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

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 $(-1,1)$

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 $\cos(7x)$.
• 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 $\sin(4x)$. – André Nicolas Sep 3 '14 at 4:21
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 $${1+z^{14}\over 2z^7}\ .$$