Trigonometry Question: find Value of.....

Question

Find value of $3 + \cos2x + \cos4x + \cos6x - 4\cos x\cos2x\cos3x$. I tried with $\cos A + \cos B$ identity but it was not simplifying.... Help..

Answer 1

Try the other way - $2\cos a\cos b=\cos(a+b)+\cos(a-b)$

Answer 2

Using that $2\cos\alpha\cos\beta=\cos(\alpha+\beta)+\cos(\alpha-\beta)$, we can write $$4\cos x\cos 2x\cos 3x=2\cos 2x\cdot \left(\cos4x+\cos 2x\right)=\cos6x+\cos2x+\cos4x+1,$$ which immediately gives $2$ as the final result.

Added: the second equality follows from \begin{align} 2\cos 2 x\cos 4x&=\cos(4x+2x)+\cos(4x-2x),\\ 2\cos 2x \cos 2x&=\cos(2x+2x)+\cos(2x-2x). \end{align}

Answer 3

Using Prosthaphaeresis Formula, $$\cos2x+\cos4x=2\cos3x\cos x$$ and using Double-Angle Formula, $$1+\cos6x=2\cos^23x$$

$$\implies\cos2x+\cos4x+1+\cos6x=2\cos3x\cos x+2\cos^23x =2\cos3x(\cos x+\cos3x)$$

Again using Prosthaphaeresis Formula,$$\cos x+\cos3x=2\cos2x\cos x$$