 2sin
B-C 2A - B - C B-C
2 2 2
(( (
cos – cos
( ] ( ( ]=2sin
B-C 2A - B – C + B – C B - C – 2A+ B + C
2 4 4
(( 2sin sin
( ( (= 2sin
B-C 2A - 2C 2B - 2A
2 4 4
(( (. sin . sin
( ( (= 4sin
B-C A-C B-A (
2 2 2
((. sin . sin ( sin(–T) = –sin T
( ( (= 4 sin
B-C C-A A-B
2 2 2
. sin((( . sin = RHS. Proved.
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Example 9. ( (If A + B + C = 180°, proved thatA A
Solution: tan 2 + tan B+C (= sec 2 .sec B+C (
2(( 2
Here, A + B + C = Sc, ((
( (sin
B+C (= sin Sc - C = cos C
2 2 2 2
( (and C
cos B+C = cos Sc - 2 = sin C
2 2 2
(LHS A B+C
= tan 2 + tan 2
= tan A + cotA2 = sinA/2 + cosA/2
2 cosA/2 sinA/2
sin2A/2 + cos2A/2
= sinA/2.cosA/2
= 1 = 1 . 1
sinA/2.cosA/2 sinA/2 cosA/2
(= cosecA2 . secA2
= sec A .sec B+C (= RHS. Proved.
2 2
Exercise 9.4
Very Short Questions
1. (a) Define conditional trigonometric identities with an example.
(b) If A,B and C are angles of a triangle ABC, then show that.
(i) sin(A + B) - sinC = 0
(ii) cos (B + C) + cosA = 0
(iii) tan(B + C) + tanA = 0
(c) If A + B + C = 180°, Show that :
(sin(B + 2C) + sin(C + 2A) + sin(A + 2B) = sin (A - C) + sin (B - A) + sin(C - B) C
(d) If A, B, C are angles of a triangle, show that tan A+B ( = cot 2
2
Short Questions
2. If A + B + C = Sc, prove that
(a) tanA + tanB + tanC - tanA tanB.tanC = 0
(b) cotA.cotB + cotB. cotC + cotC . cotA - 1 = 0
(c) tan A ..ctoatnB2B2. + tan B . tan C + tan C . tan A =1
(d) cot A2 2 2 2 2
2 C = A + B C
cot 2 cot 2 cot 2 + cot 2
(e) tan2A + tan2B + tan2C = tan2A . tan2B . tan2C
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Long Questions
3. If A + B + C = Sc, prove the following :
(a) sin2A + sin2B - sin2C = 4cosA.cosB.sinC
(b) sin2A - sin2B + sin2C = 4cosA.sinB.cosC
(c) cos2A + cos2B - cos2C = 1 - 4 sinA.sinB.cosC
4. If A + B + C = Sc, prove the following :
(a) sinA - sinB + sinC = 4sin A .cos B . sin C
2 2 2
A B C
(b) sinA + sinB - sinC = 4sin 2 .sin 2 . cos 2
(c) sinA - sinB - sinC = – 4cos A .sin B .sin C
2 2 2
A B C
(d) - sinA + sinB + sinC = 4cos 2 .sin 2 .sin 2
5. If A + B + C = 180° then prove that the following :
(a) cosA - cosB + cosC = 4cos A .sin B .cos C -1
2 2 2
A B C
(b) -cosA + cosB + cosC = -1 + 4 sin 2 .cos 2 .cos 2
(c) cosA - cosB - cosC = 1 - 4sin A .cos B . cos C
2 2 2
6. If A + B + C = Sc, prove the following :
(a) sin2A + sin2B - sin2C = 2sinA.sinB.cosC
(b) sin2A - sin2B - sin2C = - 2cosA.sinB.sinC
(c) sin2A - sin2B + sin2C = 2sinA.cosB.sinC
7. If A + B + C = Sc, prove the following :
(a) cos2A - cos2B + cos2C = 1 - 2sinA.cosB. sinC
(b) cos2A - cos2B - cos2C = - 1 + 2cosA.sinB.sinC
(c) cos2A + cos2B - cos2C = 1 – 2sinA.sinB.cosC
8. If A + B + C = Sc, prove the following :
(a) sin2 A + sin2 B + sin2 C = 1 - 2sin A .sin B .sin C
2 2 2 2 2 2
A B C A B C
(b) sin2 2 - sin2 2 + sin2 2 = 1 - 2cos 2 .sin 2 .cos 2
(c) sin2 A - sin2 B - sin2 C = 2sin A .cos B .cos C - 1
2 2 2 2 2 2
9. If A + B + C = Sc, prove the following :
( )(a) A B C A B C
cos2 2 + cos2 2 + cos2 2 = 2 1 + sin 2 . sin 2 . sin 2
(b) cos2 A - cos2 B - cos2 C = - 2sin A . cos B .cos C
2 2 2 2 2 2
A B C A B C
(c) cos2 2 - cos2 2 + cos2 2 = 2cos 2 . sin 2 .cos 2
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10. If A + B + C = Sc, prove that
(a) sin A + sin B + sin C = 1 + 4sin Sc- A . sin Sc- B . sin Sc- C .
2 2 2 4 4 4
A B C Sc-A Sc-B Sc-C
(b) cos 2 + cos 2 + cos 2 = 4cos 4 . cos 4 . cos 4
(c) cosA + cosB + cosC = 1 + 4cos Sc-A cos Sc-B cos Sc-C
2 2 2
11. If A + B + C = Sc, prove that
sin (B + C - A) + sin (C + A - B) + sin (A + B - C) = 4sinA.sinB.sinC
12. If A + B + C = Sc, prove that
( ) ( ) ( )(a) cos(B+ 2C) + cos(C + 2A) + cos(A + 2B) = 1 - 4cos
A -B .cos B-C .cos C -A
2 2 2
( ) ( ) ( )(b) sin(B+ 2C) + sin(C + 2A) + sin(A + 2B) = 4sin A -B C -A
2 .sin B-C .sin 2
2
(c) sin3A + sin3B + sin3C = – cos 32A.cos32B.cos32C
(d) cos4A + cos4B +cos4C = – 1 + 4cos2A . cos2B . cos23C
13. If D + E + J = Sc , prove that
2
(a) sin2D + sin2E + sin2J = 1 - 2 sinD. sinE.sinJ
(b) tanE.tanJ. + tanJ . tanD + tanD . tanE = 1
(c) cos(D - E - J) + cos (E - J - D) + cos(J - D - E) = 4cosD.cosE.cosJ
14. If A + B + C = Sc, prove that
(a) sin2A + sin2B + sin2C = 8 sinA2 sinB2 sin C
4cosA2 .cosB2 .cosC2 2
(b) cosA + cosB + cosC = 2
sinB.sinC sinC.sinA sinA.sinB
(c) sinA + sinB + sinC = 2tanA. tanB.tanC
cosB.cosC cosC.cosA cosA.cosB
(d) cosA.sinB.sinC + cosB.sinC.sinA + cosC.sinA.sinB = 1 + cosA.cosB.cosC.
(e) sinA.cosB.cosC + sinB.cosC.cosA + sinC.cosA.cosB = sinA.sinB.sinC
15. If A + B + C = Sc, prove that
tanA + tanB + tanC + tanB + tanC + tanA
tanB tanC tanA tanA tanB tanC
= cosA.secB.secC + cosB.secC.secA + cosC.secA.secB
16. If A + B + C = 2S, prove that
sin(S – A).sin(S – B) + sinS.sin(S – C) = sinA.sinB
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9.5 Trigonometric Equations
Let us consider the following equalities:
(a) sin2T + cos2T = 1 (b) 4sin2T = 1
(c) sec2T = 1 + tan2T (d) tanT = 1
3
Discuss the following questions from above statements.
(a) Which are identities ?
(b) Which are equations ?
(c) What are differences between identities and equations ?
Equalities like sin2T + cos2T = 1, sec2T = 1 + tan2T are satisfied by all values of the
angle T. So they are called identities.
But 4sin2T = 1 and tanT = 1 etc. are satisfied by only some values of the angle T. So they
3
are called trigonometric equations.
Definition: An equation containing the trigonometrical ratios of an unknown angle is called
trigonometric equation.
Example : sinT = 3 , 0° ≤ T ≤ 360°
2
For 0° ≤ T ≤ 360°,
sinT = 3
2
or, sinT = sin60° or sin120°
? T = 60° or 120°
To solve such trigonometric equations, we need some limitations for the value of variable.
This limitation is called the range of the variables.
We note that the values of sinT and cosT repeat after an interval Sc. If the equation involves a
variable T, 0 ≤ T ≤ 2Sc, then the solutions are called principal solutions. A general solution
is one which involves the integer 'n' and gives all solutions of trigonometric equation. But
we study only about principal solutions.
Example : Find the solution of cosT = 1 , 0° ≤ T ≤ 360°
2
We know that
cosT = cos(360° - T) = cosT
or cosT = 1 = cos30°
2
? T = 30°
or, cosT = 1 = cos(360° - 30°) = cos30° or cos330°
2
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Note :
Let T be any angle. Then we note the following :
i. The value of sinT lies between -1 and +1. Y
i.e. -1 ≤ sinT ≤ 1, otherwise there is no
solution. (180° – T),S (360° + T),
ii. The values of cosT lies between -1 and (sin and cosec positive) (All positive)
X' X
+1. i.e. -1≤ cosT ≤ 1, otherwise there is TO C
no solution. (tan and cot positive) ( cos and sec positive)
iii. The value of tanT is infinitely positive or (180° + T) (360° – T)
infinitely negative. it has no boundary of Y'
its values. i.e. - ∞ < tanT < + ∞
iv. Find the quadrants where the sign of value of trigonometric ratio of angle falls by
using the rule CAST.
Some Basic Ideas
i. If sinT = sinA, then T = A or 180° - A
Example : Solve : sinT = 1
2
1
sinT = 2 = sin45°
Since sine is positive in the first and the second quadrants, we write,
sinT = sin45° or sin(180° - 45°)
? T = 45° or 135°
ii. If sinT = - sinA, then T = 180° + A or, 360° - A
Example : Solve sinT = - 1
2
sinT = - sin45°
Since sine is negative in the third and the fourth quadrant we write,
sinT = sin (180° + 45°) or sin(360° - 45°)
? T = 225° or 315°
iii. If cosT = cosA, thenT = A or (360° - A)
Example : Solve cosT = 3
2
? cosT = cos30°
Since cos is positive in the first and the fourth quadrant, we write,
cosT = cos30° or cos(360° - 30°)
? T = 30° or, 330°
iv. If cosT = -cosA, then T = 180° - A or, 180° + A
Example : Solve : cosT = - 3 , cosT = - cos30°
2
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Since cos is negative in the second quadrant and third quadrant, we write,
cosT = cos (180° - 30°) or cos (180° + 30°)
? T = 150° or 210°
Working steps for solution of Trigonometric equations
The following are some hints that will be helpful in solving a trigonometric equations :
(i) Express all the trigonometric functions in terms of a single trigonometric function of
same angle if possible.
(ii) Transfer every terms of the equation to the left hand side.
(iii) If equation is quadratic, factorize the left hand side and equate each factor to zero.
(iv) If the equation is quadratic in certain trigonometric function use formulae for the
solution (in case not factorizable).
Note :
In the process of solving trigonometric equations if there is involvement of squaring
or cubing, which give rise to some additional equations and consequently some
additional roots. All the solutions thus obtained may not satisfy the given equation.
The values which do not satisfy are discarded: only the values which satisfy be given
equations are accepted. Each is to be checked. Only accepted values are the solution
of the given equation.
Example : sinT - cosT = 1 0 ≤ T ≤ 360°
Solution:
We can write, sinT = 1 + cosT
246
Squaring on both sides, we get
sin2T = 1 + 2cosT + cos2T
or, cos2T + 2cosT + 1 - 1 + cos2T = 0
or, 2cos2T + 2cosT = 0
or, 2cosT (cosT + 1) = 0
Either, 2cosT = 0 ................ (i)
cosT + 1 = 0 ........... (ii)
From equation (i), cosT = 0
or, cosT = cos90°, cos270°, ? T = 90°, 270°
or, cosT = -1 =cos180°
? T =180°
Hence, T = 90°, 180°, 270°
on checking, put T = 90°, in given equation
sin90° - cos90° = 1
sin90° - cos90° = 1
or, 1 = 1 (It is true.)
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Putting T = 180°,
sin180° - cos180° = 1
or, 0 + 1 = 1
? 1 = 1 (true)
Putting T = 270.
sin270° - cos270° = 1
or, -1 -0 = 1
or, -1 = 1 (false)
Hence, 270° is rejected.
? T = 90°, 180°
Some special cases for 0° ≤ T ≤ 360°
(i) If sinT = 0, then T= 0°, 180°, 360°
(ii) If cosT= 0, thenT = 90°, 270°
(iii) If tanT = 0, thenT = 0, 180°, 360°
(iv) If sinT = 1, then T = 90°
(v) if sinT = -1, then T = 270°
(vi) If cosT = -1, then T = 180°
(vii) If tanT = ∞, then T = 90°, 270°
(viii)90° × n ± , when n = odd number (i.e. 1, 3, 5, .....)
sin cos, tan cot, sec cosec
when n = even number (i.e. 2, 4, 6, .............)
sin sin, cos cos, tan tan
Worked Out Examples
Example 1. Solve the equation: (0° ≤ T ≤ 180°)
Solution:
(a) sinT = 3 , (b) tanT = 3 (c) cosT = 1
(a) Here, 2 = 2
3
sinT 2
since, 0 ≤ T ≤ 180°
sinT = 3 = sin60° or sin(180° - 60°)
2
? T = 60° 120°
(b) Here, tanT = 3
since 0° ≤ T ≤ 180°
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tanT = 3 = tan60°
? T = 60°
(c) Here, cosT = 1
2
since, 0° ≤ T ≤ 180°
cosT = 1 = cos45°
2
? T = 45°
Example 2. Solve the equation: 0° ≤ T ≤ 360°.
Solutions:
(a) 3 tanT = 1, (b) cosec2T = 2 (c) 3tan2T - 1 = 0
(a) Here, 3 tanT = 1
or, tanT = 1
3
Since tanT is positive, T lies in the first or the third quadrant.
? tanT = tan30° or, tan (180° + 30°)
? T = 30° or, 210°
(b) cosec2T = 2
or, 1 =2 or, sin2T = 1
sin2T 2
1
? sinT = ± 2
Taking positive sign, we get
sinT = 1
2
Since, sinT is positive, T lies in the first or second quadrant,
sinT = sin45° or, sin(180° - 45°)
? T = 45° or, 135°
Taking negative sign, we get
sinT = - 1
2
Since sinT is negative it lies in the third or fourth quadrant.
? sinT = sin(180° + 45°) or, sin(360° - 45)
? T = 225° or, 315°
Hence, the required values of T are 45°, 135°, 225° or 315°
(c) Here, 3 tan2T - 1 = 0
( )or, 1 2
tan2T = 3
or, tanT = ± 1
3
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Taking positive sign,
tanT = 1
3
Since tanT is positive, T lies in the first or third quadrant
? tanT = tan30° or, tan (180° + 30°)
? T = 30° or 210°
Taking negative sign, we get
tanT = - 1
3
Since tanT is negative, T lies in the second or fourth quadrant.
? tanT = tan(180° - 30°) or, tan (360° - 30°)
? T = 150° or, 330°
? T = 30°, 150°, 210°, 330°.
Example 3. Solve : tan 9T = cotT, 0° ≤ T ≤ 90°
Solution:
Here, tan 9T = cotT
or, tan9T = tan(90° - T), [cotT = tan(90° - T)]
tan(270° - T), tan(7×90° - T), tan(5×90° - T), tan(9×90° - T)
? 9T = 90° - T, 270° - T, 450° - T, 630° - T, 810° - T
or, 10T = 90°, 270°, 450°, 630°, 810°
? T = 9°, 27°, 45°, 63°, 81°
Example 4. 4sin2T + 3cosT = - 3, 0° ≤ T ≤ 360°
Solution:
Here, 4sin2T + 3cosT = - 3
or, 4 - 4cos2T + 3cosT + 3 = 0
or, - 4cos2T + 3cosT + 7 = 0
or, 4cos2T – 3cosT – 7 = 0
or, 4cos2T - 7cosT + 4cosT - 7 = 0
cosT (4cosT - 7) + 1 (4cosT - 7) = 0
or, (4cosT - 7) (cosT + 1) = 0
Either, 4cosT - 7 = 0 ........... (i)
cosT + 1 = 0 ............... (ii)
From equation (i) cosT = 7 >1
4
since, - 1 ≤ cosT ≤ 1, it has no solution.
Hence, cosT = 7 is rejected.
4
From equation (i), we get,
cosT = - 1
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or, cosT = cos180°
? T = 180°
Hence, the required solution is T = 180°.
Example 5. 2cosT = 3 cotT, 0° ≤ T ≤ 360°
Solution:
Here, 2cosT = 3 cotT
Example 6.
Solution: or, 2cosT = 3 cosT (Note: Do not cancel cosT)
sinT
or, 2cosT.sinT - 3 cosT = 0
or, cosT (2sinT - 3 ) = 0
Either, cosT = 0 .............(i)
or, 2sinT - 3 = 0 .............. (ii)
From equation (i), cosT = 0
or, cosT = cos90°, cos270° ? T = 90°, 270°
From equation (ii), sinT = 3
2
or, sinT = sin60° or, sin (180° - 60°)
? T = 60° or, 120°
Hence, the required solutions are x = 60°, 90°, 120°, 270°
Solve : 3 sinT - cosT = 1, 0° ≤ T ≤ 360°
Here, 3 sinT - cosT = 1............ (i)
coefficient of sinT = 3
coefficient of cosT = - 1
( 3)2 + (-1)2 = 3 + 1 = 2
Dividing equation (i) by '2' on both sides, we get,
3 sinT - 1 cosT = 1
2 2 2
1
or, cos30° . sinT - sin30° cosT = 2
or, sin(T - 30°) = sin30° or, sin (180° - 30°)
? T - 30° = 30° or, 150°
or, T = 30° + 30° or, 150° + 30°
? T = 60°, 180° are required solutions.
Alternative Method
Here, 3 sinT = 1 + cosT
Squaring on both sides, we get,
3sin2T = 1 + 2cosT + cos2T
or, 3(1 - cos2T) = 1 + 2cosT + cos2T
250 Approved by Curriculum Development Centre, Sanothimi, Bhaktapur
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Vedanta Opt. Maths Book 10 Final (2078)
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