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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.

enter image description here

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    $\begingroup$ What do you mean by "here"? Are we supposed to recognize the source from that picture? $\endgroup$
    – jjagmath
    Jun 26 at 5:07
  • $\begingroup$ I still have no idea what the book is saying. $\endgroup$
    – Toby Mak
    Jun 26 at 5:40

2 Answers 2

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No it's false, try it with $x=0$.

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If the given result is valid both sides should have same range. Since Cosine of angles are between -1 and 1 , range of the left side is [-1,1] but because of 1/4 in the right side range of the right side is [ - 1/4 , 1/4 ] . Therefore two sides are not identical and the given result is wrong.
Edited ,
Since we are using same x in all terms in the product of left side we need to check the existence of-1 as the minimum and 1 as the maximum. You can verify those by substituting π and 0 respectively to the product.

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  • $\begingroup$ "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. $\endgroup$
    – jjagmath
    Jun 26 at 9:34
  • $\begingroup$ @jjagmath but in this case my argument is valid, to verify you can substitute 0 then you get 1 and when you substitute π , you get -1 for the left side. When you consider the product of Sinx and Cosx the scenario is different because when one value is 1 other value can not be 1 or - 1 , therefore that argument is not valid there. Since you are using same x in all terms you need to check whether limiting situations are exist or not. $\endgroup$ Jun 26 at 10:06
  • $\begingroup$ 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 $\pi$. $\endgroup$
    – jjagmath
    Jun 26 at 10:11
  • $\begingroup$ @jjagmath actually I verified because of your objection. $\endgroup$ Jun 26 at 10:15
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    $\begingroup$ 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 $\endgroup$
    – jjagmath
    Jun 26 at 10:18

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