How do I prove the identity sin(4x) = 4sin(x)cos^3(x)-4sin^3(x)cos(x)?

[Slicktacker]

Hi all, I have a question proving an identity:

$$sin(4x) = 4sin(x)cos^3(x)-4sin^3(x)cos(x)$$

I can't seem to figure it out. I know I should be using the known identities:

$$sin(2x) = 2sin(x)cos(x)$$

and probably:

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

but I'm stuck. Please help!

Thanks!

 

[shmoe]

First hint, apply your double angle formula for sin to:

$$sin(2(2x))$$

what do you get?

Second hint, try factoring the right hand side to help see what identities you'll need:

$$sin(4x) = 4sin(x)cos(x)(cos^2(x)-sin^2(x))$$

How does this compare to your answer above? Can you see where the factors come from?

 

[gravenewworld]

Mathematicians never memorize trig identities. Trig identities are proven using complex numbers. You can if you wish try to prove your identity with complex numbers, you may find it may be easier. If I can remember correctly cos(x)=(e^(ix) +e^(-ix))/2 and sin(x) = (e^(ix))-e^(-ix))/2i (someone correct me if I am wrong). From there all you have to do is algebra instead of working with identities to get the left side= to the right side.

 

[TenaliRaman]

Rather one may also use D'Moivre,
(cos4x+isin4x) = (cosx+isinx)^4
expand RHS and compare imaginary parts ... as a bonus u also get cos4x :P

 

[sunny86]

let me try
sin (4x) = 4 sin x cos *3 x - 4 sin *3 ( x ) cos ( x)

let 2x = X
sin 2X = 2 sin X cos X
=2 sin 2x cos 2x
=2 [( 2sinx cosx ) ( cos*2 x - sin *2 x ) ]
=2 [ 2sinxcos*3 x - 2sin*3cos x ]
= 4sinx cos*3 x - 4sin*3 x cos x