Verify the identity: $$\sin 3x + \sin x = 4\sin x - 4\sin^3 x$$
All I've done is this, and I don't know where to go from here:
$$4\sin x - 4\sin^3 x$$
$$ = 4(\sin x - \sin^3 x)$$
$$ = (4\sin x)(1 - \sin^2 x)$$
$$ = (4/\sin x)(\cos^2 x)$$
Am I on the right track??
On
$$\sin 3x + \sin x = 4\sin x - 4\sin^3 x$$
which is the same as writing,
$$\sin 3x = 3\sin x - 4\sin^3 x$$
You can either assume this standard identity, and we are done. Or we can try to prove this.
We have $e^{ix}=\cos(x)+i\sin(x)\implies e^{3ix}=\cos(3x)+i\sin(3x)$
But $e^{3ix}=(e^{ix})^3$
$$(\cos(x)+i\sin(x))^3 = \cos(3x)+i\sin(3x)$$
$$\cos^3(x)+3i\sin(x)\cos^2(x)+3i^2\sin^2(x)\cos(x)+i^3\sin^3(x)= \cos(3x)+i\sin(3x)$$
The imaginary parts must be equal, which gives us
$$3\sin(x)(1-\sin^2(x))-\sin^3(x)=\sin(3x)$$
$$3\sin(x)-4\sin^3(x)=\sin(3x)$$
as desired.
On
Using Prosthaphaeresis Formulas
$$\sin3x+\sin x=2\sin2x\cos x=2\cos x(2\sin x\cos x)=4\sin x\cos^2x$$ $$=4\sin x(1-\sin^2x)$$
You will likely need the "angle-addition formula" for sine, since you probably aren't expected to know "multiple-angle formulas". (You will need the "double-angle formulas"...)
$$ \sin 3x \ + \ \sin x \ = \ \sin(2x + x ) \ + \ \sin x $$
$$ = \ (\sin 2x \ \cos x \ + \ \cos 2x \ \sin x) \ + \ \sin x $$
$$ = \ [ \ (2 \sin x \ \cos x \ \cdot \cos x) \ + \ (1 \ - \ 2 \sin^2 x) \ \sin x ] \ + \ \sin x $$
$$ = \ [ \ (2 \sin x \ \cos^2 x ) \ + \ (1 \ - \ 2 \sin^2 x) \ \sin x ] \ + \ \sin x \ = \ [ \ (2 \sin x \cdot [1 - \sin^2 x] \ ) \ + \ (1 \ - \ 2 \sin^2 x) \ \sin x ] \ + \ \sin x \ . $$
You can probably take it from there.