Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I wonder how does a WolframAlpha get this relation where input is a LHS and output is RHS:

$$\cos^2(x)\cos(2x) = \tfrac{1}{4}\cos(4x) + \tfrac{1}{2}\cos(2x) + \tfrac{1}{4}$$

share|cite|improve this question
Are you asking how to prove that equality? – Git Gud Aug 2 '13 at 16:50
Express $\cos^2(x)$ in terms of $\cos(2x), \sin(2x)$ and then use product-to-sum. – Scaramouche Aug 2 '13 at 16:52
Yes. I need to know how to get RHS out of LHS. I tried to use the double angle trigonometric identity and it only got more complicated... – 71GA Aug 2 '13 at 16:53
How Alpha gets it is an interesting question. I don't know, but it is likely not the way a human should approach it. – André Nicolas Aug 2 '13 at 17:02
up vote 2 down vote accepted

\begin{align*} \cos^2(x)\cos(2x) &= \frac{1}{2}(1+\cos(2x))\cos(2x)\\ &= \frac{1}{2}\cos(2x) +\frac{1}{2}\cos^2(2x) \\ &= \frac{1}{2}\cos(2x) + \frac{1}{4} + \frac{1}{4}\cos(4x) \end{align*} by two applications of the double angle formula.

share|cite|improve this answer


$$cos^2x=\frac{1+cos(2x)}{2}$$ $$cos^2(2x)=?$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.