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 & = \sin(2x+x)\\ & = \sin 2x \cos x + \cos 2x \sin x\\ & = 2\sin x \cos^2x + (1-2\sin^2 x)\sin x\\ & = 2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x\\ & = 2\sin x - 2\sin^3 x + \sin x - 2\sin^3 x\\ & = 3\sin x - 4\sin^3 x \end{align*}
b) \begin{align*} \cos 3x & = \cos(2x)\cos(x) - \sin(2x)\sin x\\ & = (2\cos^2x - 1) \cos x - (2\sin x \cos x) \sin x\\ & = 2\cos^3x - \cos x - 2\sin x \cos x\\ & = 2 \cos^3 x - \cos x - 2(1 - \cos^2x) \cos x\\ & = 2\cos^3x - \cos x- (2\cos x - 2\cos^3x)\\ & = 4\cos^3x - 3\cos x \end{align*}
Is b) also correct?