How do you solve $cos3x=cos^3x-3sin^2xcosx$ ?

Question

21027 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-16T01:03:03

It's true for all values of $x$ . We are to prove it as an identity.

You can prove it using the formula for the sine and cosine of a sum.

Recall that

$sin(a+b)=sin a cos b+cos a sin b$ (sine of a sum).

$cos(a+b)=cos a cos b-sin a sin b$ (cosine of a sum).

First put in $a=x, b=x$ , then

$sin(2x)=sin(x+x)=sin x cos x+ cos x sin x=2 sin x cos x$ (double angle identity for sine).

$cos(2x)=cos(x+x)=cos^2 x- sin^2 x$ (double angle identity for cosine).

Now, repeat but this time set $a=2x, b=x$ . Use the double angle identities we obtained above for $sin 2x$ and $cos 2x$ . So

$sin(3x)=sin(2x+x)=(2 sin x cos x)cos x+(cos^2 x-sin^2 x)sin x=3 sin x cos^2 x-sin^3 x$ (triple angle identity for sine).

And the identity ee originally set out to prove:

$cos(3x)=cos(2x+x)=(cos^2 x-sin^2 x)cos x-(2 sin x cos x)sin x=cos^3 x-3 sin^2 x cos x$ (triple angle identity for cosine).

How do you solve cos3x=cos^3x-3sin^2xcosxcos3x=cos^3x-3sin^2xcosx?