# Simplification of trigonometric expression using double and compound angle identities

This expression:

$$\cos(4x)\cos(3x) - 4\sin(x) \sin(3x)\cos(x)\cos(2x)$$

I have struggled for hours trying to simplify without success.

Is it legitimate to simplify the right side using the Double Angle identity thusly?

$$\cos(4x)\cos(3x) - \sin(4x)\sin(3x)\cos(2x)$$

That is, if $\sin(2x) = 2\sin(x)\cos(x)$ is it legitimate to say $\sin(4x) = 4\sin(x)\cos(x)$?

But even if I get this far, I want to say it looks like a cosine sum identity

$$\cos(4x + 3x) = \cos(4x)\cos(3x) - \sin(4x)\sin(3x)$$

but I cannot see what to do with that pesky $\cos(2x)$ on the right side.

• Equation? I see only an expression....And no, $sin4x$ is not $4sinxcosx$ Commented Nov 26, 2015 at 4:26
• The idea is fine, details not quite right. We have $4\sin x\sin 3x\cos x\cos 2x=\sin 4x\sin 3x$. Commented Nov 26, 2015 at 4:32
• imranfat, yes I confused the terminology - I will amend it if I can. Thank you. Commented Nov 26, 2015 at 4:36

Since $\sin2x=2\sin x\cos x$, the second term in your expression is $$(2\sin2x\cos2x)\sin3x\ ,$$ and then the bit in brackets is...?
Notice, use the trig identity $\color{blue}{2\sin A\cos A=\sin 2A}$ as follows
$$\cos( 4x)\cos(3x)-4\sin (x)\sin(3x)\cos(x)\cos (2x)$$ $$=\cos( 4x)\cos(3x)-2(\underbrace{2\sin (x)\cos(x)})\sin(3x)\cos (2x)$$ $$=\cos( 4x)\cos(3x)-2\sin(2x)\sin(3x)\cos (2x)$$ $$=\cos( 4x)\cos(3x)-\underbrace{2\sin(2x)\cos (2x)}\sin (3x)$$ $$=\cos( 4x)\cos(3x)-\sin(4x)\sin(3x)$$ Applying $\color{blue}{\cos A\cos B-\sin A\sin B=\cos(A+B)}$ $$=\cos( 4x+3x)$$$$=\color{red}{\cos 7x}$$