1
$\begingroup$

How would I solve the following trig problem.

$$\cos^5x = \frac{1}{16} \left( 10 \cos x + 5 \cos 3x + \cos 5x \right)$$

I am not sure what to really I know it involves the sum and difference identity but I know not what to do.

  • 0
    hint: rewrite $\cos x$ with exponentials and expand the fifth power : the binomial coefficients are [here](http://en.wikipedia.org/wiki/Binomial_theorem)2012-08-06
  • 0
    I am not sure how I would that theorem.2012-08-06
  • 0
    $\cos(x)=\frac 12 (e^{ix}+e^{-ix})$ so that $\cos(x)^5=\frac 1{2^5} (e^{ix}+e^{-ix})^5$, using binomial theorem is faster for the term at the right but it is not an obligation!2012-08-06
  • 0
    I have not studied e^ix or complex numbers unfortunately.2012-08-06
  • 1
    you may try it in the other direction $\cos(5x)=\cos(4x+x)=\cos(4x)\cos(x)-\sin(4x)\sin(x)$ and so on if you know the trigonometric rules for addition and no rules for powers...2012-08-06
  • 0
    Yes I know the ones for addition but the ones for powers I will learn one day.2012-08-06
  • 0
    Hmm, I don't think you're trying to **solve** that problem - this would mean finding those values of $x$ for which it is true. You're actually trying to **prove** a trigonometric identity, not solve an equation. If you don't ask your questions carefully, you could lead potential answerers down a garden path.2012-08-07

3 Answers 3