5 a Multiple and Submultiple Angles

Trigonometry






























Multiple and Sub-Multiple Angles

Multiple Angles Sub-multiple Angles
1. sin 2 = 2 sin  cos   
1. sin  = 2 sin cos 
2 2
2 tan  2 tan /2
2. sin 2 =  2. sin  = 
1 + tan  1 + tan /2
2
2
3. cos 2 = cos  – sin   
2
2
2
2
3. cos  = cos – sin 
2 2
2
4. cos 2 = 2 cos  – 1 
2
4. cos  = 2 cos – 1
2
2
5. cos 2= 1 – 2 sin  
5. cos  = 1 – 2 sin 
2
2
1 – tan  1 – tan /2
2
2
6. cos 2 =  6. cos  =
1 + tan  1 + tan /2
2
2
2 tan  2 tan /2
7. tan 2 = 1 – tan   7. tan = 1 – tan /2 
2
2
2
2
cot  – 1 cot /2 – 1
8. cot 2 =  8. cot  = 
2 cot  2 cot /2
2
9. 1 + cos 2 = 2 cos  
9. 1 + cos  = 2 cos 
2
2
10. 1 – cos 2= 2 sin  
2
10. 1 – cos  = 2 sin 
2
2
11. sin 3 = 3 sin  – 4 sin   
3
11. sin  = 3 sin – 4 sin 
3
3 3
3
12. cos 3 = 4 cos  – 3 cos   
3
12. cos  = 4 cos – 3 cos 
3 3
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3 tan  – tan   
3
3
13. tan 3=  3 tan – tan
2
1 – 3 tan  3 3
13. tan  =  
2
1 – 3 tan
3

















































































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Transformation of Trigonometric Formulae


Product to Sum or difference Sum or difference to product
(a) sin (A + B) + sin (A – B) = 2 sin A cos B C + D C – D
(a) sin C + sin D = 2 sin cos
2 2
(b) sin (A + B) – sin (A – B) = 2 cos A sin B C + D C – D
(b) sin C – sin D = 2 cos sin
2 2
(c) cos (A + B) + cos (A – B) = 2 cos A cos B C + D C – D
(c) cos C + cos D = 2 cos cos
2 2
(d) cos (A – B) – cos (A + B) = 2 sin A sin B C + D D – C
Or (d) cos C – cos D = 2 sin 2 sin 2
cos (A + B) – cos (A – B) = – 2 sin A sin B
C + D C – D
= – 2 sin sin
2 2


Question for Knowledge Level (1 mark)


1. Express Cos2A in terms of sinA. 6. What is the formula for sin3A?
2. Express Cos2A in terms of CosA. 7. Write Cos3A in terms of cosA
3. Express Cos2A in terms of TanA. 8. What is the formula for tan3A?


4. Write Sin in terms of Cosθ 9. Express Sinθ + Sinα in terms of product.
2
10. Express 2Sinα Sinβ in terms of sum or difference.

5. Write Cos in terms of Cosθ
2

Question for Comprehension Level (2 marks)

3
1. (a) If sin = , find the values of cos 2 and sin 3.
4
12 4
(b) If cos  = and sin = , find the values of sin 2 and cos 2.
13 5
 3
2. If  is an acute angle and sin = , find the value of sin  
2 5
 3
3. (a) If cos = find the value of cos .
3 5
 3
(b) If sin = , find the value of sin 
3 5
 1
(c) If cos = , then show that cos = 1.
3 2
1 45 o
o
(d) If sin45 = , show that tan = 3  2 2
2 2
b
(e) If tan = , show that acos2 + bsin2 = a.
a
3 3 +1
o
o
(f) If cos330 = , then show that cos165 = 
2 2 2
1 1
(g) If tanA = and tanB = , then show that cos2A = sin4B.
7 3

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4. Find the value of :
3
(a) cos 2 when 5 sin  = 4 (b) sin 2A when tan A =
4
5. If tan A/2 = 3/4, find the numerical value of tan A.
6 Prove that :
2 sin + sin 2  cos  1+ tan /2
(a) = cot 2 (b) =
2 sin  – sin 2 2 1– sin  1– tan /2

1  A A   c    c   
2
2
(c) cot A = cot – tan  (d) cos  –  –sin  –  = sin
2  2  2  4  4  4  4 2
sin 2A – sin A
2
2
(e) sin (45° – ) + sin (45° + ) =1 (f) = tan A
1 – cos A + cos 2A
 c   
2
1 – tan  –   4  sin – 1 + sin  
2
 4
(g)  c   = sin (h)  = cot
2
2
2
1 + tan  –  cos – 1 + sin 
 4  4 2
2
(i) 1 – sin 2A = 2sin (45° – A) (j) 1 + cos 2A = 2 sin (90° – A)
2
1 + cos  
(k) = cot
sin  2
7. Prove that:
1  sin2A 1 + tanA 1 + sin2A cosA + sinA
(a) = (b) =
cos2A 1 – tanA cos2A cosA  sinA
sin + sin2
2
(c) 1  cos2 + 2cos  = 2 (d)
1 + cos + cos2
1  cos2A + sin2A
(e) tan  cot = 2cot2 (f) = tanA
1 + cos2A + sin2A
3
3
sin A  cos A 1
(g) = 1 + sin2A. (h) tan(45 + ) = sec2 + tan2.
sinA  cosA 2
cosA  1 + sin2A
(i) (2cos + 1) (2cos 1) = 2cos2 + 1. (j) = tanA.
sinA  1 + sin2A
2
6
2
6
(k) (sin + cos)  (sin  cos) = 2sin2 . cos2. (l) 4(sin  + cos ) = 4  3sin2.
2
2
2
(m) (1 + sin2 + cos2) = 4cos (1 + sin2) . (n) (sec2 + 1) sec   1 = tan2.
8. Prove that:
1 A A sin2A cosA A
(a) (cot  tan ) = cotA. (b) . = tan
2 2 2 1 + cos2A 1 + cosA 2
A A
cos + sin
1 + sinA 2 2 2tan 
(c) = . (d) = sec 2
cosA A A tan + sin 2
cos  sin
2 2
    1 1 + sinA + cosA A
3
3
(e) sin . cos + cos . sin = sin. (f) = cot .
2 2 2 2 2 1 + sinA  cosA 2

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
1 + cos + cos
      2 
(g) sec  +  . sec    = 2sec. (h) = cot .
4  2 4  2  2
sin + sin
2
9. Find the value without using calculator or table:
(a) sin 75° sin 15° (b) cos15°– cos75°
(c) cos 105° cos 15° (d) sin 105° sin 15°
(e) sin 70° – cos 80° + cos 140° (f) 4 sin 105° sin15°
(g) 4 cos 75° sin 105° (h) sin 75° + sin 15°

10. Without using calculator or table, prove that :
(a) cos 70° + cos 40° = 2 cos 55° cos 15°
3
(b) cos 75° + cos 15° =
2
(c) sin 50° + sin 70° = 3 cos 10°

11. Prove the followings:
(a) sin (45+) - cos(45) = 1 + sin2 (b) sin50º + sin70º = 3 cos10º
(c) cos40° – cos100º = sin 70° (d) cos40° – sin40º = 2 sin5°
1
(e) cos105° + sin105° = (f) sin42° + cos72º = cos12º
2
(g) cos20º + cos100º + cos140º = 0 (h) cos28º  sin58 ° sin2º = 0

12. Show that:
sinA + sinB  A + B  cos40° –cos60º
(a) = tan   (d) = tan50º
cosA + cosB  2  sin60º – sin40º

sinA + sinB  A + B   A – B  sin5 – sin3
(b) = tan   . cot   (e) = tan
sinA – sinB  2   2  cos5 + cos3
cosA + cosB  A + B   A – B 
(c) = cot   cot   (f) cos10º + cos20º + cos40º + cos50º = 2 3 cos15º cos5º
cosBcosA  2   2 
1 sin(40 + A) c sin(40 –A)
(g) cos(45°+ A) – cos(45° –A) = cos2A (h) = tanA
2 cos(40 + A) + cos(40 – A)

Question for Application Level (4 marks)

1. Prove that) M
1 3 1
6
6
(a) – = 4 (f) cos  + sin  = (5 + 3 cos 4)
sin 10° cos 10° 8
3 1 8 8 2 1 4
(b) – = 4 (g) cos  + sin  = 1 – sin 2 + sin 2
8
sin 20° cos 20°
1

(c) cos – sin = cos 2 1– sin 2   (h) cosec 2A + cot 4A = cot A – cosec4A
2
6
6
2
2
 4  (i) (1 + sin 2A + cos 2A) = 4 cos A(1 + sin 2A)
1 cos 4A 1 + cos 2 1 + cos  
(d) + = cot A – cosec 4A (j) sin 2 × cos  = cot
2
sin 2A sin 4A
(e) 3 cosec 20° – sec 20° = 4
2. Prove that:

2
2
2
1 1 (c) 2cos  + cos 2  2cos  . cos2 = 1.
(a)  = cot2A.
tan3A  tanA cot3A  cotA (d) cosec4A + cot8A = cot2A  cosec 8A.
2
2
2
2
(b) cos A + sin A . cos2B = cos B + sin B . cos2A.
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3 2  4  1
2
2
2
(e) sin (  120) + sin  + sin ( + 120) = . (n) cos . cos  +   . cos +   = cos3.
2  3   3  4
o
o
o
3
3
o
(f) 4(cos 20 + sin 50 ) = 3(cos20 sin50 ). 4  4 3 4 5 4 7 3
(g) 16sin . cos . cos2 . cos4 . cos8 = sin16. (o) sin + sin 8 + sin 8 + sin 8 = .
8
2
3
3
3
(h) cos  . cos3+ sin . sin3 = cos 2.     3   5   7  1
3
3

 8 
 8 
(i) 4sin  . cos3 + 4cos  . sin3 = 3sin4. (p) 1 + cos  1 + cos  1 + cos  1 + cos  =
 8 
 8
8
(j) (2cosA + 1) (2cosA  1) (2cos2A  1) = 2cos4A + 1  9 3 5
6
3
6
(k) 4(cos   sin ) = cos 2 + 3cos2. (q) 2cos . cos + cos 13 + cos = 0.
13
13
13

 
 (l) 2cos = 2 + 2+ 2+ 2+2cos16 . (r) sec  +   . sec      = 2sec.
sec4  1 tan2 4  2 4  2
(m) =      
sec8  1 tan8 (s) tan  +  + tan    = 2sec.
4  2 4  2
(t) tanA + 2tan2A + 4tan4A + 8cot8A = cotA.
sin 2
3. (a) If 2tan  = 3 tan , prove that: tan ( – ) = 5 – cos 2
1  1  1 1 
 3
(b) If cosA = a +  , show that cos3A = a + 3 
2   a 2  a 
A 1  1  1 1 
 3
(c) If sin = p +  , show that sinA = p + 3  .
3 2   p 2  p 

4. Prove the following trigonometric identities:

2
2
sin –sin  (e) 4 sin  sin (60° – ) sin (60° + ) = sin 3
(a) = tan (+)
sin a cos –sin  cos   c    c  
 4
cos  – sin  (f) sec  +  sec  –  = 2 sec 
 2
 2
 4
2
2
(b) = cot ( + ) c c c c

sin  cos  + sin  cos  (g) 2 cos cos 9 + cos 3 + cos 5 = 0
1 13 13 13 13
(c) cos (60° + ) cos (60° – ) cos  = cos 3 2 2
4 (h) 4 cosec 2A cot 2A = cosec A – sec A
(d) 4 cos (60° + ) sin (30° + ) cos  = cos 3

5. Prove the following:

cos2A . cos3A  cos2A . cos7A sin7A + sin3A (f) 16cos20 . cos40 . cos60 . cos80 . = 1
o
o
o
o
(a) =
sin4A . sin3A – sin2A . sin5A sinA
o
o
o
(g) 8cos40 . cos100 . cos160 = 1.
sin5A + sin4A + sin2A + sinA
(b) = tan3A. 3 2 1
cos5A + cos4A + cos2A + cosA (h) cos x sin x = (2 cos x – cos 3x – cos 5x)
16
sin2A + sin5A  sinA
(c) = tan2A. 2 2 2  A – B 
cos2A + cos5A + cosA (i) (cos A + cos B) + (sin A + sin B) = 4 cos   2 

(sin4 + sin2) (cos4  cos8) (j) cos10º + cos20º + cos40º + cos50º = 2 3 cos15º . cos5º.
(d) = 1
(sin7 + sin5) (cos  cos5)
3
o
o
o
o
(e) sin20 . sin40 . sin60 . sin80 =
16
tan
6. a) If tan = k, then show that (k + 1) sin(  ) = (k  1) sin( + ).
1 1  A + B  1
b) If cosA + cosB = and sinA + sinB = , prove that tan   =
2 4  2  2
c) An angle  is divided into two parts A and B such that tanA : tanB = x : y show that. sin(A  B) =  x  y   sin.
x + y 
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