# Is the statement true: $\cos x\cos 2x\cos 4x=1/4\cos 3x$? [closed]

How to show that $$\cos x\cos 2x\cos 4x=1/4\cos 3x$$?

I have tried by $$\cos x\cos 2x\cos 4x=\frac{1}{2}\cos x[2\cos 2x\cos 4x]$$. $$=\frac{1}{2}\cos x[\cos 6x+\cos 2x]$$. $$=\frac{1}{4}[2\cos x\cos 6x+2\cos x\cos 2x]$$ But it is not going to the required result.

Updated: I found it here. • What do you mean by "here"? Are we supposed to recognize the source from that picture? Jun 26 at 5:07
• I still have no idea what the book is saying. Jun 26 at 5:40

No it's false, try it with $$x=0$$.
• "Since Cosine of angles are between -1 and 1 , range of the left side is [-1,1]“ This argument is wrong (although the range is $[-1,1]$). Take for example $\sin(x)$ and $\cos(x)$. They are both between $-1$ and $1$ but the range of $\sin(x)\cos(x)$ is $[-1/2,1/2]$. Another counterexample: the range of $\cos(x)\cos(2x)\cos(3x)$ is not $[-1,1]$, even when the range of the three factors is. Jun 26 at 9:34
• Your original argument is not valid (even if it's conclusion is right in this case), otherwise you wouldn't need to verify that you get $1$ and $-1$ evaluating at $0$ and $\pi$. Jun 26 at 10:11
• And if you don't include that in your original proof, it's incomplete, so not valid. In your original answer, your wording suggest that the reason of the left side has range $[-1,1]$ is because each of the factors have range $[-1,1]$, which is false Jun 26 at 10:18