Can you factorize this trigonometric expression?

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Discussion Overview

The discussion revolves around the factorization of the trigonometric expression $\cos^2 x+\cos^2 2x+\cos^2 3x+\cos 2x+\cos 4x+\cos 6x$. Participants explore various methods to factor this expression, engaging in a collaborative effort to arrive at a solution.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • Multiple participants attempt to factor the same trigonometric expression, indicating a shared interest in the problem.
  • Some participants express agreement with each other's approaches to factorization, suggesting a collaborative atmosphere.

Areas of Agreement / Disagreement

While there is some agreement among participants regarding their methods, the discussion does not resolve into a single consensus on the factorization itself.

Contextual Notes

The discussion lacks detailed steps or specific methodologies for the factorization, and the assumptions underlying the approaches are not fully articulated.

anemone
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Factorize $\cos^2 x+\cos^2 2x+\cos^2 3x+\cos 2x+\cos 4x+\cos 6x$.
 
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anemone said:
Factorize $\cos^2 x+\cos^2 2x+\cos^2 3x+\cos 2x+\cos 4x+\cos 6x$.

$=\frac{1}{2}(2\cos^2 x+2\cos^2 2x+2\cos^2 3x+2\cos 2x +2\cos 4x + 2\cos 6x)$
$= \frac{1}{2}(\cos 2x + 1 + \cos 4x + 1 + \cos 6x + 1 + 2\cos 2x +2\cos 4x + 2\cos 6x)$
$=\frac{3}{2} (\cos 2x+ \cos 4x+ \cos 6x+ 1)$
$= \frac{3}{2}(\cos 2x+ \cos 6x+ \cos 4x+ 1)$
$=\frac{3}{2}(2 \cos 2 x \cos 4x + 2\cos^2 2x)$
$=3(\cos 2 x \cos 4x + \cos^2 2x)$
$=3\cos 2 x(\cos 4x + \cos 2x)$
$=3\cos 2x(2\cos 3x\cos\, x)$
$=6\cos\,x \cos 2x\cos 3x$
 
Thanks for participating, kaliprasad! I factored the sum the same way you did!(Cool)
 
anemone said:
Factorize $\cos^2 x+\cos^2 2x+\cos^2 3x+\cos 2x+\cos 4x+\cos 6x$.

$$\cos^2 x+\cos^2 2x+\cos^2 3x+\cos 2x+\cos 4x+\cos 6x$$
$$=3\cos^2x+3\cos^22x+3\cos^23x-3$$
$$=3(\cos^2x+(2\cos^2x-1)\cos2x+(4\cos^3x-3\cos x)\cos3x-1)$$
$$=3(\cos^2x+2\cos^2x\cos2x-\cos2x+4\cos^3x\cos3x-3\cos x\cos3x-1)$$
$$=3(\cos^2x+2\cos^2x\cos2x+4\cos^3x\cos3x-3\cos x\cos3x-\cos2x-1)$$
$$=3(2\cos^2x\cos2x+4\cos^3x\cos3x-3\cos x\cos3x-\cos^2x)$$
$$=3\cos x(2\cos x\cos2x+4\cos^2x\cos3x-3\cos3x-\cos x)$$
$$=3\cos x(\cos x(2\cos2x-1)+\cos3x(4\cos^2x-3))$$
$$=3\cos x(\cos x(2\cos2x-1)+\cos3x(2(1+\cos2x)-3))$$
$$=3\cos x(\cos x(2\cos2x-1)+\cos3x(2\cos2x-1))$$
$$=3\cos x(2\cos2x\cos3x+2\cos2x\cos x-\cos x-\cos3x)$$
$$=6\cos x\cos2x\cos3x$$
 
Last edited:
greg1313 said:
$$\cos^2 x+\cos^2 2x+\cos^2 3x+\cos 2x+\cos 4x+\cos 6x$$
$$=3\cos^2x+3\cos^22x+3\cos^23x-3$$
$$=3(\cos^2x+(2\cos^2x-1)\cos2x+(4\cos^3x-3\cos x)\cos3x-1)$$
$$=3(\cos^2x+2\cos^2x\cos2x-\cos2x+4\cos^3x\cos3x-3\cos x\cos3x-1)$$
$$=3(\cos^2x+2\cos^2x\cos2x+4\cos^3x\cos3x-3\cos x\cos3x-\cos2x-1)$$
$$=3(2\cos^2x\cos2x+4\cos^3x\cos3x-3\cos x\cos3x-\cos^2x)$$
$$=3\cos x(2\cos x\cos2x+4\cos^2x\cos3x-3\cos3x-\cos x)$$
$$=3\cos x(\cos x(2\cos2x-1)+\cos3x(4\cos^2x-3))$$
$$=3\cos x(\cos x(2\cos2x-1)+\cos3x(2(1+\cos2x)-3))$$
$$=3\cos x(\cos x(2\cos2x-1)+\cos3x(2\cos2x-1))$$
$$=3\cos x(2\cos2x\cos3x+2\cos2x\cos x-\cos x-\cos3x)$$
$$=6\cos x\cos2x\cos3x$$

Well done, greg1313! And thanks for participating!(Cool)
 

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