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 $(-1,1)$
[Math] How do i simplify $\cos(4x)\cos(3x)−4\sin(x)\sin(3x)\cos(x)\cos(2x)$
trigonometry
Best Answer
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 $\cos(7x)$.