Multiple angles

General

Optional

Mathematics

Class 10

GOM

Binod Kasula

1

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sin(A + B) = sinA.cosB + cosA.sinB
sin(A + A) = sinA.cosA + cosA.sinA
sin2A = 2sinA.cosA
sin2B = 2sinB.cosB
sin(A + B) = sinA.cosB + cosA.sinB
sin(B + B) = sinB.cosB + cosB.sinB
sin2B = 2sinB.cosB
cos2A
cos(A + B) = cosA.cosB - sinA.sinB
cos(A + A) = cosA.cosA - sinA.sinA
cos2A = cos2A - sin2A
cos2B = cos2B - sin2B
sin2A = 2sinA.cosA
cos2A= cos2A - sin2A
Formula of cos2A in term of sinA
cos2A = cos2A - sin2A
or,cos2A = 1- sin2A - sin2A
or, cos2A = 1 - 2sin2A

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Formula of cos2A in term of cosA.

cos2A = cos2A - sin2A

or, cos2A = cos2A -( 1 - cos2A)

or, cos2A = cos2A - 1 + cos2A

cos2A = 2cos2A - 1

cos2A = 1-2sin2A

or, 2sin2A = 1 - cos2A

or, sin2A = 1 - cos2A
2

cos2A = 2cos2A - 1

or, cos2A +1 = 2cos2A

or, 2cos2A = 1 + cos2A

cos2A = 1 +cos2A
2

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1.sin2A = 2sinA.cosA

2.cos2A = cos2A - sin2A

3.cos2A = 2cos2A - 1

4. cos2A = 1 - 2sin2A

5. 2sin2A = 1 - cos2A

6.2cos2A = 1 + cos2A

7. sin2A = 1 - cos2A
2

8.cos2A = 1 + cos2A
2

tanA + tanB
tan (A+B) = 1 - tanA.tanB

tanA + tanA
tan(A+A) = 1 - tanA.tanA

2tanA
9. tan2A = 1 -tan2A

cotB.cotA - 1
cot(A + B) = cotB + cotA

cotA.cotA - 1
cot(A + A) = cotA + cotA

cot2A - 1
10.cot2A = 2cotA

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Formula of sin2A in term of tanA.

sin2A = 2sinA.cosA

sin2A = 2sinA.cosA . cosA

cosA

sin2A = 2 sinA .cos2A
.cosA

1
sin2A = 2 tanA. sec2A

1
sin2A = 2 tanA .1+ tan2A

2tanA
sin2A = 1+tan2A

HW

formula of sin2A in term of cotA

formula of cos2A in term of tanA

formula of cos2A in term of cotA

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cos2A = cos2A - sin2A

or, cos2A =(cos2A - sin2A).cos2A

cos2A

cos2A sin2A 1
or, cos2A = ( cos2A - cos2A ) . sec2A

or, cos2A = ( 1 - tan2A) . 1
1 + tan2A

 cos2A = 1 - tan2A
1 + tan2A

1 - tan2A
The formula Cos2A in term of tanA is 1 + tan2A

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1. sin2A = 2sinA.cosA
2tanA

2. sin2A = 1 + tan2A
2cotA

3. sin2A = 1 + cot2A
4. sin3A = 3sinA - 4sin3A
5. cos3A = 4cos3A - 3cosA
6. cos2A = cos2A - sin2A
7. cos2A = 2cos2A - 1
8. cos2A = 1 - 2sin2A
9. 2cos2A = 1 + cos2A
10. 2sin2A = 1 - cos2A

1 - tan2A
11. cos2A = 1 + tan2A

cot2A - 1
12. cos2A = cot2A + 1
13. cos3A = 4cos3A - 3cosA

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2tanA
14. tan2A = 1 - tan2A

cot2A - 1
15. cot2A = 2cotA

3tanA - tan3A
16. tan3A= 1- 3 tan2A

cot3A - 3cotA
17. cot3A = 3cot2A - 1

18. Prove that tan2A = 1- cos2A
1+ cos2A

19. Prove that cot2A = 1+ cos2A
1 - cos2A

we have

cot2A - 1
cos2A = cot2A + 1

or, cos2A ( cot2A + 1 ) = cot2A - 1

or, cos2A . cot2A + cos2A = cot2A - 1

or, 1 + cos2A = cot2A - cos2A . cot2A

or, 1 + cos2A = cot2A( 1 - cos2A)

or, cot2A = 1+ cos2A
1 - cos2A

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18. Prove that tan2A = 1- cos2A
1 + cos2A

we have

1 - tan2A
cos2A = 1 + tan2A

or, cos2A( 1 + tan2A) = 1 - tan2A

or, cos2A + cos2A. tan2A = 1 - tan2A

or, cos2A. tan2A + tan2A = 1 - cos2A

or, tan2A( cos2A + 1 ) = 1 - cos2A

or, tan2A = 1 - cos2A
1 + cos2A

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4
1. If cosA = 5 , find the vaules of sin2A, cos2A and tan2A.

find the value of sinA.

sinA = 1 - cos2A

4
here, cosA = 5

sinA = 1-cos2A = 2 16 25-16 93
1- 25 = 25 = 25 = 5
4
1- 5 =

we have, sin2A = 2sinA .cosA

3 4 24
or, sin2A = 2. 5 .5 = 25

4 22
5 3 16 97
again, cos2A = cos2A - sin2A = - 5 = 25 - 25 = 25

24
sin2A 25 24
here, again for, tan2A = cos2A = 7 = 7

25

Exercise :A

4
1.If sinA = 5 then find cos2A, sin2A and tan2A.

1
2.If sinA = 2 then find sin3A and cos3A.

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2.Prove that 1- cos2A = 2cos2(90 - A)
LHS

1- cos2A
= 2sin2A
= 2cos2(90 - A) RHS Proved .

2.Prove that 1+ cos2A = 2sin2(90 - A)
LHS, 1+ cos2A
= 2cos2A
=2sin2(90-A) RHS, proved. here cosA= sin(90 - A)
Prove that 1 + sin2A = 2cos2(45 - A)

3. Prove that 1 - sin2A = 2sin2(45 - A)
LHS , 1 - sin2A
= 1 - cos(90 - 2A)
= 1 - cos2(45-A)
= 2sin2(45-A) RHS proved

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4. Prove that 1 + sin2A = 2cos2(45 - A)

LHS, 1 + sin2A

= 1 + cos(90 -2A)

= 1 + cos2(45-A)

= 2cos2(45 - A) RHS, proved

sin2A
5.Prove that 1 + cos2A = tanA

sin2A
LHS, 1 + cos2A

2sinA.cosA
= 2cos2A

sinA
= cosA

= TanA RHS proved

1 - cos2A
6. Prove that sin2A = tanA

1 - cos2A
LHS, sin2A

2sin2A sinA
= 2sinA.cosA = cosA = tanA RHS proved

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7. Prove that 2cos2(45- A) = 1+ sin2A
LHS, 2cos2(45- A)
= 1+ cos2(45-A)
= 1+ cos(90- 2A)
= 1+ sin2A RHS proved

8. Prove that sin2(45 - A ) + sin2(45 + A ) = 1

LHS, sin2(45 - A ) + sin2(45 + A )

1- cos2(45-A) 1- cos2(45+A) here, sin2 = 1- cos2θ
= 2+ 2 /2

1- cos(90 - 2A) + 1 - cos(90 + 2A)
=2

here,{cos( 90-2A) = sin2A and cos( 90+2A) = - sin2A}

1- sin2A + 1 + sin2A
=2

2
=2

= 1 RHS proved

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1
9. Prove that sinA. cos2A = 4 sin4A .secA

LHS, sinA. cos2A
2sinA.cosA. cos2A

= 2cosA
sin2A.cos2A

= 2cosA
2sin2A.cos2A

= 2×2cosA
sin2.2A

= 4cosA
11

= sin4A. 4 . cosA
1

=4 sin4A.secA RHS proved

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Exercise:B

1+ cos2A = sin2(90 - A)
1.Prove that 2

1 - sin2A = sin2(45 - A)
2. Prove that 2

1 + sin2A = cos2(45 - A)
3. Prove that 2

sin2A
4.Prove that 1 + cos2A = tanA

1 - cos2A
5. Prove that sin2A = tanA

1+ cos2A
6.Prove that sin2A = cotA

1 - cos2A
7.Prove that sin2A = tanA

sin2A
8.Prove that 1 + cos2A = tanA

sin2A
9.Prove that 1 - cos2A = cotA
10. Prove that 2cos2(45- B) -1 = sin2A

11. Prove that sin2(45 - A ) = 1 - sin2(45 + A )

sinA sin4A
13.Prove that secA = 4cos2A

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10. Prove that sin4A + cos4A = 1 - 1 sin22A
2

LHS, sin4A + cos4A

= ( sin2A)2 + (cos 2A)2

= (sin2A+ cos2A)2 - 2 sin2A cos 2A { a2 + b2 = (a +b)2 - 2ab}

= 12 - 1 × 2 × 2 sin2A cos 2A
2

= 1 - 1 (2sinA.cosA)2
2

= 1 - 1 (sin2A)2
2

= 1 - 1 sin22A RHS proved
2

sinB + sin2B
11.Prove that 1+ cosB + cos2B = tanB

sinB + sin2B
LHS, 1+ cosB + cos2B

sinB + 2sinB.cosB
= cosB + 1 + cos2B

sinB + 2sinB.cosB
= cosB + 2cos2B

sinB(1 + 2.cosB)
= cosB(1 + 2.cosB)
= tanB RHS Proved

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1 + sin2A - cos2A
12.Prove that 1+ sin2A + cos2A = tanA.

1 + sin2A - cos2A
LHS, 1+ sin2A + cos2A

1 - cos2A + sin2A 2sin2A +2 sinA.cosA
= 1 + cos2A+ sin2A = 2cos2A+2 sinA.cosA

2sinA(sinA + cosA) sinA
=2cosA(cosA + sinA) = cosA = tanA RHS proved

cos2A 1 - tanA
13.Prove that 1 + sin2A = 1+ tanA .

cos2A
LHS, 1 + sin2A

1- tan2A 1- tan2A
1+ tan2A here, (cos2A = 1+ tan2A and sin2A
= 2tanA
1+ 1+ tan2A

2tanA 1-tan2A
= 1+ tan2A 1+tan2A
= 1+ tan2A +2tanA
1+ tan2A

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

1- tanA
= 1 + tanA RHS proved

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Exercise:C

1- cos2A + sin2A
1.Prove that 1+cos2A + sin2A = tanA.

1+cos2A + sin2A
2.Prove that 1- cos2A + sin2A = cotA.

1+cosB + cos2B
3.Prove that sinB + sin2B = cotB

sinA + sin2A
4. Prove that 1+cosA + cos2A = tanA

1- cosB + cos2B
5.Prove that sin2B - sinB = cotB

sin2B - sinB
6.Prove that 1- cosB + cos2B = tanB

7.Prove that sin4B + cos4B = 1 - 1 sin22B
2

8.Prove that cos4A+ sin4A = 1 - 1 sin22A
2

cos2B 1 - tanB
9.Prove that 1 + sin2B = 1 + tanB .

1 + sin2B 1 + tanB
10.Prove that cos2B = 1 - tanB

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cosA - 1 + sin2A
4. Prove that sinA - 1 + sin2A = tanA

cosA - 1 + sin2A
LHS , sinA - 1 + sin2A

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

cosA - (sinA + cosA)2
= sinA - (sinA + cosA)2

cosA - ( sinA +cosA)
= sinA - (sinA + cosA)

cosA - sinA - cosA
= sinA - sinA - cosA

- sinA
= - cosA

sinA
= cosA
= tanA RHS proved

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1 - sin2A cos2A
15.Prove that 1 + sin2A = 1 + sin2A

1 - sin2A
LHS, 1 + sin2A

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

(cosA - sinA)2
= (cosA +sinA)2

cosA - sinA
= cosA + sinA

cosA - sinA cosA + sinA
= cosA + sinA . cosA + sinA

cos2A - sin2A
= (cosA + sinA)2

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

cos2A
= 1 + sin2A RHS proved

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16. If cosA = 1 x + 1  then show that cos2A = 1 x2 + 1 
2 x  2 x2 

Given, cosA 1 x + 1 
=2 x 

We have,

cos2A = 2cos2A - 1

2

or, cos2A = 2. 1 x + 1   -1
2 x  

1  x2 + 2.x . 1 +  1 2  -1
or, cos2A = 2 . 4  x  x  
  

or, cos2A = 1  x2 + 2 + 1  -1
2  x2 

or, cos2A = 1 . x2 + 1 + 11 -1
2 2.2 2 . x2

or, cos2A = 1 x2 + 1 + 1 . 1 - 1
2 2 x2

1 x2 + 1 1
or ,cos2A = 2 2. x2

Therefore, cos2A = 1 x2 + 1  RHS proved
2 x2 

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17. If SinA = 1 x + 1  then show that cos2A = - 1 x2 + 1  .
2 x  2 x2 

we have , cos2A = 1 - 2 sin2A

2

or, cos2A = 1 - 2. 1 x + 1  
2 x  

or, cos2A = 1  x2 + 2.x . 1 +  1 2 
1-2.4  x  x  
  

or, cos2A = 1 - 1  x2 + 2 + 1 
2  x2 

or, cos2A = 1- 1 . x2 - 1 - 11
2 2.2 2 . x2

or, cos2A = 1- 1 x2 -1- 11
2 2 . x2

or ,cos2A = - 1 x2 - 1 1
2 2. x2

Therefore, cos2A = - 1 x2 + 1  RHS proved
2 x2 

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18.If cosA = 1 x + 1  then show that cos3A = 1 x3 + 1 
2 x  2 x3 

here , Given

cosA = 1 x + 1 
2 x 

now, cos3A = 4 cos3A - 3cosA

3

or, cos3A = 4 1 x + 1   - 3. 1 x + 1 
2 x   2 x 

1 x 1 3 3 x 1 
8 x  2 x 
or,cos3A = 4. + - +

1 x 1 3 3 x 1 
2 x  2 x 
or, cos3A = + - +

1 x + 3 - 3x + 1  
or, cos3A = 2 x  
1  
x 

 cos3A = 1 x3 + 1 
2 x3 

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Exercise:D
cosA - 1 + sin2A
1.Prove that sinA + 1 - sin2A = - tanA.

1 + sin2A 1 + sin2A
2.Prove that 1 - sin2A = cos2A

3.If SinA = 1 x + 1  then show that cos2A = - 1 x2 + 1  .
2 x  2 x2 

4.If cosA = 1 x + 1  then show that cos2A = 1 x2 + 1  .
2 x  2 x2 

5.If sinA = 1 x + 1  then show that cos3A = - 1 x3 + 1  .
2 x  2 x3 

6.If cosA = 1 x + 1  then show that cos3A = 1 x3 + 1  .
2 x  2 x3 

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17. Prove that cotA ( 1- cos2A) = sin2A

LHS, cotA ( 1- cos2A)

= cosA . 2sin2A
sinA

= 2sinA.cosA

= sin2A RHS proved

sin 3A - cos3A 1
18. Prove that sinA - cosA = 1+ 2 sin2A

sin 3A - cos3A
LHS, sinA - cosA

(sinA - cosA)(sin2A + sinAcosA + cos2A)
= sinA -cosA

= sin2A + cos2A + 1 . 2sinA.cosA
2

1
= 1 +2 sin2A RHS proved

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sin5A cos5A
19. Proved that sinA - cosA = 4 cos2A

sin5A cos5A
LHS, sinA - cosA

sin5A.cosA - cos5A.sinA
= sinA.cosA

sin(5A-A)
= sinA.cosA

sin4A
=1

2 . 2sinA.cosA
2. 2sin2A.cos2A
= sin2A
= 4cos2A RHS Proved

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cosA - sinA
20. Prove that cosA + sinA = sec2A - tan2A

cosA - sinA
LHS, cosA + sinA

cosA - sinA cosA - sinA
= cosA + sinA . cosA - sinA

(cosA - sinA)2
= cos2A - sin2A

cos2A + sin2A - 2sinA.cosA
= cos2A

1 - sin2A
= cos2A

1 sin2A
= cos2A - cos2A
= sec2A - tan2A RHS proved

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11
21. Prove that tan7A - tan4A - cot7A - cot4A = cot3A

11
LHS, tan7A - tan4A - cot7A - cot4A

11
=1 1 - cot7A - cot4A
cot7A - cot4A

11
= cot4A - cot7A - cot7A - cot4A

cot7A .cot4A

cot7A .cot4A 1
= cot4A - cot7A - cot7A - cot4A

cot7A .cot4A 1
= cot4A - cot7A + cot4A - cot7A

cot4A .cot7A + 1
= cot4A - cot7A

= cot(7A - 4A)

= cot3A RHS proved

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Exercise:E

1. Prove that cosec2A + cot2A = cotA

2. Prove that cosec2A - cot2A = tanA

1 - cos2A = tan2A
3. prove that 1 + cos2A

4. Prove that cotA - tanA = 2cot2A

cotA + tanA
5. Prove that cotA - tanA = sec2A

6.Prove that tanA(1 + cos2A) = sin2A

sin6A cos6A
7. Prove that sin2A - cos2A = 2

tan2A
8. Prove that tanA - 1 = sec2A

sin 3A + cos3A 1
9. Prove that sinA + cosA = 1- 2 sin2A

11
10.Prove that tan3A + tanA - cot3A + cotA = cot4A

11
11. Prove that tan3A - tanA - cot3A - cotA = cot2A

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22.Prove that cosec10° - 3 sec10°= 4

LHS, cosec10° - 3 sec10°

1 - tan60° 1
= sin10° cos10°

1 sin60° 1
= sin10° - cos60° . cos10°

cos60°.cos10° - sin60°. sin10°
= sin10°. cos60°. cos10°

=1 cos(60°+10°)
2
sin10°.cos10°

2cos70°
= sin10° cos10°

2 .2 cos(90-20) °
= 2sin10° cos10°

4sin20°
= sin20°

= 4 RHS Proved

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23.Prove that 2 + 2 + 2cos4A = 2cosA

LHS, 2 + 2 + 2cos4A

= 2 + 2 (1+ cos4A)

= 2 + 2 (2cos22A)

= 2 + (2cos2A) 2

= 2 + 2cos2A

= 2(1 + cos2A)

= 2.2 cos2A

= (2 cosA) 2

= 2cosA RHS Proved

24. Prove that 4( cos310° + sin320°) = 3 (cos 10° + sin20°)

LHS, 4( cos310° + sin320°)

= 4cos310° + 4sin320°

= {3cos10° + cos( 3×10) ° } + { 3sin20° - sin(3 ×20°)}

= 3cos10° + cos30 ° + 3sin20°- sin60°

33
= 3cos10° + 2 + 3sin20°- 2

= 3cos10°+ 3sin20° = 3(cos10°+ sin20°) RHS Proved

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Exercise:F
1. 4( cos320° + sin350°) = 3 (cos 20° + sin50°)
2. 4( cos320° + sin310°) = 3 (cos 20° + sin10°)

3. Prove that 2+ 2 + 2 + 2cos8A = 2cosA

4. Prove that 2+ 2+ 2 + 2 + 2cos16A = 2cosA
5. Prove that cosec50° + 3 sec50°= 4
6. Prove that sec80° - 3 cosec80°= 4
7. Prove that 3 sin75° - cos75° = 2
25. Prove that cos4A = 8cos4 A - 8cos 2A + 1

LHS , Cos4A
= cos2.2A
= 2cos2 2A - 1
=2(cos2A)2 - 1
= 2(2cos2A -1)2 - 1
=2{(2cos 2A)2 - 2 .2cos2 A.1 + 12} - 1
=2(4cos4A - 4cos2A + 1) - 1
=8cos4A - 8cos2A +2 - 1
= 8cos4A - 8cos2A + 1 RHS Proved.

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26. Prove that Sin4A = 4sinA.cos3A - 4sin3A.cosA
LHS , Sin4A
=Sin2.2A
=2sin2A.cos2A
=2.2sinA.cosA(cos2 A - sin2 A)
=4sinA.cosA(cos2 A - sin2 A)
=4sinA.cos3A - 4sin3A.cosA RHS proved .

Alternative method
Prove that Sin4A = 4sinA .cos3A - 4sin3A.cosA
LHS, Sin4A

= sin(3A + A)
= sin3A.cosA + cos3A.sinA
= (3sinA - 4sin3A).cosA + (4cos3A - 3cosA).sinA
= 3sinA.cosA - 4sin3A.cosA + 4 cos3A .sinA - 3cosA.SinA
= 4 cos3A .sinA - 4sin3A.cosA

RHS proved

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27. Prove that Sin5A = 16sin5 A - 20sin3 A + 5sinA

LHS, sin5A

= sin(3A + 2A)

= sin3A.cos2A + cos3A.sin2A

= (3sinA - 4 sin3A)(1- 2 sin2A) + (4cos3A - 3cosA).2sinA.cosA

= 3sinA - 6sin3A - 4sin3A + 8sin5A + 8cos4A.sinA - 6sinA.cos2A

= 3sinA - 6sin3A - 4sin3A + 8sin5A + 8(cos2A)2.sinA - 6sinA(1-
sin2A)

= 3sinA - 10 sin3A + 8sin5A + 8(1- sin2A) 2 . sinA - 6sinA +
6sin3A

= -3sinA - 4 sin3A + 8sin5A + 8sinA{1 - 2.1. sin2A + (sin2A)2 }

= -3sinA - 4 sin3A + 8sin5A+ 8sinA - 16sin3A + 8sin5A

= 16 sin5A - 20 sin3A + 5sinA RHS Proved.

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28. Prove that cos6A - sin6A =  1 sin22A
4
cos2A1- 



LHS , cos6A - sin6A

= {cos2A}3 - {sin2A}3

= (cos2A - sin2A){(cos2A)2 + cos2A.sin2A + (sin2A)2}

= cos2A{ (sin2A)2 + (cos2A)2 + cos2A.sin2A}

= cos2A{( sin2A + cos2A)2 - 2 sin2A.cos2A + sin2A.cos2A}

= cos2A{(1)2 - sin2A cos2A}

 1 .4 sin2A.cos2A
4
=cos2A1 - 



=  1 (2sinA.cosA)2
4
cos2A1- 



 1 sin22A RHS proved.
4
=cos2A1- 



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29. Prove that (2cosA+1)(2cosA -1)(2cos2A - 1 ) = 2cos4A + 1
LHS, (2cosA + 1 ) ( 2cosA -1) (2cos2A - 1 )

= { (2cosA)2 - 12}(2cos2A - 1 )
= (4cos2A - 1) (2cos2A - 1 )
= (2.2cos2A - 1) (2cos2A - 1 )
={2.(1+cos2A) - 1 }(2cos2A - 1 )
= (2+2cos2A -1) (2cos2A - 1 )
= (2cos2A +1) (2cos2A - 1 )
= ( 4cos22A - 1)
= (2.2cos22A - 1)
={2 (1+cos4A) - 1 }
= 2 + 2cos4A -1
= 2cos4A + 1 RHS Proved

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30.Prove that
(2cosA + 1)(2cosA -1)(2cos2A -1)(2cos4A -1) = (2cos8A + 1)
LHS, (2cosA + 1 ) ( 2cosA -1) (2cos2A - 1 ) ( 2cos4A - 1 )

= { (2cosA)2 - 12}(2cos2A - 1 ) ( 2cos4A - 1 )
= (4cos2A - 1) (2cos2A - 1 ) ( 2cos4A - 1 )
= (2.2cos2A - 1) (2cos2A - 1 ) ( 2cos4A - 1 )
={2.(1+cos2A) - 1 }(2cos2A - 1 ) ( 2cos4A - 1 )
= (2+2cos2A -1) (2cos2A - 1 ) ( 2cos4A - 1 )
= (2cos2A +1) (2cos2A - 1 ) ( 2cos4A - 1 )
= ( 4cos22A - 1) ( 2cos4A - 1 )
= (2.2cos22A - 1) ( 2cos4A - 1 )
={2 (1+cos4A) - 1 }( 2cos4A - 1 )
= (2 + 2cos4A -1) ( 2cos4A - 1 )
= (2cos4A + 1 ) ( 2cos4A - 1 )
= (4cos24A - 1)
= ( 2.2cos24A -1)
={2(1+cos8A) -1}
= (2+2cos8A -1)
= (2cos8A + 1 ) RHS proved.

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Exercise:G

1.Prove that Cos6A + sin6A = 1 ( 1 + 3cos22A)
4

2.Prove that cos6A - sin6A = 1 (cos3A + 3cos2A)
4

3. Prove that Cos6A + sin6A = 1 ( 5 + 3cos4A)
8

4. Prove that Cos6A + sin6A = 1 ( 4 - 3sin22A)
4

5.Prove that Cos8A + sin8A = 1 - sin22A + 1 sin42A
8

2cos4A + 1
6. Prove that ( 2cosA -1) (2cos2A - 1 ) = 2cosA + 1

2cos8A+1
7.Prove that (2cosA-1)(2cos2A-1 )(2cos4A-1) = 2cosA+1

8. Prove that (2cosA-1)(2cos2A-1)(2cos4A-1)(2cos8A-1) =
2cos16A+1

2cosA+1

9. Prove that (2cosA-1)(2cos2A-1)(2cos4A-1)(2cos8A-1)(2cos16A-1) =
2cos32A+1
2cosA+1

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31.Prove that tanA + 2tan2A + 4tan4A + 8 cot8A = cotA

LHS, tanA + 2tan2A + 4tan4A + 8 cot8A

1
= tanA + 2 tan2A + 4tan4A + 8. tan8A

1
= tanA + 2 tan2A + 4tan4A + 8. tan2.4A

1
= tanA + 2 tan2A + 4tan4A + 8. 2tan4A

1-tan24A

2tanA
here,tan2A = 1 - tan2A

1-tan24A
= tanA + 2 tan2A + 4tan4A + 8 . 2tan4A

= tanA + 2 tan2A + 8tan24A + 8( 1-tan24A)

2tan4A

= tanA + 2 tan2A + 8tan24A + 8 - 8 tan24A

2tan4A

8
= tanA + 2 tan2A + 2tan4A

1
= tanA + 2 tan2A + 4. tan4A

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1
= tanA + 2 tan2A + 4 . tan2.2A

1- tan22A
= tanA + 2 tan2A + 4 . 2tan2A

= tanA + 4tan22A + 4 - 4tan22A

2tan2A

2
= tanA + tan2A

1-tan2A
= tanA + 2 2tanA

2tan2A + 2 - 2tan2A
= 2tanA

1
= tanA

= cotA RHS proved.

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32. Prove that Cos6A + sin6A = 1 ( 5 + 3cos4A)
8

LHS, Cos6A + sin6A

= (cos2A)3 + (sin2A)3

= ( cos2A + sin2A) { (cos2A)2 - cos2A. sin2A + (sin2A)2}

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

= { (cos2A + sin2A)2 - 2 cos2A. sin2A - cos2A. sin2A}

= 1 - 3 sin2A .cos2A

= 1 - 3 . 1 .(2sinA.cosA)2
4

=1 - 3 . sin22A
4

= 1 - 3 . 1 . 2sin22A
4 2

3 here, 2sin2A = 1- cos2A
= 1 - 8 (1 - cos4A)

8 - 3+3cos4A
=8

1
= 8 (5 + 3cos4A) RHS proved

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Exercise : H
2
1.Prove that tanA + tan2A = cotA
2
2. Prove that tan2A + tan4A = cot2A
2
3. Prove that tan4A + tan8A = cot4A
2
4. Prove that tan8A + tan16A = cot8A

2
5. Prove that tan16A + tan32A = cot16A
6. Prove that tanA + 2tan2A + 4tan4A + 8 tan8A +16cot16A =
cotA
7. Prove that tanA + 2tan2A + 4tan4A + 8 tan8A

+ 16tan16A + 32cot32A = cotA

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33. Prove that cos2A + sin2A. cos2B = cos2B + sin2B cos2A.

LHS, cos2A + sin2A. cos2B

= cos2A + ( 1 - cos2A) cos2B

= cos2A + cos2B - cos2A. cos2B

= cos2B + cos2A - cos2A. cos2B

= cos2B + cos2A( 1- cos2B)

= cos2B - sin2B + cos2A.2 sin2B

= cos2B + cos2A.2 sin2B - sin2B

= cos2B + sin2B(2cos2A -1)

= cos2B + sin2B.cos2A RHS proved

34.Prove that sin2A – cos2A cos2B = sin2B – cos2B cos2A

LHS, sin2A – cos2A. cos2B

= sin2A - cos2A(2cos2B - 1)

= sin2A - cos2A. 2cos2B + cos2A

= sin2A + cos2A - 2cos2A. cos2B

= 1 - ( 1+ cos2A).cos2B

= 1 - cos2B - cos2B.cos2A

= sin2B - cos2B.cos2A RHS proved

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Sec4A - 1 tan4A
35. Prove that sec2A - 1 = tanA

Sec4A - 1
LHS, sec2A - 1

1
cos4A - 1
=1
cos2A - 1
1-cos4A
cos4A
= 1-cos2A
cos2A

1- cos4A cos2A
= 1- cos2A . cos4A

1- cos2.2A cos2A
= 1- cos2A . cos4A

2sin22A cos2A
= 2sin2A . cos4A

2sin2A .sin2A cos2A
= 2sin2A . cos4A

2sin2A .cos2A sin2A
= 2sin2A .cos4A

sin4A sin2A
= 2sin2A . cos4A

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2sinA.cosA sin4A
= 2.sinA.sinA . cos4A
= cotA. tan4A

1
= tanA . tan4A

tan4A
= tanA RHS proved

Exercise I
1.Prove that Sin2B + sin2A cos2B = sin2A + sin2B cos2A
2.Prove that Sin2A + cos2A cos2B = cos2B - sin2B cos2A

Sec8A - 1 tan8A
3.Prove that sec4A - 1 = tan2A

Sec16A - 1 tan16A
4.Prove that sec8A - 1 = tan4A

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36.Prove that cosec2A + cot4A = cotA - cosec4A

LHS, cosec2A + cot4A
1 cos4A

= sin2A + sin4A

1 cos4A
= sin2A + 2sin2A.cos2A

2cos2A + cos4A
= 2sin2A.cos2A

2cos2A + cos2.2A
= 2sin2A.cos2A

2cos2A + 2cos22A - 1
= 2sin2A.cos2A

2cos2A( 1 + cos2A) -1
= 2sin2A.cos2A

2cos2A( 1+ cos2A) 1
= 2sin2A.cos2A - 2sin2A.cos2A

1+ cos2A 1
= sin2A - sin4A

2cos2A
= 2sinA.cosA - cosec4A

cosA
= sinA - cosec4A

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= cotA - cosec4A RHS proved.

37.Prove that 2sinA 2sin3A 2sin9A = tan27A – tanA.
cos3A + cos9A + cos27A

2sinA 2sin3A 2sin9A
LHS , cos3A + cos9A + cos27A

2sinA. cosA 2sin3A.cos3A 2sin9A.cos9A
= cos3A.cosA + cos9A.cos3A + cos27A.cos9A

sin2A sin6A sin18A
= cos3A.cosA + cos9A.cos3A + cos27A.cos9A

sin(3A - A) sin(9A-3A) sin(27A -9A)
= cos3A.cosA + cos9A.cos3A + cos27A.cos9A

sin3A.cosA - cos3A.sinA sin9A.cos3A -cos9A.sin3A
= cos3A.cosA + +
cos9A.cos3A

sin27A.cos9A- cos27A.sin9A
cos27A.cos9A

sin3A.cosA cos3A.sinA sin9A.cos3A cos9A.sin3A
= cos3A.cosA - cos3A.cosA + cos9A.cos3A - cos9A.cos3A +

sin27A.cos9A cos27A.sin9A
cos27A.cos9A - cos27A.cos9A

sin3A sinA sin9A sin3A sin27A sin9A
= cos3A - cosA + cos9A - cos3A + cos27A - cos9A

= tan3A - tanA + tan9A - tan3A + tan27A - tan9A

= tan27A - tanA RHS proved

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 2  3 1
38. Prove that cos 7 . cos 7 . cos 7 = 8

 2  3
LHS, cos 7 . cos 7 . cos 7

  2  3
2sin 7 cos 7 . cos 7 . cos 7
=

2sin 7
2  2  3
sin 7 . cos 7 . cos 7
=

2sin 7
2  2  3
2 sin 7 . cos 7 . cos 7
=

2. 2sin 7
2  3
sin2. 7 cos 7
=
4 sin 7
4  3
sin 7 cos 7
=
4sin 7

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sin(  - 3 3
7 ) cos 7
=
4sin 7

3 3
sin 7 .cos 7
=

4sin 7

3 3
2 sin 7 .cos 7
=

2.4sin 7

6
sin 7
=
8sin 7

sin( - 
7)
=
8sin 7

 1
sin 7 = 8 RHS proved
=
8sin 7

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Exercise :J

1. Prove that Cosec4A + cot8A = cot2A - cosec8A

2. Prove that Cose2A + cosec4A = cotA - cot4A

3. Prove that cosec8A + cot16A = cot4A - cosec16A

4.Prove that 2sin2A 2sin6A 2sin18A = tan54A – tan2A.
cos6A + cos18A + cos54A

5.Prove that 2 cos A 2 cos 3A 2 cos 9A = cotA – cot27A
sin3A + sin 9A + sin 27A

6.Prove that 2 cos 2A 2 cos 6A 2 cos 18A = cot2A –
sin6A + sin 18A + sin 54A

cot54A
 2  4 1

7. Prove that cos 7 . cos 7 . cos 7 = - 8

 5 7 1
8. Prove that sin 18 .sin 18 . sin 18 = 8

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