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.

I am stuck on trying to prove a trig identity using De Moivre's theorem.

I have to prove, $$\cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta)$$

I am not sure where to even start, I broke the LHS down to $$\cos(3\theta) + i\sin(3\theta)$$

but I have no idea where to go from here, or if this is fully correct.

If I could get some pointers or a simple worked example that I could follow it would be great.

Thanks

share|improve this question

3 Answers 3

up vote 5 down vote accepted

De Moivre's formula reads $$(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)$$ Of course this identity implies the real part should be also equality. That is $$\cos(n\theta)=\Re\{(\cos\theta+i\sin\theta)^n\}$$ Hence we have $$\cos(3\theta)=\Re\{\cos^3\theta+3i\cos^2\theta\sin\theta-3\cos\theta\sin^2\theta-i\sin^3\theta\}=\cos^3\theta-3\cos\theta\sin^2\theta$$

share|improve this answer
    
I have reached the last line in your answer but where do I go from here? –  user2352274 Feb 17 at 15:26
    
@user2352274 Note that $\sin^2+\cos^2=1$ –  Shuchang Feb 17 at 15:29
    
I got it, thanks for all your help. –  user2352274 Feb 17 at 16:00

Hint: Using de Moivre's identity: $$ \cos 3\theta = \mathrm{Re}\left(e^{i3\theta}\right) $$ Now, $e^{i3\theta} = \left(e^{i\theta}\right)^3$ and (by definition of $e^{i\theta}$) $$\left(e^{i\theta}\right)^3 = \left(\cos\theta+i\sin \theta\right)^3 $$ Expand the RHS (it's only a cube, so it's straightforward), and simplify the $i^3$, $i^2$; then take the real part to get $\mathrm{Re}\left(e^{i3\theta}\right)$.

share|improve this answer

Hint: $$\cos(n \phi) = \Re((\cos(\phi) + i\sin (\phi))^n)$$ Then use the Binomial Theorem to expand the expression.

(Where $\Re$ is the real part of the expression.)

share|improve this answer

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.