Show that $\frac {\sin(3x)}{ \sin x} + \frac {\cos(3x)}{ \cos x} = 4\cos(2x)$ Show that 

$$\frac{\sin(3x)}{\sin x} + \frac{\cos(3x)}{\cos x} = 4\cos(2x).$$

 A: Use this fact that $$\sin(x+3x)=\sin(3x)\cos(x)+\cos(3x)\sin(x)$$
A: Here is a start
$$ \frac{\sin(3x)}{\sin x} + \frac{\cos(3x)}{\cos x}=2 \frac{\cos(x)\sin(3x)+\sin(x)\cos(3x)}{\sin 2x}  = \dots. $$
A: $$\dfrac{\sin(3x)}{\sin x} + \frac{\cos(3x)}{\cos x}$$
$$\dfrac{\sin(3x)\cos x + \cos(3x)\sin x}{\cos x{\sin x}}$$
$$\dfrac {\sin (3x+x)}{\sin x\cos x}$$
$$\dfrac {\sin 4x}{\sin x\cos x}$$
$$\dfrac {2\sin 2x\cos 2x}{\sin x\cos x}$$
$$\dfrac {4\sin x\cos x\cos 2x}{\sin x\cos x}$$
$$4\cos 2x$$
A: We have,
$$\sin(3x) = 3\sin(x) - 4\sin^{3}(x)$$
and
$$\cos(3x) = 4\cos^{3}(x) - 3 \cos(x).$$
Thus, your expression reduces to 
$$3 - 4\sin^{2}(x) + 4\cos^{2}(x) - 3$$
which is equal to
$$4 \cos(2x),$$
as $\cos(2x) = \cos^{2}(x) - \sin^{2}(x)$.
A: HINT:
$\sin (3x) = 3\sin x - 4 \sin^3 x$
and
$\cos(3x) = 4\cos^3x - 3\cos x$
A: 
$$\frac {\sin3x} {\sin x} = 3-4\sin^2x$$

as $\sin3x = 3\sin x-4\sin^3x$

$$\frac {\cos3x} {\cos x} = 4\cos^2x - 3$$

as $\cos3x = 4\cos^3x -3\cos x$
adding the two above we get $4(\cos^2x - \sin^2x)=4\cos2x.$
