MHB 33. Express sin 4x in terms of sin x and cos x

[karush]

Express function as a trigonometric function of x
$$\sin(4x)$$
use $\sin2a=2\sin a\cos a$ then
$$\sin4x=2\sin 2x\cos 2x$$
with $\cos(2x) = \cos^2(x)-\sin^2(x)$ replace again
$$\sin 4x=4\sin x\cos x+\cos^2(x)-sin^2(x)$$

ok not real sure if this is what they are asking for
and if I should go further with it even if the steps are ok

 

[skeeter]

Express function as a trigonometric function of x
$$\sin(4x)$$
use $\sin2a=2\sin a\cos a$ then
$$\sin4x=2\sin 2x\cos 2x$$
with $\cos(2x) = \cos^2(x)-\sin^2(x)$ replace again
$$\color{red}{\sin 4x=4\sin x\cos x+\cos^2(x)-sin^2(x)}$$

ok not real sure if this is what they are asking for
and if I should go further with it even if the steps are ok

$\color{red}{\sin(4x) = 2\sin(2x)\cos(2x) = (4\sin{x}\cos{x})(\cos^2{x}-\sin^2{x})}$