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# Sin (5x)=16 sin^5 x-20 sin^3 x-5 sin x How Do I Prove this equation?

Any math genius, please help me out

### 6 Answers

- 1 decade agoFavorite Answer
u are gonna need the foll :

sin(a+b) = sinacosb +cosasinb

cos(a+b) = cosacosb-sinasinb

sin2a=2sinacosa

cos2a=1-2sin^2a

cos3a= cosa(1-4sin^2a)

sin3a =3sina-4sin^3a

sin(5x) = sin(3x +2x) = sin3xcos2x + sin2xcos3x

= (3sinx-4sin^3x)(1-2sin^2x) + 2sinxcosx*cosx(1-4sin^2x)

= (3sinx-4sin^3x)(1-2sin^2x) + 2sinx(1-sin^2x)(1-4sin^2x)

= 16sin^5x - 20sin^3x -5sinx

- Mein Hoon NaLv 71 decade ago
This cab be proved as

sin 5A = SIN(A+4A)

= sin A cos 4A + cos A sin 4A

= sin A (1- 2 Sin^2A) + cos A 2. Sin 2A cos 2A

= sin A(1- 8 Sin^2 A cos^2 A) + Cos A 4 Cos A sin A (1- 2 Sin^2 A)

= sin A - 8 sin ^3A (1-sin^2 A) + 4 sin A Cos ^2 A(1- 2 sin ^2 A)

= sin A - 8 sin ^3 A + 8 Sin ^5 A + 4 sin A(1-sin^2A)(1-2 SIn ^2 A)

= sin A - 8 sin ^3 A + 8 sin ^5 A + 4 sin A ( 1- 3 sin ^2A + 2 Sin ^5 A)

= 5 SIn A - 20 sin ^3 A + 16 sin ^5 A

please follow the above step

by putting A = pi/2 I find my ans correct 5 in place of -5 .

please check the same.

- PretzelsLv 51 decade ago
The form of this suggests that you should be familiar with the Euler identity:

e^(i*x) = cos(x) + i * sin(x)

Where i = sqrt(-1)

If you arent familiar with this, the problem is very difficult. Using the Euler identity, notice that:

cos(5*x) + i * sin(5*x) = e^(i*5*x) = (e^(i*x))^5 = (cos(x) + i * sin(x))^5

If you carry out the expansion and equate the real and immaginary parts separately, the identity you are trying to prove will be obvious.

By the way, the identity as you typed it is incorrect. The last term needs a + sign instead of a - sign. The Euler identity is so powerful it allowed me to spot the error immediately without having to calculate anything.

- 1 decade ago
Listen dear it is too long..

u have 2 do it urself

hints are here

sin(5x)=sin(4x+1)

sin(x+y)=sinxcosy+cosxsiny

sin(2x)=2sinxcosx

cos(2x)=cos^2(x)-sin^2(x)

cos^2(x)=1-sin^2(x)

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- 1 decade ago
think of sin(5x) as sin[(2x) + ((2x) + x )]

sin(2x)=2sinxcosx

sin(a + b)=sinacosb +sinbcosa

cos(2x)=2(cos^2)(x) - 1

just replace and cancel until you get the final answer.

- rajLv 71 decade ago
sin5A=sin(3A+2A)=sin3Acos2A

+cos3Asin2A)

(3sinA-4sin^3A)(1-2sin^2A)

+(4cos^3A-3cosA)(2sinAcosA)

=3sinA-6sin^3A-4sin^3A

+8sin^5A+8sinAcos^4A

-6sinAcos^2A

8sin^5A-10sin^3A+3sinA

+8sinA(1-sin^2A)^2-6sinA

(1-sin^2A)

8sin^5A-10sin^3A+3sinA+8sinA

+8sin^5A-16sin^3A-6sinA+6sin^3A

16sin^5A-20sin^3A+5sinA