Vedanta Optional Math Teacher's guide

6. Prove that:
1

a. cos4θ + sin4θ = 4 (3 + cos4θ)

Solution
LHS = cos4θ + sin4θ

= (cos2θ – sin2θ)2 + 2 sin2θ . cos2θ

= (cos2θ)2 + 1 (2 sinθ . cosθ)2
2

= cos22θ + 1 sin22θ
2

= 1 (2 cos22θ + sin22θ)
2

= 1 (cos22θ + sin22θ + cos22θ)
2

= 1 (1 + cos22θ)
2

= 1 (2 + 2 cos22θ)
4

= 1 (2 + 1 + cos4θ)
4

= 1 (3 + cos4θ) ( 2 cos2x = 1 + cos2x)
4

= RHS proved

b. cos6θ – sin6θ = cos2θ 1 – 1 sin22θ 
4 

Solution
LHS = cos6θ – sin6θ

= (cos2θ – sin2θ) (cos4θ + cos2θ . sin2θ + sin4θ)
= cos2θ {(cos2θ – sin2θ)2 + 2 sin2θ . cos2θ + sin2θ . cos2θ}

= cos2θ (cos22θ + 3 sin2θ . cos2θ)

= cos2θ cos22θ + 3 (2 sinθ . cosθ)2
4

= cos2θ 4 sins22θ + 3 sin22θ
4

= cos2θ 4 – 4 cos22θ + 3 sin22θ
4

= cos2θ 1 – 1 sin22θ
4

= RHS proved

Vedanta Optional Mathematics Teacher's Guide ~ 10 251

1
c. sin4θ = 8 (3 – 4 cos2θ + cos4θ)

Solution

RHS = 1 (3 – 4 cos2θ + cos4θ)
8

= 1 {3 – 4(1 – 2 sin2θ) + (1 – 2sin22θ)}
8

= 1 {3 – 4 + 8 sin2θ + 1 – 2 sin22θ}
8

= 1 {8 sin2θ – 2(2 sinθ . cosθ)2}
8

= 1 {8 sin2θ – 8 sin2θ . cos2θ}
8

= 1 . 8 sin2θ(1 – cos2θ)
8

= sin2θ . sin2θ

=sin4θ = LHS proved

1
d. cos8θ + sin8θ = 1 – sin22θ + 8 cos42θ

Solution

LHS = cos8θ + sin8θ

= (cos4θ + sin4θ)2 – 2 sin4θ . cos4θ

= {(cos2θ – sin2θ)2 + 2 sin2θ . cos2θ}2 – 2 sin4θ . cos4θ

=cos22θ + 1 sin22θ 2 – 1 sin42θ
2  8
1 1
=cos42θ + cos22θ . sin22θ + 4 sin42θ – 8 sin42θ

= (1 – sin22θ)2 + (1 – sin22θ) . sin22θ + 1 sin42θ
8
1
= 1 – 2 sin22θ + sin42θ + sin22θ – sin42θ + 8 sin42θ

= 1 – sin22θ + 1 sin42θ
8
= RHS proved

6. Prove that:
31

a. sin40° + cos40° = 4

Solution

LHS = 3 + 1
sin40° cos40°

252 Vedanta Optional Mathematics Teacher's Guide ~ 10

= 3 cos40° – sin40°
sin40° . cos40°

2  3 + 1 sin40°
 2 cos40° 2
=
sin40° . cos40°

= 2(cos30° . cos40° + sin30° . sin40°)
sin40° . cos40°

= 4 cos(40° – 30°)
2 sin40° . cos40°

= 4 cos10°
sin80°

= 4 cos10°
sin(90° – 10°)

= 4 cos10°
cos10°

= 4 = RHS proved

b. cosec10° – 3 sec10° = 4

Solution
LHS = cosec10° – 3 sec10°

= 1 – 3
sin10° cos10°

= cos10° – . 3 sin10°
sin10° cos10°

4  1 cos10° – 23sin10°
2
=
2 sin10° . cos10°

= 4 sin30° . cos10° – cos30° . sin10°
sin20°

= 4 sin(30° – 10°)
sin20°

= 4 sin20°
sin20°

= 4 = RHS proved

7. Prove that:

a. (2 cosθ + 1) (2 cosθ – 1) = 2 cos2θ + 1

Vedanta Optional Mathematics Teacher's Guide ~ 10 253

Solution
LHS = (2 cosθ + 1) (2 cosθ – 1)
= 4 cos2θ – 1
= 2(2 cos2θ – 1) + 1
= 2 cos2θ + 1
RHS proved

b. sec8θ – 1 = tan8θ
sec4θ – 1 tan2θ

Solution

LHS = sec8θ – 1
sec4θ – 1

= 1 – 1
cos8θ – 1

1
cos4θ

= 1 – cos8θ × cos4θ
cos8θ – cos4θ
1

= 2 sin24θ . cos4θ
cos8θ . 2 sin22θ

= (2 sin4θ . cos4θ) sin4θ
cos8θ . 2 sin22θ

= sin8θ . 2 sin2θ . cos2θ
cos8θ 2 sin22θ

= tan8θ
sin2θ

cos2θ

= tan8θ = RHS proved
tan2θ

c. tanθ + 2 tanθ + 4 tan4θ + 8 cot8θ = cotθ

Solution
Example (14) text book

d. sin2α – cos2α . cos2β = sin2β – cos2β . cos2α

Solution
LHS = sin2α – cos2α . cos2β
= sin2α – cos2α(2 cos2β – 1)

= sin2α – 2 cos2α . cos2β + cos2α

= sin2α + cos2α – 2 cos2β . cos2α

254 Vedanta Optional Mathematics Teacher's Guide ~ 10

= 1 – 2 cos2α . cos2β
= sin2β + cos2β – 2 cos2α . cos2β
= sin2β – cos2β(2 cos2α – 1)
= sin2β – cos2β . cos2α = RHS proved

e. 2 + 2 +   2 + 2 cos8θ= 2 cosθ
Solution
LHS = 2 + 2 +   2 + 2 cos8θ

= 2 + 2 +   2(1 + cos8θ)
= 2 + 2 +   2 . 2 cos24θ
= 2 + 2 + 2 cos4θ

= 2 + 2(1 + cos4θ)

= 2 + 4 cos22θ

= 2 + 2 cos2θ

= 2(1 + cos2θ)

= 2 . 2 cos2θ
= 2 cosθ = RHS proved

sin2α – sin2β
f. sinα . cosα – sinβ . cosβ = tan(α + β)

Solution

RHS = tan(α + β)

= 1 tanα – tanβ
– tanα . tanβ

= 1 tanα – tanβ
– tanα . tanβ

sinα + sinβ
cosα cosβ
=
sinα sinβ
1 – cosα . cosβ

= sinα . cosβ + sinβ . cosα × sinα . cosβ – sinβ . cosα
cosα . cosβ – sinα . sinβ sinα . cosβ – sinβ . cosa

Vedanta Optional Mathematics Teacher's Guide ~ 10 255

= sinα . cosa . cos2β – sin2α . cos2β – sin2β . cos2α cosβ + sin2β . sina . cosa
cos2α . sinb . cosβ – sin2α . sinb .

= sinα . sin2α(1 – sin2β) – sin2β(1 – sin2α) + sin2a)
cosa(cos2β + sin2b) – sinb . cosβ(cos2α

= sinα sin2α – sin2β cosβ
. cosα – sinβ .

= LHS proved

8. a) 4(cos310° + sin220°) = 3(cos10° + sin20°)

Solution
LHS = 4(cos310° + sin220°)

= 4 cos310° + 4 sin220°

= 3 cos10° + cos(3.10°) + 3 sin20° – sin(3.20°)

= 3(cos10° + sin20°) + 3(cos30° – sin60°)

= 3(cos10° + sin20°) + 3 3 – 23 
2

= 3(cos10° + sin20°) + 3.0

= 3(cos10° + sin20°) = RHS proved

b) sin310° + cos320° = 3 (cos20° + sin10°)
4

Solution

LHS = sin310° + cos320°

By using formula,
1

cos3θ = 4 (3 cosθ + cos3θ)

1
LHS =si14n3[θ3=si4n1(30°si–nθsi–n(s3in.13θ0)°) + 3 cos20° + cos(3.20°)]
1
= 4 [3 sin10° – sin30° + 3 cos20° + cos60°]

= 3 (sin10° + cos20°) + 1  –12 + 1 
4 4  2 

= 3 (sin10° + cos20°)
4

= RHS proved

9. Prove that: 2 cos2A
1 + sin2A
a) cot(A + 45°) – tan(A – 45°) =

256 Vedanta Optional Mathematics Teacher's Guide ~ 10

Solution

LHS = cot(A + 45°) – tan(A – 45°)

= tan 1 45°) – tan(A – 45°)
(A +

= 1 – 1 tanA – tan45°
tanA + tan45° + tanA . tan45°

1 – tanA . tan45°

= 1 – tanA – tanA – 1
1 + tanA 1 + tanA

= 1 – tanA – tanA + 1
1 + tanA

= 2(1 – tanA)
1+ tanA

1 – sinA
cosA
= 2
sinA
1 + cosA

= 2 cosA – sinA × cosA + sinA
cosA + sinA cosA + sinA

= 2 cos2A + cos2A – sin2A . cosA
sin2A + 2 sinA

= 2 cos2A = RHS proved
1 + sin2A

b) tan(A + 45°) + tan(A – 45°) = 2 tan2A

Solution

LHS = tan(A + 45°) + tan(A – 45°)

= tanA + tan45° + 1 tanA – tan45°
1 – tanA . tan45° + tanA . tan45°

= tanA + 1 + tanA – 1
1 – tanA 1 + tanA
(1 + tanA)2 + (tanA – 1) (1 – tnaA)
= 1 – tan2A

= 1 + 2 tanA + tan2A + tanA – tan2A – 1 + tanA
1– tan2A

= 4 tanA

1 – sin2A
cos2A

= 4 sinA × cos2A
cosA cos2A – sin2A

Vedanta Optional Mathematics Teacher's Guide ~ 10 257

= 2 2 sinA cosA
cos2A
sin2A
= 2 cos2A

= 2 tan2A = RHS proved

c) tan(A + 45°) – tan(A – 45°) = 2 sec2A

Solution

LHS = tan(A + 45°) – tan(A – 45°)

= tanA + tan45° – 1 tanA – tan45°
1 – tanA . tan45° + tanA . tan45°

= tanA + 1 – tanA – 1
1 – tanA 1 + tanA
tanA + 1 1 – tanA
= 1 – tanA + 1 + tanA

= (1 + tanA)2 + (1 – tanA)2
1 – tan2A
1 + 2 tanA + tan2A + 1 – 2 tanA – tan2A
= 1 – tan2A

= 2 (1 + tan2A)
1 – tan2A
1
= 2 1 – tan2A

1 + tan2A

= 2
cos2A

= 2 sec2A = RHS proved

d) tanA + tan π + A  – tan π – A = 3 tan3A
3 3

Solution

LHS = tanA + tan π + A  – tan π – A
3  3

= tanA + tan23π + tanA – tan3π – tanA
1 – tanπ3 . tanA + tan3π . tanA
1

= tanA + 1 3 + tanA – 1 3 – tanA
– 3 tanA + 3 tanA

= tanA + ( 3 + tanA) (1 + 3 tanA) – ( 3 – 3tanA) (1 – 3 tanA)

1 – 3 tan2A

258 Vedanta Optional Mathematics Teacher's Guide ~ 10

= tanA + 3 + 3 tanA + tanA + 3 tan2A – 3 + tanA + tanA – 3 tan2A

1 – 3 tan2A

= tanA + 1 8 tanA
– 3tan2A

tanA – 3 tan3A + 8 tanA

= 1 – 3 tan2A

9 tanA – 3 tan3A
= 1 – 3 tan2A

3(3 tanA – tan3A)
= 1 – 3 tan2A
= 3 tan3A = RHS proved

sin2β
10. a) If 2 tanα = 3 tanβ, prove that : tan(α – β) = 5 – cos2β

Solution

Given, 2 tanα = 3 tanβ

or, tanα = 3 tanβ
2

LHS = tan(α – β)

= 1 tanα – tanβ
+ tanα . tanβ

3 tanβ – tanβ
2
=
3
1 + 2 tanβ . tanβ

= 3 tanb – 2 tanβ
2 + 3 tan2b

sinβ

= cosβ

2 + 3 sin2β
cos2β

= sinβ × 2 cos2β sin2b
cosb cos2b + 3

= 4 2 sinβ . cosβ
cos2b + 6 sin2b

= 4 – sin2β 6 sin2b
4sin2b +

= 4 sin2β
+ 2 sin2b

= 5 – sin2β
1 + 2 sin2b

Vedanta Optional Mathematics Teacher's Guide ~ 10 259

= 5 – sin2β
(1 – 2 sin2b)

= 5 sin2β = RHS proved
– cos2β

11
b) tanθ = 7 and tanb = 3, prove that : cos2θ = sin4b

Solution

Here, tanθ = 1 , tanb = 1
7 3

LHS = cos2θ = 1 – tan2q
1 + tan2q

= 1 –  1 2
1 + 7 

 1 2
7 

1 – 1
49
=
1
1 + 49

= 48 × 49
49 50

= 24
25

RHS = sin4b = sin2(2b)

= 2 sin2b . cos2b

= 2 1 2 tanb × 1 – tan2b
+ tan2b 1 + tan2b

= 4 × 1 1 × 1 – 1
3 9 1 + 9
1 1
+ 9

= 4 × 9 × 8 × 9
3 10 9 10

= 24
25

 LHS = RHS proved

11. Prove that :

a) cosA – 1 + sin2A
sinA – 1 + sin2A = tanA

260 Vedanta Optional Mathematics Teacher's Guide ~ 10

Solution

LHS = cosA – 1 + sin2A
sinA – 1 + sin2A

= cosA – sin2A + cos2A + 2 sinA cosA
sinA – sin2A + cos2A + 2 sinA cosA

= cosA – (sinA + cosA)2
sinA – (sinA + cosA)2

= cosA – sinA – cosA
sinA – sinA – cosA

= –sinA
–cosA

= tanA = RHS proved

b. 1 – 1 cotθ= cot4θ
tan3θ + tanθ cot3θ +

Solution

LHS = 1 tanθ – 1 cotθ
tan3θ + cot3θ +

= 1 tanθ – 1 1 1
tan3θ + tan3θ + tanθ

= 1 tanθ – tanθ . tan3θ
tan3θ + tanθ + tan3θ

= 1 – tanθ . tan3θ
tan3θ + tanθ

1
= tan3θ+ tanθ

1 – tanθ . tan3θ

= 1 + θ)
tan(3θ

= cot4θ = RHS proved

c. cotA – tanA = 1
cotA – cot3A tan3A – tanA

Solution

LHS = cotA – tanA
cotA – cot3A tan3A – tanA

1

= tanA – tanA
tan3A – tanA
1 – 1
tanA tan3A

Vedanta Optional Mathematics Teacher's Guide ~ 10 261

= 1 tanA . tan3A – tanA
tanA tan3A – tanA tan3A – tanA

= tanA – tanA
tan3A – tanA tan3A – tanA

= tan3A – tanA
tan3A – tanA

= 1 = RHS proved

12. Prove that:

a) 8 1 + sin π  1 + sin 3π  1 – sin 5π  1 – sin 7π  = 1
8  8  8  8 

Solution

LHS = 8 1 + sin π  1 + sin 38π  1 – sin 58π  1 – sin 7π 
8 8

= 8  1 + sin π  1 + sin 38π  1 – sin  π – 3π  1 – sin π – π 
 8   8  8 

= 8 1 + sin π  1 + sin 38π  1 – sin 38π  . 1 – sin π8
8

= 8 1 – sin2 π  . 1 – sin2 3π 
8 8

= 8 cos2 π . cos2 3π  ∴cos2 3π = cos2  π – π  = sin2 π 
8 8  8  2 8  8 
π π
= 8 cos2 8 . sin2 8

= 2  4 sin2 π . cos2 π 
 8 8 

= 2 2 sin π . cos π 2
8 8 

= 2 sin2 π
4

= 2  1 2
 2

= 1 = RHS proved

b) sin4  π  + sin4  3π  + sin4  5π  + sin2  7π  = 3
8 8 8 8 2

Solution

LHS = sin4  π  + sin4  3π  + sin4  5π  + sin2  7π 
8 8 8 8

= sin4  π  + sin4  3π  + sin4 2π + 8π + sin4 2π + 3π 
8 8 8

= sin4  π  + sin4  3π  + cos4 8π + cos4  3π 
 8   8   8 

= sin4  π  + sin4  π – 8π + cos4  π  + cos4  π – π 
 8   2  8   2 8 

262 Vedanta Optional Mathematics Teacher's Guide ~ 10

= sin4  π  + cos4  π  + cos4  π  + sin4  π 
8 8 8 8

=2 sin4  π  + cos4  π 
 8   8 

=2 sin2  π  + cos2  π  2 – 2 sin2  π  . cos2  π 
 8   8   8   8 

=2 1 – 1 4 sin2  π  . cos2  π 
2 8 8

=2 1 – 1 sin2 π
2 4

= 2  1 – 1 . 12
 2

= 2 1 – 14

= 2 . 3
= 4
3 RHS proved
2 =

Questions for practice

1. If cosθ = 1 , then find the values of sin2θ, cos2θ and tan2θ.
2

1+cos2θ
2. Prove that cotθ = ± 1– cos2θ

3. Prove the following :

(a) 2sin2  π – A= (1– sinA)
 4

(b) tanα + cotα = 2cosec2α

(c) tan 4π + θ = cos2θ
1–sin2θ
π
(d) 1–tan2  4 – θ = sin2θ

1 + tan2  π – θ
 4

(e) 1+sin2A =  cotA+1 2
1–sin2A  cotA–1 

4. Prove the following:

(a) cos2θ + sin2θ.cos2ß = cos2ß + sin2ß . cos2θ

(b) cosec 20° + cot 40° = cot 10° – cosec 40°

(c) 4 cosec2θ . cot2θ = cosec2θ – sin2θ

5. Prove that : tanθ + 2tan2θ + 4cot4θ = cotθ

6. Prove that : tanθ + tan 3π + θ + tan  2π – θ = 3 tan3θ.
3

Vedanta Optional Mathematics Teacher's Guide ~ 10 263

Sub multiple angles

Estimated Periods: 7

1. Objectives Objectives
S,N. Level
To define sub-multiple angles of an angle.
(i) Knowledge(k) To tell the formulae of trigonometric ratios of sub-multiple angles.

(ii) Understanding(U) To explain to derive the formulae of sub-multiple angles by
using compound angle formulae.
(ii) Application(A)
To solve problems of trigonometric identities of sub-multiple angles.
(iv) Higher Ability
(HA) To solve very long question of trigonometric ratios multiple
angles.

2. Teaching Materials
Formula chart of trigonometric ratios sub-multiple angles.

Teaching Strategies

– Review the formulae of trigonometric ratios of compound angles and multiple angles.
q q q
– Define sub-multiple angles of q as 2 , 3 , 4 etc.

– Show how to derive the formulae of trigonometric ratios of sub-multiple angles by
using compound angle formulae.

– Compare formula of multiple and sub-multiple angle formulae of rigonometry like

sin2q = 2 sinq . cosq

sinq = sin2  q  = 2 sin q . cos q .
 2  2 2

cos2q = cos2q – sin2q = 2 cos2q – 1 = 1 – tan2q
1 + tan2q

 q  cos22q q 1 – tan2 q
 2  2 2
cosq = cos2 – – sin2 =
1 + tan22q
12.
5. Discuss how to evaluate values of sinq, cosq, tanq if sinq =

6. Discuss to solve problems related to trigonometric ratios of sub-multiple questions

given in exercises.

List of formulae: 2 tan q
2
1. sinθ = 2 sin q . cos q =
2 2 1 – tan22q

264 Vedanta Optional Mathematics Teacher's Guide ~ 10

cos22q q cos22q q 1 – tan2 q
2 2 1 2
2. cosθ = – sin2 = 2 – 1 = 1 – 2 sin2 =
+ tan22q

3. sinθ = 3 sin3q – 4 sin3 q
3

cosθ = 4 cos3 q – 3 cos q
3 3

3 tan q – tan3 q
1– 3 tan23q 3
4. tanθ =
3

cot2 q – 1
2
5. cotθ =
q
2 cot 2

cot3 q – 3 cot q
3 3
6. cotθ =
3 cot23q – 1
Some solved problems

Prove the following :

cot3  x  – 3 cot 3x
3
1. cotx =
 x 
3 cot2 3 – 1

Solution

LHS = cotx = cot  x + cot23x 
 3 

cot x . cot 2x – 1
3 3
=
2x x
cot 3 + cot 3

x cot2 x – 1
3 3
= cot . – 1
x
2 cot 3

cot2 x – 1 x
3 3
= + cot
x
2 cot 3

cot3  x  – 3 cot  x 
 3   3 
=
 x  1
3 cot2  3  –

Vedanta Optional Mathematics Teacher's Guide ~ 10 265

= RHS proved

2. a) If cos q = 1 p + p1  , then prove that : cosθ = 1 p3 + 1 
3 2 2 p3

Solution

Here, cos q = 1 p + 1 
3 2 p

LHS = cosθ = 4 cos3 q – 3 cos q
3 3

= 4 . 1 p + 1 3 – 3 p + 1 
8 p  2 p

= 1 p + 1   p + 1 2 – 3
2 p p 

= 1 p + 1  p2 + 2 . p . 1 + 1  – 3
2 p p p2

= 1 p + 1   p2 + 2 + 1 – 3
2 p p2

= 1 p + 1   p2 – 1 + 1 
2 p p2

= 1 p3 + 1 
2 p3

= RHS proved

b) If sin q = 1 p + 1  , then prove that : cosθ = –21 p2 + 1 
2 2 p p2

Solution

LHS = cosθ = 1 – 2 sin2 q
2

= 1 – 2 . 1 p + 1 2
4 p 

= 1 – 1 p2 + 2 . p . 1 + 1 
2 p2 p2

= 1 – 1 p2 + 1 + 2
2 p2

= 1 2 – p2 – 1 – 2
2 p2

= –12 p2 + 1 
p2

266 Vedanta Optional Mathematics Teacher's Guide ~ 10

= RHS proved

3. Prove the following

1 – sinA 1 – tan A
cosA 2
a) =
1 + tanA2

Solution

LHS = 1 – sinA
cosA

1 – 1 2 tan A/2
+ tan2 A/2
=
1 – tan2A/2

1 + tan2A/2

1 + tan2 A – 2 tan A
1– 2 2
=
A
tan2 2

1 – tan A 2
2 
= tan
 A2  1 – tan A2 
1 +

1 – tan A
1 2
=
+ tanA2
= RHS proved

b) 1 – 2 sin2  p – θ  = sinθ
4 2

Solution

LHS = 1 – 2 sin2  p – θ 
4 2

= cos2  p – θ 
 4 2 

= cos  p – θ
2

= sinθ = RHS proved

1– tan2  p – θ  θ
1+  4 4  2
c) = sin
 p θ 
tan2  4 – 4 

Vedanta Optional Mathematics Teacher's Guide ~ 10 267

Solution

1– tan2  p – θ 
1+  4 – 4 
LHS =
 p θ 
tan2  4 4 

= cos2  p – θ 
 4 4 

= cos  p – θ 
 2 2 

= sin θ = RHS proved
2

4. Prove that

a) cos4  θ  – sin4  θ  = cosθ
2 2

Solution

LHS = cos4  θ  – sin4  θ 
2 2

= cos2 θ + sin2 θ   cos2 θ – sin2 θ 
2 2 2 2

= 1 . cosθ

= cosθ = RHS proved.

b) 2 sinθ – sin2θ = tan
2 sinθ + sin2θ

Solution

LHS = 2 sinθ – sin2θ
2 sinθ + sin2θ

= 2 sinθ – 2 sinθ . cosθ
2 sinθ + 2 sinθ . cosθ

= 2 sinθ (1 – cosθ)
2 sinθ (1 + cosθ)

= 2 sin2 θ
2 cos2 2

θ
2

= tan2 θ = RHS proved.
2

268 Vedanta Optional Mathematics Teacher's Guide ~ 10

c) 1 sin2θ . 1 cosθ = tan θ
+ cos2θ + cosθ 2

Solution

LHS = 1 sin2θ . 1 cosθ
+ cos2θ + cosθ

= sin2θ cosθ . 1 cosθ
2 cos2θ + cosθ

= 1 sinθ
+ cosθ

2 sin θ . cos θ
2 2
=
θ
2 cos2 2

= tan θ = RHS proved
2

d) 1 + sinθ = tan2  p + θ 
1 – sinθ  4 2 

Solution

RHS = tan2  p + θ 
4 2

= tan  p + θ  2
4 2

tan p + tan θ 2
4 2
=
p θ
1 – tan 4 . tan 2

sin θ 2
cos 2
1 +
θ
= 2

sin θ 2
1– 2

cos θ
2

= cos θ + sin θ 2
2 2 

cos θ – sin θ 2
2 2 

= cos2 θ + sin2 θ + 2 cos θ . sin θ
2 2 2 2

cos2 θ + sin2 θ – 2 cos θ . sin θ
2 2 2 2
Vedanta Optional Mathematics Teacher's Guide ~ 10 269

= 1 + sinθ = LHS proved
1 – sinθ

cos θ – 1 + sinθ θ
e) 2 1 + sinθ 2
= tan
θ
sin 2 –

Solution

LHS = cos θ – 1 + sinθ
2 1 + sinθ

sin θ –
2

cos θ – sin2 θ + cos2 θ + 2 sin θ . cos θ
2 2 2 2 2
=

sin θ – sin2 θ + cos2 θ + 2 sin θ . cos θ
2 2 2 2 2

cos θ – sin θ + cos θ 2
2 2 2
=

sin θ – sin θ + cos θ 2
2 2 2

= cos θ – sin θ – cos θ
2 2 2

sin θ – sin θ – cos θ
2 2 2

= –sin θ
2

–cos θ
2

= tan θ = RHS proved.
2

5. Prove the following

a) tan  p + θ  = secθ + tanθ
4 2

Solution

LHS = tan  p + θ 
 4 2 

= tan p + tan θ
4 2

1 – tan p . tan θ
4 2

270 Vedanta Optional Mathematics Teacher's Guide ~ 10

= 1 + tan θ
1 – tan 2

θ
2

= 1 + sin q/2
cos q/2

1 – sin q/2
cos q/2

= cos θ + sin θ × cos θ + sin θ
2 2 2 2

cos θ – sin θ cos θ + sin θ
2 2 2 2

= cos θ + sin θ 2
2 2 

cos2 θ – sin2 θ
2 2

= 1 + sinθ
cosθ

= 1 + sinθ
cosθ cosθ

= secθ + tanθ = RHS proved.

b) tan  p – θ  = 1 – sinθ
4 2 1 + sinθ

Solution

RHS = 1 – sinθ
1 + sinθ

sin2 θ + cos2 θ – 2 sin θ . cos θ
2 2 2 2
=

sin2 θ + cos2 θ + 2 sin θ . cos θ
2 2 2 2

cos θ – sin θ 2
2 2 
=

cos θ + sin θ 2
2 2 

= cos θ – sin θ
2 2

cos θ + sin θ
2 2

Dividing numerator and denominator by cos θ , we get
2

Vedanta Optional Mathematics Teacher's Guide ~ 10 271

= 1 – tan θ
2

1 + tan θ
2

= tan p – tan θ
4 2

1 + tan p . tan θ
4 2

= tan  p – θ  = LHS proved
4 2

c) sec  p + θ  . sec  p + θ  = 2 secθ
4 2 4 2

Solution

LHS = sec  p + θ  . sec  p + θ 
 4 2   4 2 

11
= –
p θ p θ
cos  4 – 2  cos  4 + 2 

= 11 .

cos p . cos θ + sin p . sin θ cos p . cos θ + sin p . sin θ
4 2 4 2 4 2 4 2

= 11 .

1 . cos θ + 1 . sin θ 1 . cos θ + 1 . sin θ
2 2 2 2 2 2 2 2

= 2

cos θ + sin θ  cos θ – sin θ 
2 2 2 2
2
=
θ θ
cos2 2 – sin2 2

= 2
cosθ

= 2 secθ = RHS proved.

d) tan  p – θ  = cosθ
4 2 1 + sinθ

Solution

LHS = tan  p – θ 
4 2

= tan p – tan θ
4 2

1 + tan O4pp.ttioanna2θl Mathematics Teacher's Guide ~ 10
272 Vedanta

= 1 – sin q/2
cos q/2

1 + sin q/2
cos q/2

= cos θ – sin θ
2 2

cos θ + sin θ
2 2

= cos θ – sin θ cos θ + sin θ
2 2 × 2 2

cos θ + sin θ cos θ + sin θ
2 2 2 2

= cos2 θ – sin2 θ
2 2

sin θ + cos θ 2
2 2 

= cosθ

sin2 θ + cos2 θ + 2 sin θ . cos θ
2 2 2 2

= 1 cosθ = RHS proved.
+ sinθ

e) cot  θ + p  – tan  θ – p  = 2 cosθ
2 4 2 4 1 + sinθ

Solution

LHS = cot  θ + p  – tan  θ – p 
2 4 2 4

= cot θ . cot p – 1 – tan θ – tan p
2 4 2 4

cot θ + cot p 1 + tan θ . tan p
2 4 2 4

= cos q/2 – 1 – sin q/2 –1
sin q/2 cos q/2
sin q/2
cos q/2 + 1 1+ cos q/2
sin q/2

= cos θ – sin θ – sin θ – cos θ
cos 2 + sin 2 cos 2 + sin 2

θ θ θ θ
2 2 2 2

= cos θ – sin θ + cos θ – sin θ
2 2 2 2

cos θ + sin θ cos θ + sin θ
2 2 2 2

Vedanta Optional Mathematics Teacher's Guide ~ 10 273

= 2 cos θ – sin θ 
2 2

cos θ + sin θ 
2 2

=2 cos θ – sin θ × cos θ + sin θ
2 2 2 2

cos θ + sin θ cos θ + sin θ
2 2 2 2
θ θ
=2 cos2 2 – sin2 2

cos θ + sin θ 2
2 2 
2 cosθ
=
θ θ θ θ
cos2 2 + sin2 2 + 2 sin 2 . cos 2

= 2 cosθ = RHS proved.
1 + sinθ

f) tan  p + θ  + tan  p – θ  = 2 secθ
4 2 4 2

Solution

LHS = tan  p + θ  + tan  p – θ 
 4 2   4 2 

= tan p + tan θ + tan p – tan θ
4 2 4 2
p
1 – tan 4 . tan θ 1 + tan p . tan θ
2 4 2

= 1+ sin q/2 + 1 – sin q/2
1– cos q/2 cos q/2
sin q/2
cos q/2 1 + sin q/2
cos q/2

= cos θ + sin θ + cos θ – sin θ
2 2 2 2

cos θ – sin θ cos θ + sin θ
2 2 2 2

= cos θ + sin θ 2 + cos θ – sin θ 2
2 2  2 2 

cos θ – sin θ  cos θ + sin θ 
2 2 2 2

= 1 + sinθ + 1 – sinθ
θ θ
cos2 2 – sin2 2

= 2
cosθ

= 2 secθ = RHS proved.

6. Prove that

274 Vedanta Optional Mathematics Teacher's Guide ~ 10

a) (cosa – cosb)2 + (sina – sinb)2 = 4 sin2 a – b
 2 

Solution

LHS = (cosa – cosb)2 + (sina – sinb)2

= cos2a – 2 cosa . cosb + cos2b + sin2a – 2 sina . sinb + cos2b

= 2(cos2a + sin2a) + (cos2b + sin2b) – 2(cosa . cosb + sina . sinb)

= 1 + 1 – 2 cos(a – b)

= 2 – 2 cos(a – b)

= 2[1 – cos(a – b)]

= 2 . 2 sin2  a – b 
 2 

= 4 sin2  a – b  = RHS proved
2

b) (sina + sinb)2 + (cosa + cosb)2 = 4 cos2 a – b
2

Solution

LHS = (sina + sinb)2 + (cosa + cosb)2

= sin2a + sin2b + 2 sina . sinb + cos2a + cos2b – 2 cosa . cosb

= (sin2a + cos2a) + (sin2b + sin2a) + 2(sina . sinb + cosa . cosb)

= 1 + 1 + 2 cos(a – b)

= 2 + 2 cos(a – b)

= 2[1 + cos(a – b)]

= 2 . 2 cos2  a – b
2 

= 4 cos2  a – b = RHS proved
2

7. Prove the following

a) cos  2p  . cos  4p  . cos  8p  . cos  1165p  = 1
15 15 15 16

Solution

LHS = cos  2p  . cos  4p  . cos  8p  . cos  16p 
15 15 15 15

= 1 2 sin 2p . cos 2p  . cos  4p  . cos  8p  . cos  1165p 
15 15 15 15
2 sin 2p
15

Vedanta Optional Mathematics Teacher's Guide ~ 10 275

= 1 sin  4p  . cos  4p  . cos  8p  . cos  1165p 
15 15 15
2 sin 2p
15

= 1 sin  8p  . cos  8p  . cos  16p 
 15   15   15 
4 sin 2p
15

= 1 sin  16p  . cos  16p 
15 15
8 sin 2p
15

= 1 2p sin  32p 
sin 15  15 
16

= 1 2p . sin 2p + 2p 
sin 15 15
16

= 1 2p . sin 2p
sin 15 15
16

= 1 = RHS proved.
16

8. Proved that:

1 + cos 8p  1 + cos 3p  1 + cos 5p  1 + cos 7p  = 1
8 8 8 8

Solution

LHS = 1 + cos 8p  1 + cos 3p  1 + cos 5p  1 + cos 7p 
8 8 8

= 1 + cos 8p  1 + cos 3p  1+ cos p – 3p  1 + cos p – p 
 8  8  8 

= 1 + cos p8 1 + cos 3p  1 – cos 3p  1 – cos 8p
8 8

= 1 – cos2 p8 1 – cos2 3p 
8

= sin2 p . sin2 3p
8 8

= sin2 p . sin2 p2 – p8
8

= sin2 p . cos2 p
8 8

276 Vedanta Optional Mathematics Teacher's Guide ~ 10

= 1 2 sin p . cos p 2
4 8 8 

= 1 sin2  p  2
4 8

= 1 sin p 2
4 4 

= 1  1 2
4 2

= 1 = RHS proved.
8

8. Prove that:

tan 712° = 6 – 3 + 2 – 2

Solution

LHS = tan 712°

= tan 15°
2

= sin 15° × 2 sin 15°
cos 2 2 sin 2
15°
15° 2

= 2 sin2 15°
2

sin 15°

= 1 – cos 15°
sin 15°

= 1 – cos(45° – 30°)
sin(45° – 30°)

= 1 – cos 45° . cos 30° – sin 45° . sin 30°
sin 45° . cos 30° – cos 45° . sin 30°

1 – 1 3 – 1 . 1
2 .2 2 2
=
1 3 1 1
2 . 2 – 2 . 2

= 2 2 – 3 – 1 × 3 + 1
3–1 3+1

= 2 6 – 3 – 3+2 2– 3 –1
3–1

Vedanta Optional Mathematics Teacher's Guide ~ 10 277

= 2 6 – 2 3 + 2 2 – 4
2

= 6 – 3 + 2 – 2 = RHS proved

Questions for practice

1. Prove that: tan A =± 1 – cosA
2 1 + cosA

2. If cos 330° = 23, prove that : sin 165° = 1 2– 3
2

3. Prove that: 1 + cosθ + sinθ = cot θ
1 – cosθ + sinθ 2

4. Prove that: secθ + tanθ = tan  p + θ 
 4 2 

5. If sin θ = 1 a + 1 , prove that: sinθ = 1 a3 + 1 
3 2 a 2 a3

6. Prove that: tan  p + θ  = 1 + sinθ = secθ + tanθ
 4 2  1 – sinθ

7. Prove that: 1 sin2A × 1 cosA = tan A
+ cos2A + cosA 2

8. Prove that: cot 712° = 2 + 3 + 4 + 6

278 Vedanta Optional Mathematics Teacher's Guide ~ 10

Transformation of Trigonometric Formula
of sine and cosine

Estimated Periods: 5

1. Objectives Objectives
S,N. Level
To tell the formulae of transformation of trigonometric ratios
(i) Knowledge(k) of sine and cosine.

(ii) Understanding(U) To explain to derive the formulae of trannsformation of
trigonometric ratios.
(ii) Application(A)
To solve the problems of transformation of trigonometric formulae.
(iv) Higher Ability
(HA) To solve harder problems of transformation of trigonometric
formulae.

2. Required teaching materials
Formula chart of trigonometric ratio of compound angles and transformation formula of
trigonometric ratios.

Teaching strategies
– Review the formulae of trigonometric ratios of compound angles.
– List the trigonometric ratios of compound angles as given below:

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 and subtracting above identities we get the required formulae.

Example adding (i) and (ii), we get
2 sinA . cosB = sin(A + B) + sin(A – B)
Subtracting (ii) from (i), we get
2 cosA . sinB = sin(A + B) – sin(A – B)
Similarly, explain to get the following results
2 cosA . cosB = cos(A – B) + cos(A – B)
2 sinA . sinB = sin(A – B) – cos(A + B)
Again, discuss how to derive the following formulae
–

sinC + sinD = 2 sin C + D  . cos C – D
 2  2 
C + D C – D
sinC – sinD = 2 cos 2 . sin 2 

cosC + cosD = 2 cos C + D . cos C – D 
2 2
C + D C – D
cosC – cosD = 2 sin  2  . sin  2 

Vedanta Optional Mathematics Teacher's Guide ~ 10 279

List of formula:

1. 2 sinA . cosB = sin(A + B) + sin(A – B)

2. 2 cosA . sinB = sin(A + B) – sin(A – B)

3. 2 cosA . cosB = cos(A + B) + cos(A – B)

4. 2 sinA . sinB = cos(A – B) – cos(A + B)

5. sinC + sinD = 2 sin C + D  . cos C – D
2  2

6. sinC – sinD = 2 cos C + D . sin C – D
 2   2 

7. cosD – cosC = 2 sin  C + D . sin C – D
 2   2 

8. If a = c then a+b = c+d is called componendo and dividendo.
b d a–b c–d

Some solved problems

1. Prove the following: 3
a) cos75° + cos15° = 2

Solution
LHS = cos75° + cos15°

= 2 cos  75° + 15° cos  75° – 15° 
2  2

= 2 cos45° . cos30°

=2. 1 . 3
2 2

= 3 = RHS proved.
2

b) sin75° – sin15° = 1
2

Solution
LHS = sin75° – sin15°

= 2 cos  75° + 15° . sin  75° – 15° 
 2   2 

= 2 cos45° . sin30°

280 Vedanta Optional Mathematics Teacher's Guide ~ 10

=2. 1 . 1
2 2

1
= 2 = RHS proved.

c) cos52° + cos68° + cos172° = 0

Solution
LHS = cos52° + cos68° + cos172°

= 2 cos  52° + 68° . cos  52° – 68°  + cos(180° – 8°)
 2  2 

= 2 cos60° . cos(–8°) – cos8°

= 2 . 1 . cos8° – cos8°
2

= cos8° – cos8°

= 0 = RHS proved.

2. Prove that:

a) cos5A + sin3A = cotA
sin5A – sin3A
Solution

LHS = cos5A + sin3A
sin5A – sin3A

= 2 cos 5A + 3A . cos  5A – 3A 
2 cos  2  . sin  5A 2 3A 
5A + –
 2 3A 2


= cos4A . csoinsAA cos(–θ) = cosθ
cos4A .

= cotA = RHS proved.

b) cos40° – cos60° = tan50°
sin60° – sin40°
Solution

LHS = cos40° – cos60°
sin60° – sin40°

= 2 sin  40° + 60°  . sin  40° – 60° 
2 cos  2  . sin  40° 2 60° 
 40° +  – 
 2 60°   2 


= sin50° . sin10°
cos50° . sin10°

= tan50° = RHS proved.

Vedanta Optional Mathematics Teacher's Guide ~ 10 281

c) cos80° + cos20° = 3
sin80° – sin20°

Solution

LHS = cos80° + cos20°
sin80° – sin20°

= 2 cos 80° + 20° . cos  80° – 20° 
2 cos 80° 2 20° . sin  80° 2 20° 
+ –
2 2

= cos30°
sin30°

3

= 2
1

2

= 3 × 2
2

= 3 = RHS proved.

d) cos8° + sin8° = tan53°
cos8° – sin8°

Solution

LHS = cos8° + sin8°
cos8° – sin8°

= cos8° + sin(90° – 82°)
cos8° – sin(90° – 82°)

= cos8° + cos82°
cos8° – cos82°

= 2 cos  8° + 82°  . cos  8° – 82° 
2 sin  8° 2  . sin  2 
+
2 82°  82° –8°
 2

= cos37°
sin37°

= cot37°

= cot(90° – 53°)

= tan53° = RHS proved.

e) cos10° – sin10° = cot55°
cos10° + sin10°

282 Vedanta Optional Mathematics Teacher's Guide ~ 10

Solution

LHS = cos10° – sin10°
cos10° + sin10°

= cos10° – sin(90° – 80°)
cos10° + sin(90° – 80°)

= cos10° – cos80°
cos10° + cos80°

2 sin 10° + 80° . sin  80° – 10° 
 2   2 
=
10° + 80°  10° – 80° 
2 cos  2  . cos  2 

= sin45° . sin35°
cos45° . cos35°

= tan35°

= tan(90° – 55°)

= cot55° = RHS proved.

f) cos(40° + A) + cos(40° – A) = cotA
sin(40° + A) – sin(40° – A)
Solution

LHS = cos(40° + A) + cos(40° – A)
sin(40° + A) – sin(40° – A)

= 2 cos 40° + A + 40° – A . cos 40° + A – 40° + A
2 cos  + A 2 40° –  . sin  + A 2 40° + 
40° + 40° –
cosA  2 A  2 A
sinA  

=

= cotA = RHS proved.

3. Prove the following.

a) sinA . sin2A + sin3A . sin6A = tan5A
sinA . cos2A + sin3A . cos6A
Solution

LHS = sinA . sin2A + sin3A . sin6A
sinA . cos2A + sin3A . cos6A

= 2 sinA . sin2A + 2 sin3A . sin6A
2 sinA . cos2A + 2 sin3A . cos6A

= cos(A – 2A) – cos(A + 2A) + cos(3A – 6A) – cos(3A + 6A)
sin(A + 2A) + sin(A – 2A) + sin(3A + 6A) + sin(3A – 6A)

Vedanta Optional Mathematics Teacher's Guide ~ 10 283

= cosA – cos3A + cos3A – cos9A
sin3A – sinA + sin9A – sin3A

= cosA – cos9A
sin9A – sinA

= 2 sin A + 9A . sin A + 9A
2 cos  2  . sin  2 
A + A +
 2 9A  2 9A
 

= tan5A = RHS proved.

b) cos2A . cos3A – cos2A . cos7A = sin7A + sin3A
sin4A . sin3A – sin2A . sin5A sinA
Solution

LHS = cos2A . cos3A – cos2A . cos7A
sin4A . sin3A – sin2A . sin5A

= 2 cos2A . cos3A – 2 cos2A . cos7A
2 sin2A . sin3A – 2 sin2A . sin5A

= cos(2A + 3A) + cos(2A – 3A) – cos(2A + 7A) – cos(2A – 7A)
cos(4A – 3A) – cos(4A + 3A) – cos(2A – A) + cos(2A + 5A)

= cos5A + cosA – cos9A – cos5A
cosA – cos7A – cos3A + cos7A

= cosA – cos9A
cosA – cos3A

2 sin  A + 9A . sin  9A – A
 2   2 
=
 A + 3A  3A – A
2 sin  2  . sin  2 

= sin5A . sin4A
sin2A . sinA

= sin5A . 2 sin2A . cos2A
sin2A . sinA

= 2 sin5A . cos2A
sinA

RHS = sin7A + sin3A
sinA

2 sin 7A + 3A . cos  7A – 3A 
 2  2
=
sinA

= 2 sin5A . cos2A
sinA

 LHS = RHS proved

284 Vedanta Optional Mathematics Teacher's Guide ~ 10

c) cos7A + cos3A – cos5A – cosA = cot2A
sin7A – sin3A – sin5A + sinA
Solution

LHS = cos7A + cos3A – cos5A – cosA
sin7A – sin3A – sin5A + sinA

= (cos7A – cosA) + (cos3A – cos5A)
(sin7A + sinA) – (sin3A + sin5A)

= 2 sin 7A + A . sin  7A – A  + 2 sin 3A + 5A . sin 5A – 3A
2 sin  2  . cos  2 A  – 2  2 cos 3A 2 
7A + sin 3A + 5A –
 2 A 7A – 2  . 2 5A 
 2 

= 2 sin4A [sin3A + sinA]
2 sin4A [cos3A – cosA]

= sinA – sin3A
cos3A – cosA

2 cos  A + 3A . sin  A – 3A 
2 sin 2  . sin  A 2 
= – 3A
 3A + A 2
2 

= cos2A
sin2A

= cot2A = RHS proved.

d) sinA + sin3A + sin5A + sin7A = tan4A
cosA + cos3A + cos5A + cos7A
Solution

LHS = sinA + sin3A + sin5A + sin7A
cosA + cos3A + cos5A + cos7A

= (sinA + sin7A) + (sin3A + sin5A)
(cosA + cos7A) + (cos3A + cos5A)

2 sin  A + 7A . cos  A – 7A  + 2 sin  3A + 5A . cos 3A – 5A
2 cos  2 7A . cos  A 2  + 2 cos  2 . cos 3A 2
= A +  – 7A  3A + 5A – 5A
2  2  2  2 
 

= 2 sin4A [cos3A + cosA]
2 cos4A [cos3A + cosA]

= tan4A = RHS proved.

e) sin5A – sin7A – sin4A + sin8A = cot6A
cos4A – cos5A – cos8A + cos7A
Solution

LHS = sin5A – sin7A – sin4A + sin8A
cos4A – cos5A – cos8A + cos7A

Vedanta Optional Mathematics Teacher's Guide ~ 10 285

= (sin5A – sin7A) – (sin4A – sin8A)
(cos4A – cos8A) – (cos5A – cos7A)

2 cos 5A + 7A . sin  5A – 7A  – 2 cos 4A + 8A  . sin  4A – 8A 
2 sin 4A 2 8A . sin  8A 2 4A  – 2 sin 5A 2  . sin  7A 2 5A 
= +  –  + 7A  – 
2  2  2  2 

= cos6A [–sinA + sin2A]
sin6A [sin2A – sinA]

= cot6A = RHS proved.

f) sin(p + 2)θ – sinpθ = cot(p + 1)θ
cospθ – cos(p + 2)θ

Solution

LHS = sin(p + 2)θ – sinpθ
cospθ – cos(p + 2)θ

= 2 cos p + 2 + p θ . sin p +2 – p θ
2 sin p + 2 + 2 θ . sin p 2 – p θ
p
2 +2
2

= cos(p + 1)θ . sinθ
sin(p + 1)θ . sinθ

= cot(p + 1)θ = RHS proved.

g) (sin4A + sin2A) . (cos4A – cos8A) = 1
(sin7A + sin5A) . (cosA – cos5A)

Solution

LHS = (sin4A + sin2A) . (cos4A – cos8A)
(sin7A + sin5A) . (cosA – cos5A)

2 sin  4A + 2A  . cos  4A – 2A  . 2 sin 4A + 8A  . sin  8A – 4A 
2 2 . 2 sin A 2  . sin  2
= 2 sin + A
7A + 5A  7A – 5A  2 5A  5A – 
 2  . cos 2  2

= sin3A . cosA . sin6A . sin2A
sin6A . cosA . sin3A . sin2A

= 1 = RHS proved.

4. Prove that:

a) sin20° . sin40° . sin60° . sin80° = 3
16

286 Vedanta Optional Mathematics Teacher's Guide ~ 10

Solution

LHS = sin20° . sin40° . sin60° . sin80°

= 3 . 1 (2 sin20° . sin40°) . sin80°
2 2

= 3 cos(20° – 40°) – cos(20° + 40°) . sin80°
4

= 3 {cos20° – cos60°} . sin80°
4

= 3 cos20° . sin80° – 3 . 1 . sin80°
4 4 2

= 3 sin(20° + 80°) – sin(20° – 80°) – 3 sin80°
8 8

= 3 sin100° + 3 sin60° – 3 sin80°
8 8 8

= 3 sin(180° – 80°) – 3 sin80° + 3 . 3
8 8 8 2

= 3 + 3 sin80° – 3 sin80°
16 8 8

= 3 = RHS proved.
16

b) sin10° . sin30° . sin50° . sin70° = 1
16

Solution

LHS = sin10° . sin30° . sin50° . sin70°

= sin10° . 1 . sin50° . sin70°
2

= 1 (2 sin10° . sin50°) sin70°
4

= 1 {cos(10° – 50°) – cos(10° + 50°)} . sin70°
4

= 1 {cos40° – cos60°} . sin70°
4

= 1 cos40° . sin70° – 1 . 1 . sin70°
4 4 2

= 1 (2 cos40° . sin70°) – 1 sin70°
8 8

= 1 {sin110° – sin(–30°)} – 1 sin70°
8 8

= 1 sin110° + 1 . 1 – 1 sin70°
8 8 2 8

Vedanta Optional Mathematics Teacher's Guide ~ 10 287

= 1 sin(180° – 70°) + 1 – 1 sin70°
8 16 8

= 1 sin70° – 1 sin70° + 1
8 8 16

= 1 = RHS proved.
16

d) cos40° . cos100° . cos160° = 1
8

Solution

LHS = cos40° . cos100° . cos160°

= 1 [2 cos40° . cos100°] . cos160°
2

= 1 [cos40° + 100°) + cos(40° – 100°)] . cos160°
2

= 1 [cos140° + cos60°] . cos160°
2

= 1 cos140° . cos160° + 1 . 1 . cos160°
2 2 2

= 1 (cos300° + cos20°) + 1 . cos160°
4 4

= 1 . 1 + 1 cos20° – 1 cos20°
4 2 4 4

= 1 = RHS proved.
8

e) tan20° . tan40° . tan80° = 3

Solution

LHS = tan20° . tan40° . tan80°

= sin20° . sin40° . sin80°
cos20° cos40° cos80°

Numerator = sin20° . sin40° . sin80°

Simplify it to get, 3
8

Again, denominator = cos20°, cos40°, cos80°

Simplify it to get, 1
8

3

Then, LHT = 8 = 3 = RHS proved.
1

8

288 Vedanta Optional Mathematics Teacher's Guide ~ 10

5. Prove that:

a) sin(45° + θ) . sin(45° – θ) = 1 cos2θ
2

Solution

LHS = sin(45° + θ) . sin(45° – θ)

= 1 {2 sin(45° + θ) . sin(45° – θ)}
2
1
= 2 {cos(45° + θ – 45° + θ) – cos(45° + θ + 45° – θ)}

= 1 {cos2θ – cos90°}
2
1
= 2 cos2θ = RHS proved.

b) cos(45° + θ) . cos(45° – θ) = 1 cos2θ
2

Solution

LHS = cos(45° + θ) . cos(45° – θ)

= 1 {cos(45° + θ + 45° – θ) + cos(45° + θ – 45° + θ)}
2
1
= 2 [cos90° + cos2θ]

= 1 cos2θ = RHS proved.
2

c) cosθ . cos(60° – θ) . cos(60° + θ) = 1 cos3θ
4

Solution

LHS = cosθ . cos(60° – θ) . cos(60° + θ)

= 1 cosθ {2 cos(60° – θ) . cos(60° + θ)}
2
1
= 2 cosθ {cos(60° – θ + 60° + θ) + cos(60° – θ – 60° – θ)}

= 1 cosθ {cos120° + cos2θ}
2

= 1 cosθ  –1  + 1 cosθ . cos2θ
2  2  2

= – 1 cosθ + 1 {cos3θ . cosθ}
4 4

=– 1 cosθ + 1 cos3θ + 1 cosθ
4 4 4
1
= 4 cos3θ = RHS proved.

Vedanta Optional Mathematics Teacher's Guide ~ 10 289

6. Prove that: cos(36° – θ) . cos(36° + θ) + cos(54° + θ) . cos(54° – θ) = cos2θ

Solution

LHS = cos(36° – θ) . cos(36° + θ) + cos(54° + θ) . cos(54° – θ)

= 1 {cos(36° – θ + 36° + θ) + cos(36° – θ – 36° – θ)}
2
1
+ 2 {cos(54° + θ + 54° – θ) + cos(54° + θ – 54° + θ)}

= 1 {cos72° + cos2θ + cos108° + cos2θ}
2

= 1 . 2 cos2θ + 1 {cos72° + cos(180° – 72°)}
2 2

= cos2θ + 1 {cos72° – cos72°}
2

= cos2θ + 1 . 0
2

= cos2θ = RHS proved.

7. Prove that:

a) tan  p + θ + tan  p – θ = 2 sec2θ
4 4

Solution

LHS = tan  p + θ + tan  p – θ
4 4

= tan p + tanθ – tan p – tanθ
4 . tanθ 4
p
1 – tan 4 1 + tan p . tanθ
4

= 1 + sin q + 1– sin q
cos q 1+ cos q
sin q
1 – sin q cos q
cos q

= cosθ + sinθ + cosθ – sinθ
cosθ – sinθ cosθ + sinθ

= (cosθ + sinθ)2 + (cosθ – sinθ)2
cos2θ – sin2θ

= cos2θ + sin2θ + 2 sinθ. cosθ + cos2θ + sin2θ – 2 sinθ cosθ
cos2θ

= 1+1
cos2θ

= 2 1
cos2θ

= 2 sec2θ = RHS proved.

290 Vedanta Optional Mathematics Teacher's Guide ~ 10

b) tan  p + θ – tan  p – θ = 2 tan2θ
4 4

Solution

LHS = tan  p + θ – tan  p – θ
 4  4

= tan p + tanθ – tan p – tanθ
4 4

1 – tan p . tanθ 1 + tan p . tanθ
4 4

= 1 + sin q – 1 – sin q
cos q cos q

1 – sin q 1 + sin q
cos q cos q

= cosθ + sinθ – cosθ – sinθ
cosθ – sinθ cosθ + sinθ

= (cosθ + sinθ)2 – (cosθ – sinθ)2
cos2θ – sin2θ

= cos2θ + sin2θ + 2 sinθ. cosθ – cos2θ – sin2θ + 2 sinθ cosθ
cos2θ

= sin2θ + cos2θ
cos2θ

= 2 sin2θ
cos2θ

= 2 tan2θ = RHS proved.

8. Prove the following.

a) (cosA + cosB)2 + (sinA + sinB)2 = 4cos2  A – B 
 2 

Solution

LHS = (cosA + cosB)2 + (sinA + sinB)2

= 2 cos A + B . cos A – B  2+ 2 sin  A + B  . cos  A – B 2
2 2   2   2

= 4 cos2  A – B  cos2  A + B  + sin2  A + B 
2 2 2

= 4 cos2 A – B = RHS proved.
2 

b) (CosB – cosA)2 + (sinA – sinB)2 = 4 sin2 A – B 
2
Solution

LHS = (CosB – cosA)2 + (sinA – sinB)2

= 2 sin  A + B  . sin  A – B  2 + 2 cos  A + B  . sin  A – B  2
2 2 2 2

Vedanta Optional Mathematics Teacher's Guide ~ 10 291

= 4 sin2  A + B  sin2  A – B  + 4 cos2 A + B  . sin2  A – B 
2 2  2 2

= 4 sin2  A – B  sin2  A + B  + cos2  A + B 
 2   2   2 

= 4 sin2  A – B  . 1
 2 

= 4 sin2 A – B = RHS proved.
2 

c) sin2 π + A – sin2 π – A = 1 sinA
8 2 8 2 2

Solution

LHS = sin2 π + A  – sin2 π – A 
8 2 8 2

= 1 1 – cos2 π + A  – 1 1 – cos2 π – A 
2 8 2  2 8 2 

= 1 – 1 cos  π + A – 1 + 1 cos π4 – A
2 2 4 2 2

= 1 cos π – A  – cos  π – A
2 4 4

1 π – A + π + A  π + A – π + A
2 4 4 4 4
= 2 sin . sin
22

= sin π . sinA
4
= 1 sinA = RHS proved.
2
3
d) sin2A + sin2(A + 120°) + sin2(A – 120°) = 2

Solution

LHS = sin2A + sin2(A + 120°) + sin2(A – 120°)

= sin2A + 1 {1 – cos2 (A + 120°) + 1 – cos2 (A – 120°)}
2

= 1 + 1 + sin2A – 1 {cos(2A + 240°) + cos(2A – 240°)}
2 2 2

= 1 + sin2A – 1 . 2 cos 2A + 240° + 2A – 240° cos  2A + 240° – 2A – 240° 
2  2  2

= 1 + sin2A – cos2A . cos240°

= 1 + sin2A – cos2A cos240°

= 1 + sin2A – (1 – 2 sin2A) .  –1 
2
1
= 1 + sin2A + 2 – sin2A

= 1 + 1 = 3 = RHS proved.
2 2

292 Vedanta Optional Mathematics Teacher's Guide ~ 10

9. a) If sinα + sinb = 1 and cosα + cosb = 12, prove that: tan a + b  = 1
4  2 2

Solution

Here, sinα + sinb = 1
4
1
or, 2 sin a + b . cos  a – b  = 4 ... ... ... (i)
 2  2
1
and cosα + cosb = 2

or, 2 cos  a + b . cos  a – b  = 1 ... ... ... (ii)
 2   2  2

Dividing (i) by (ii), we get,

 tan(α + b) = 1 proved.
2

b) If sinx = k siny, then prove that: tan  x – y  = k – 1 tan  x + y 
 2  k + 1  2 

Solution

Here, k = sinx
siny

RHS = k–1 tan x + y
k+1  2 

= sinx – 1 tan  x + y 
siny + 1 2
sinx
siny

= sinx – siny × siny tan  x + y 
siny sinx + siny  2 

= 2 cos x + y . sin  x – y  tan  x + y 
2  2  . cos  x 2 y  2
 
sin  x + y  – 
 2  2

= cot  x + y  . tan  x + y  . tan  x – y 
 2   2   2 

= tan  x – y  = RHS proved.
2

c) If sin(A + B) = k sin(A – B), then prove that: (k – 1) tanA = (k + 1) tanB.

Solution

Here, = sin(A + B) = k
sin(A – B) 1

By using componendo and dividendo

sin(A + B) + sin(A – B) = k+1
sin(A + B) – sin(A – B) k–1

Vedanta Optional Mathematics Teacher's Guide ~ 10 293

or, 2 sinA . cosB = k+1
2 cosA . sinB k–1

or, tanA = k+1
tanB k–1

 (k – 1) tanA = (k + 1) tan B proved.

10. Prove that: sin2x – sin2y = tan(x y)
sinx . cosx – siny . cosy

Solution

LHS = sinx sin2x – sin2y cosy
. cosx – siny .

= 2 sin2x – sin2y

2 sinx . cosx – 2 siny . cosy

= 1 – cos2x – 1 + cos2y
sin2x – sin2y

= cos2y – cos2x
sin2x – sin2y

= 2 sin 2y + 2x . sin 2x – 2y
 2   2 

2 cos  2x + 2y  . sin  2x – 2y 
 2   2 
= tan(x y) = RHS proved.

11. Prove that cos 2π + cos 4π + cos 6π = –12
7 7 7
Solution

LHS = cos 2π + cos 4π + cos 6π
7 7 7
1
= 2 cos 2π . sin p + 2 cos 4π . sin p + 2 cos 6π . sin p
p 7 7 7 7 7 7
2 sin 7

1 p sin 2π + p – sin 2π – p + sin 4π + p – sin  4π – p
= 7 7 7 7 7 7 7 7 7

2 sin

+ sin 67π + p7 – sin  6π – 7p
 7

= 1 sin 3π – sin p + sin 5π – sin 3π + sin π – sin 5π
7 7 7 7 7
2 sin p
7

1 –sin p + 0
= p 7
7
2 sin

= –21 = RHS proved.

294 Vedanta Optional Mathematics Teacher's Guide ~ 10

11. Prove that: x = y cot a + b , if x cosα + y sinα = x cosβ + y sinβ.
 2

Solution

Here, x cosα + y sinα = x cosβ + y sinβ

or, x(cosα – cosb) = y(sinβ – sina)

or, x . 2 sin  a + b sin  b – a  = y 2 cos b + a sin  b – a 
2  2  2  2

 x = y cot  a + b  proved.
 2 

Questions for practice

Prove the following

1. cos15° . sin75° = 2 + 3
4

2. sin5x – sin3x = tan4x
cos5x + cos3x

3. cos20° – sin20° = tan25°
cos20° + sin25°

4. cosθ – cos2θ + cos3θ = cot2θ
sinθ – sin2θ + sin3θ

5. sin8θ . cosθ – sin6θ . cos3θ = tan2θ
cos2θ . cosθ – sin3θ . sin4θ

6. cosec π – θ  . cosec  π + θ = 2
4   4 cosθ

7. sin10° . sin50° . sin60° . sin70° = 3
16
1
8. cos80° . cos140° . cos160° = 8

9. cosθ . cos(60° – θ) . cos(60° + θ) = 1 cos3θ
4

10. cos20° . cos40° . cos80° . cos240° = –116

11. cos π . cos 2π . cos 4π = –18
7 7 7

12. If sin2x + sin2y = 1 and cos2x + cos2y = 12, then show that: tan(x + y) = 32.
3

Conditional Trigonometric Identities

Vedanta Optional Mathematics Teacher's Guide ~ 10 295

Estimated Periods: 6

1. Objectives

S,N. Level Objectives

(i) Knowledge(k) To state the conditional identities related to angles of a triangle.

(ii) Understanding(U) To explain to derive the conditional trigonometric identities.

(ii) Application(A) To solve the problems related to conditional identities in
trigonometry.

(iv) Higher Ability (HA) To solve the order problems related to conditional identities.

2. Teaching Materials
List of formula of conditional trigonometric identity in a chart paper.

3. Teaching Learning Stategies:
– Review the formulae of transformation of trigonometric ratios.

– Discuss components of a triangle 3 sides and 3 angles.

– If A, B and C are angles of a triangle. Then A + B + C = π or A + B = π – C

Taking sin, cosine and tangent ratios on both side, find different identities.

Example sin(A + B) = sin(π – C) = sinC

cos(A + B) = cos (π – C) = –cosC

tan(A + B) = tan (π – C) = –tanC
A B C
– Similarly, from 2 + 2 = π – 2
2

find different conditional identities.
– Again from 2A + 2B = 2π – 2C

find different conditional identities.
– Solve some question from exercise of the text book and give guidance to the students
to solve the problems in the same exercise.

List of formula:

1. 2 sinA . cosB = sin(A + B) + sin(A – B)
2. 2 cosA . sinB = sin(A + B) – sin(A – B)
3. 2 sinA . sinB = cos(A – B) – cos(A + B)
4. 2 cosA . cosB = cos(A + B) + cos(A – B)

5. sinC + sinD = 2 sin C + D  . cos  C – D
 2   2 

6. sinC – sinD = 2 cos C + D . sin C + D
 2   2 

296 Vedanta Optional Mathematics Teacher's Guide ~ 10

7. cosC + cosD = 2 cos C + D . cos  C – D 
 2   2 

8. cosC – cosD = 2 sin  C + D . sin  D– C 
 2   2 

= –2 sin  C + D . sin  C – D 
 2  2 

9. sin(A + B) = sin(π – C) = sinC

sin A + B = sin  π – C = cos C
2 2 2 2 2

cos(A + B) = cos(π – C) = –cosC

cos A + B = cos π – C = sin C
2 2 2 2 2

tan(A + B) = tan(π – C) = –tanC

tan A + B = tan π – C = tan C
2 2 2 2 2

sin(2A + 2B) = sin(2π – 2C) = –sin2C

cos(2A + 2B) = cos(2π – 2C) = cos2C

tan(2A + 2B) = tan(2π – 2C) = –tan2C

Some solved problems

1. 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)

Solution
Here, A + B + C = 180°
or, A + B = 180° – C
 sin(A + B) = sin(180° – C) = sinC

LHS = sin(B + 2C) + sin(C + 2A) + sin(A + 2B) (∴180° = π)
= sin(B + C + C) + sin(C + A + A) + sin(A + B + B)
= sin(π – A + C) + sin(π – B + A) + sin(π – C + B)
= sin{π – (A – C)} + sin{π – (B – A)} + sin{π – (C – B)}
= sin(A – C) + sin(B – A) + sin(C – B)
= RHS proved.

2. If A + B + C = π, prove that:

a) cot A . cot B . cot C = cot A + cot B + cot C
2 2 2 2 2 2

Vedanta Optional Mathematics Teacher's Guide ~ 10 297

b) tan A . tan B + tan B . tan C + tan C . tan A = 1.
2 2 2 2 2 2
Solution

a) cot A . cot B . cot C = cot A + cot B + cot C
2 2 2 2 2 2

Here, A + B + C = πc

or, A + B = π – C
2 2 2 2

 cot A + B = cot π – C
2 2 2 2

or, cot A . cot B –1 = tan C
2 2 2
A
cot B + cot 2
2

or, cot A . cot B –1 = 1
2 2 cot
A
cot B + cot 2 C
2 2

 cot A . cot B . cot C = cot A + cot B + cot C proved.
2 2 2 2 2 2

b) tan A . tan B + tan B . tan C + tan C . tan A = 1.
2 2 2 2 2 2
Solution

Here, A + B + C = πc

or, A + B = π – C
2 2 2 2

 tan A + B = tan π – C
2 2 2 2

or, tan A + tan B = cot C
2 2 2

1 – tan A . tanB2
2

or, tan A + tan B = 1
2 2 tan
A
1 – tan 2 . tan B C
2 2

 tan A . tan B + tan B . tan C + tan C . tan A = 1 proved.
2 2 2 2 2 2

3. If A + B + C = π, prove that:

a) cos2A + cos2B – cos2C = 1 – 4 sinA . sinB . cosC

Solution

298 Vedanta Optional Mathematics Teacher's Guide ~ 10

Here, A + B + C = π
or, A + B = π – C
 cos(A + B) = cos(π – C) = –cosC

LHS = cos2A + cos2B – cos2C

= 2 cos  2A + 2B  . cos 2A – 2B  –(2 cos2C – 1)
 2   2 

= 2 cos(A + B) . cos(A – B) – 2 cos2C + 1
= 2(–cosC) . cos(A – B) – 2 cos2C + 1
= 1 – 2 cosC [cos(A – B) + cosC]
= 1 – 2 cosC [cos(A – B – cos(A + B)]
= 1 – 2 cosC . 2 sinA . sinB
= 1 – 4 sinA . sinB . cosC
= RHS proved.

4. If A + B + C = pc, prove the following:

a. SinA + sinB – sinC = 4 sinA2 . sinB2 . cosC2

Solution

Here, A + B + C = pc

A + B = pc – C
2 2 2 2

∴ sin A B = sin pc C = cosC2
2 –2 2 –2

and cos AB = cos pc C = sinC2
2–2 2 –2

Here, LHS = sinA + sinB – sinC

= 2sin A +B . cos A–B – 2sinC2 . cosC2
2 2

= 2cosC2 . cos A–B . 2sinC2 . cosC2
2

= 2cosC2 cos A – B  – sinC2
2

= 2cosC2 cos A – B  – cos A + B
2  2 

= 2cosC2 . 2sinA2 . sinB2

= 4sinA2 . sinB2 . cosC2

Vedanta Optional Mathematics Teacher's Guide ~ 10 299

= RHS proved.

b. –SinA + sinB + sinC = 4cosA2 . sinB2 . sinC

Solution

Here, A + B + C = pc

A + B = pc – C
2 2 2 2

∴ sin A + B = sin pc C = cosC2
2 2 2 –2

and cos AB = sinC2
LHS 2+2

= sinB + sinC – sinA

= 2sin B+C . cos B–C – 2sinA2 . cosA2
2 2

= 2cosA2 cos B – C  – sinA2
2

= 2cosA2 cos B – C  – cos B + C 
2 2

= 2cosA2 2sinB2 . sinC2
= 4cosA2 . sinB2 . sinC2

= RHS proved.

5. If A + B + C = pc, prove that
a. cos2A + cos2B – cos2C = 1 – 4sinA . sinB . cosC

Solution
Here, A + B + C = p
A + B = p – C
∴ cos(A + B) = cos(p – C) = –cosC

LHS = cos2A + cos2B – cos2C

= 2cos 2A + 2B . cos 2A – 2B – (2cos2C – 1)
2 2

= 2cos(A + B) . cos(A – B) – 2cos2C + 1
= –2cosC . cos(A – B) – 2cos2C + 1
= 1 – 2cosC[cos(A – B) + cosC]
= 1 – 2cosC[cos(A – B) – cos(A + B)]
= 1 – 2cosC . 2sinA . sinB
= 1 – 4sinA . sinB . cosC
= RHS proved.

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