If $k\sin x\cos x=\sin(kx)$ is an identity for all $x$, then what is the smallest value of $k$?

Question

If $k\sin x\cos x=\sin(kx)$ is an identity for all $x$, then what is the smallest value of $k$?

I tried analyzing the domain and range of the sine and cosine functions, and how this identity came to be but I could not conceptualize this problem. Any help to understand this would be appreciated

Answer 1

We know that for $k=2$

$$\sin (2x)=2\sin x \cos x$$

but for $k=1$, in general

$$\sin (x)\neq \sin x \cos x$$

and for $k=-2$

$$\sin (-2x)=-2\sin(2x)=-2\sin x \cos x$$

therefore for $k\in \mathbb{Z}$ the minimum value seems to be $k=-2$ (refer to Zachary Selk's answer for the counterexample for $|k|>2$).

Answer 2

If $|k|>2$ then:

$$|k\sin(\pi/4)\cos(\pi/4)|=\left|\frac{k}{2}\sin(\pi/2)\right|=\left|\frac{k}{2}\right|>1\ge|\sin(k\pi/2)|$$

So $|k|\le 2$. Note that $k=-2$ works because:

$$-2\sin(x)\cos(x)=-\sin(2x)=\sin(-2x)$$

So $k=-2$

Answer 3

For $k\neq 0$, $k\sin x\cos x$ is zero at, and only at, all integer multiples of $\pi/2$; as it's equivalent to a sinusoid, that sinusoid necessarily has period $\pi$. We conclude that $k$ can be only $2$, $-2$, or $0$.

Determining the smallest $k$ is left as an exercise to the reader.