1. ## Identity question

"Prove (algebraically) that: $\displaystyle cos6x+cos4x=2cos5xcosx$"

I'm not sure if I should handle this like an identity or not... I'm well out of the identity chapter, so I don't know why I'd be asked this now.

I'm thinking the composite argument property might be of use?

2. Hello, wiseguy!

Prove (algebraically) that: .$\displaystyle \cos6x+\cos4x\:=\:2\cos5x\cos x$

I'm not sure what "algebraically" means . . .
. . some Trigonometry must be used.

We have the identity: .$\displaystyle \cos(A \pm B) \;=\;\cos A\cos B \mp \sin A\sin B$

Then: . $\displaystyle \begin{array}{cccccccc} \cos(6x) &=& \cos (5x + x) &=& \cos5x\cos x - \sin5x\sin x & [1] \\ \cos(4x) &=& \cos(5x - x) &=& \cos5x\cos x + \sin5x\sin x & [2] \end{array}$

Add [1] and [2]: . $\displaystyle \cos6x + \cos 4x \;=\;2\cos5x \cos x$

3. Wow my brain must of been asleep. Thank you!!!

4. Or...

$\displaystyle 2cosAcosB=cos(A+B)+cos(A-B)$

Alternatively..

$\displaystyle cosA+cosB=2cos\left(\frac{A+B}{2}\right)cos\left(\ frac{A-B}{2}\right)$