Prove that $\sin 3x = 3 \sin x - 4\sin^2 x$

Question

Prove that $\sin 3x = 3 \sin x - 4\sin^2 x$

What I've tried:

$$\sin(2x + x) = \sin2x \cos x + \cos2x \sin x$$ We can rewrite $\cos 2x$ and $\cos(x+x)$ and $\sin 2x$ as $\sin (x+x)$ $$ \sin(x+x) \cos x + \cos(x+x) \sin x$$

Applying trig. identity:

$$ (2\sin x \cos x) \cos x + (\cos^2 x - \sin^2x) \sin x$$ $$ \sin x(2 \cos^2 x) + (\cos^2 x - \sin^2x) \sin x$$

Factoring out $\sin x$

$$ \sin x(2 \cos^2x + (\cos^2 x - \sin^2x) )$$ $$ \sin x(3 \cos^2x - \sin^2x) $$

Rewriting $\cos^2 x$ as $1 - \sin^2 x$

$$ \sin x(3 (1-\sin^2 x) - \sin^2x) $$ $$ \sin x(3-3\sin^2 x - \sin^2x) $$ $$ \sin x(3-4\sin^2 x) $$ $$ 3\sin x -4\sin^3 x$$

Since $3\sin x -4\sin^3 x ≠ 3 \sin x - 4\sin^2 x$, I must've made a mistake somewhere. Can someone tell me where the mistake is?

Answer 1

Checking on a graphing calculator, it's easy to see that $\sin(3x)$ and $3\sin x - 4\sin^2x$ are different functions, whereas $\sin(3x)$ and $3\sin x - 4\sin^3x$ look equal. I don't think you've made a mistake.

Answer 2

Your calculations are correct and the formula is $$\sin 3x=3\sin x-4\sin ^3 x$$

Answer 3

A direct computation shows that the claim you were supposed to prove is wrong for $x=\pi/3$: In fact: $\sin(3\cdot \pi/3) - \left( 3\sin(\pi/3) - 4\sin^2(\pi/3)\right) = 3-3\sqrt{2}/2 \neq 0$ so both sides are not equal. Your proof is correct.

Answer 4

Another angle for understanding the situation:

$\displaystyle \sin(x)^3=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^3=-\frac 1{8i}\left(e^{i3x}-\underbrace{3e^{i2x}e^{-ix}+3e^{ix}e^{-2ix}}_{\large{-3e^{ix}+3e^{-ix}}}-e^{-i3x}\right)=-\frac 14\left(\sin(3x)-3\sin(x)\right)$

Thus $$\sin(3x)=3\sin(x)-4\sin(x)^3$$

In the linearization process on $\sin(x)^n$ you see that you have products of $e^{ikx}$ and $e^{-i(n-k)x}$ which leads to $e^{\pm i(2k-n)x}$.

Since $2k-n$ has the same parity as $n$, then for $n$ odd, only odd terms can appear. For instance :

• $\sin(x)^3=a\sin(x)+b\sin(3x)$
• $\sin(x)^5=a\sin(x)+b\sin(3x)+c\sin(5x)$

And so on, this is the reason why a term in $\sin(x)^2$ would denote in this formula.