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How do i simplify $\cos(4x)\cos(3x) − 4\sin(x)\sin(3x)\cos(x)\cos(2x)$ ?

I tried plugging in the double angle formulas $\cos(x+3x)$ and $\cos(x+2x)$ and went nowhere please help me.

Also maybe $\cos^{-1}(t) - \sin^{-1}(t)$ for $t$ in $(-1,1)$

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We play for a while with the second term $4\sin x\cos x\cos(2x)\sin(3x)$. Note that $2\sin x\cos x=\sin(2x)$. Using similar reasoning, we conclude therefore that $4\sin x\cos x\cos(2x)=2\sin(2x)\cos(2x)=\sin(4x)$.

So our expression is equal to $$\cos(4x)\cos(3x)-\sin(4x)\sin(3x).$$ By the addition law for cosine, this is $\cos(7x)$.

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  • $\begingroup$ You are welcome. I see your comment got interrupted. That happens when one presses the Return key. Used to happen to me a lot. $\endgroup$ – André Nicolas Sep 3 '14 at 4:16
  • $\begingroup$ In, cos(4x)cos(3x)−4sin(x)sin(3x)cos(x)cos(2x) arrange as, =cos(4x)cos(3x)-2sin(x)cos(x)*2sin(3x)cos(2x) see that, 2sinxcosx=sin(2x) so, =cos(4x)cos(3x)-sin(2x)*2sin(3x)cos(2x) eq(a) But, 2sin(2x)cos(2x)=sin(4x) And eq(a) arrange as = cos(4x)cos(3x)-2sin(2x)cos(2x)*sin(3x) Hence, = cos(4x)cos(3x)-2sin(4x)sin(3x) Which is the sum angle of cos(4x+3x) = cos(7x) Tytytytytytytytytytyyt $\endgroup$ – Ivan Sep 3 '14 at 4:17
  • $\begingroup$ At the very end you have a typo. You wrote $\cos(4x)\cos(3x)-2\sin(4x)\sin(3x)$. There should not be a $2$ there, please see my answer. Both $2$'s from the $4$ get absorbed, in going from $4\sin x\cos x\cos(2x)$ to $\sin(4x)$. $\endgroup$ – André Nicolas Sep 3 '14 at 4:21
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Here is a systematic procedure: Put $e^{ix}=:z$ and use Euler's formula to rewrite the given expression in terms of $z$. E.g., $\cos(4x)=(z^4+z^{-4})/2$. When an essential simplification is possible this will be reflected in the resulting rational function of $z$. In the case at hand the result is $${1+z^{14}\over 2z^7}\ .$$

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