Multiple angles

39. cos8A + sin8A = 1 - sin22A + 1 sin42A
8

LHS, cos8A + sin8A

= (cos4A)2 + (sin4A)2

= (cos4A - sin4A)2 + 2 cos4A.sin4A

= {(cos2A)2 - ( sin2A)2}2 + 2 1 .16 cos4A.sin4A
.16

= {(cos2A + sin2A)(cos2A - sin2A)}2 + 1 (2sinAcosA)4
8

= (1 . cos2A)2 + 1 (sin2A)4
8

= cos22A + 1 sin42A
8

= 1 - sin22A + 1 sin42A RHS proved
8

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40. sin3A.sin3A + cos3A.cos3A = cos32A

LHS, sin3A.sin3A + cos3A.cos3A

3sinA - sin3A 3cosA + cos3A
= 4 .sin3A + 4 . cos3A

sin3A( 3sinA -sin3A) cos3A( 3cosA + cos3A)
=4+
4

= 1 ( 3sin3AsinA - sin23A + 3cos3AcosA + cos23A)
4

= 1 ( 3cos3AcosA +3sin3AsinA+ cos23A- sin23A)
4

1
= 4 { 3(cos3A.cosA + sin3A.sinA) + cos6A}

1
= 4 {3cos(3A -A) + cos3.2A}

= 1 (3cos2A + 4cos32A - 3cos2A)
4

= 1 .4cos32A
4

= cos32A RHS proved

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Exercise :K

1.Prove that sin3A.cos3A + cos3A.sin3A 3 sin4A
=4

2. Prove that cos3A.sin33A + sin3A.cos33A = sin4A
4

3. Prove that cos8A + sin8A = cos22A + 1 sin42A
8

4. Prove that 8cos8A + 8sin8A= 8 - 8sin22A+ sin42A

41.Prove that (1 + sin2A + cos2A)2 = 4 cos2A( 1 + sin2A)

LHS (1 + sin2A + cos2A)2

= ( 1 + cos2A+ sin2A)2

= ( 2cos2A + 2sinA.cosA)2

= {2cosA(cosA +sinA)}2

= 4cos2A ( cosA + sinA)2

= 4cos2A( cos2A + 2cosA.sinA + sin2A)

= 4cos2A(sin2A + cos2A+ 2cosA.sinA)

= 4cos2A( 1 + sin2A) RHS Proved

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1
42. Prove that sin6°.cos12°.cos24°.cos48° = 16
LHS, sin6°.cos12°.cos24°.cos48°

2sin6°.cos6°.cos12°.cos24°.cos48°
= 2cos6°

sin12°.cos12°.cos24°.cos48°
= 2cos6°

2sin12°.cos12°.cos24°.cos48°
= 2 ×2cos6°

sin24°.cos24°.cos48°
= 4cos6°

2sin24°.cos24°.cos48°
= 2×4cos6°

sin48°.cos48°
= 8cos6°

2 sin48°.cos48°
= 2×8cos6°

sin96°
= 16cos6°

sin(90+6)°
= 16cos6°

cos6° 1
= 16cos6° = 16 RHS Proved

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1
43. Prove that sinAsin(60° + A) sin(60°- A) = 4 sin3A

LHS, sinAsin(60° + A) sin(60°- A)

= sinA.(sin60°.cosA + cos60°.sinA) ( sin60°cosA- cos60°.sinA)

= sinA  3 .cosA + 1   3 .cosA - 1 
 2 2  2 2
 sinA  sinA
 
 

= sinA43 .cos2A - 1 sin2A
4

= sinA 3 .(1- sin2A) - 1 sin2A
4 4

= sin.A. 1 (3 -3sin2A - sin2A)
4

= sin.A. 1 ( 3 - 4sin2A)
4

= 1 (3sinA - 4sin3A)
4

1
= 4 sin3A RHS Proved

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44. If tan2A = 1 + 2 tan2B , show that cos2B = 1 + 2cos2A .

Given, tan2A = 1 + 2 tan2B

or, 1- cos2A =1+2  1- cos2B 
1+ cos2A  1+ cos2B 

1- cos2A 1 + cos2B + 2(1 - cos2B)
or, 1+ cos2A =
(1 + cos2B)

or, ( 1- cos2A) (1 + cos2B) = ( 1 +cos2A) ( 1 + cos2B + 2 - 2

cos2B)

or, 1 + cos2B - cos2A - cos2A.cos2B = (1 + cos2A) ( 3 - cos2B)

or, 1 + cos2B - cos2A - cos2A.cos2B = 3 - cos2B + 3cos2A -

cos2A.cos2B

or, cos2B + cos2B = 3 - 1 + 3cos2A + cos2A - cos2A.cos2B+

cos2A.cos2B

or, 2cos2B = 2 + 4 cos2A

or, cos2B = 2( 1+2cos2A)

2

or, cos2B = 1 + 2cos2A Proved

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Exercise:L

1.Prove that (1 + sin2A + cos2A)2 = (1 + sin2A)

4 cos2A

(1 + sin2A + cos2A)2 4
2. Prove that (1 + sin2A) = sec2A

1
3. Prove that cos36°.cos72°.cos108°.cos144° = 16

1
4.Prove that cos24°.cos48°.cos96°.cos168° = 16

5. Prove that sinAcosAcos2A cos4A = sin8A

6. Prove that tanAtan (60° + A) tan (60°- A) = tan3A

7.If 4tan2A = tan2B then prove that cos2A = 3 + 5cos2B
5 + 3cos2B

3cos2B - 1
8.If tanA = 2 tanB then prove that cos2A = 3 - cos2B

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sinA sin4A
1. prove that secA = 4cos2A

sin4A
RHS, 4cos2A

sin2.2A
= 4cos2A

2sin2A.cos2A
= 4cos2A

2sin2A
=4

2.2sinA.cosA
=4

= sinA.cosA LHS Proved
1

= sinA.secA
sinA

=secA

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