Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Why: $$\cos ^2(2x) = \frac{1}{2}(1+\cos (4x))$$

I don't understand this, how I must to multiply two trigonometric functions?

Thanks a lot.

share|improve this question
1  
I've changed algebra tag to algebra-precalculus, since we don't use algebra tag anymore, see meta for details. –  Martin Sleziak Oct 29 '12 at 13:34
2  
This Wikipedia article is a useful resource for various formulas about trigonometric functions: List of trigonometric identities –  Martin Sleziak Oct 29 '12 at 13:36
add comment

3 Answers

up vote 10 down vote accepted

Recall the formula $$\cos(2 \theta) = 2 \cos^2(\theta) - 1$$ This gives us $$\cos^2(\theta) = \dfrac{1+\cos(2 \theta)}{2}$$ Plug in $\theta = 2x$, to get what you want.

EDIT The identity $$\cos(2 \theta) = 2 \cos^2(\theta) - 1$$ can be derived from $$\cos(A+B) = \cos(A) \cos(B) - \sin(A) \sin(B)$$ Setting $A = B = \theta$, we get that $$\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) = \cos^2(\theta) - (1-\cos^2(\theta)) = 2 \cos^2(\theta) - 1$$

share|improve this answer
    
Thanks, this is a complete answer, now understand. –  calbertts Oct 28 '12 at 2:23
add comment

It’s just the double-angle formula for the cosine: for any angle $\alpha$, $\cos 2\alpha=\cos^2\alpha-\sin^2\alpha\;,$ and since $\sin^2\alpha=1-\cos^\alpha$, this can also be written $\cos2\alpha=2\cos^2\alpha-1$. Now let $\alpha=2x$: you get $\cos4x=2\cos^22x-1$, so $\cos^22x=\frac12(\cos4x+1)$.

share|improve this answer
    
Thanks by your time, this helped me. –  calbertts Oct 28 '12 at 2:23
add comment

$$\cos(4x) = \cos^2 (2x) - \sin^2 (2x) = 2\cos^2 (2x) - 1$$

share|improve this answer
    
Thanks by your time. –  calbertts Oct 28 '12 at 2:25
add comment

Your Answer

 
discard

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.