Proving the formulas for $\sin 3\theta$ and $\cos 3\theta$. (Check my work.) Substitute in the equation $3\theta=(2\theta+\theta)$, prove that:
a. $\sin 3\theta = 3\sin\theta-4\sin^3\theta$
b. $\cos 3\theta = 4\cos^3\theta-3\cos\theta$

Is this the answer to a)?
$$\sin 2x = 2\sin x\cos x,\quad\cos 2x = 1-2\sin^2 x$$
\begin{align*}
\sin 3x &amp; = \sin(2x+x)\\
&amp; = \sin 2x \cos x + \cos 2x \sin x\\
&amp; = 2\sin x \cos^2x + (1-2\sin^2 x)\sin x\\
&amp; = 2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x\\
&amp; = 2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x\\
&amp; = 3\sin x - 4\sin^3 x
\end{align*}
b)
\begin{align*}
\cos 3x &amp; = \cos(2x)\cos(x) - \sin(2x)\sin x\\
        &amp; = (2\cos^2x - 1) \cos x - (2\sin x \cos x) \sin x\\
        &amp; = 2\cos^3x - \cos x - 2\sin x \cos x\\
        &amp; = 2 \cos^3 x - \cos x - 2(1 - \cos^2x) \cos x\\
        &amp; = 2\cos^3x - \cos x- (2\cos x - 2\cos^3x)\\
        &amp; = 4\cos^3x - 3\cos x
\end{align*}
Is b) also correct?
Additional material:

Using complex numbers,

$$\cos3\theta+i\sin3\theta=e^{i3\theta}=(e^{i\theta})^3=(\cos\theta+i\sin\theta)^3.$$

Then by the binomial formula, and with abbreviations,

$$c^2+3ic^2s-3cs^2-is^3.$$

From this

$$\cos3\theta=c^2-3cs^2=4c^3-3c$$

and

$$\sin3\theta=3c^2s-s^3=3s-4s^3.$$
Additional material:

Using complex numbers,

$$\cos3\theta+i\sin3\theta=e^{i3\theta}=(e^{i\theta})^3=(\cos\theta+i\sin\theta)^3.$$

Then by the binomial formula, and with abbreviations,

$$c^2+3ic^2s-3cs^2-is^3.$$

From this

$$\cos3\theta=c^2-3cs^2=4c^3-3c$$

and

$$\sin3\theta=3c^2s-s^3=3s-4s^3.$$