# Identity question

• Jul 23rd 2010, 09:20 AM
wiseguy
Identity question
"Prove (algebraically) that: $cos6x+cos4x=2cos5xcosx$"

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? (Thinking)
• Jul 23rd 2010, 09:52 AM
Soroban
Hello, wiseguy!

Quote:

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

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

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

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

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

• Jul 23rd 2010, 11:03 AM
wiseguy
Wow my brain must of been asleep. Thank you!!!
• Jul 23rd 2010, 12:22 PM
$2cosAcosB=cos(A+B)+cos(A-B)$
$cosA+cosB=2cos\left(\frac{A+B}{2}\right)cos\left(\ frac{A-B}{2}\right)$