use de moivre's formula with n= 2 to find trigonometric identities cos2theta and sin 2theta
plz help me.....:(

Question

anonymous · Answer

farily simple

anonymous · Answer

fairly lol

anonymous · Answer

$$(\cos(\theta) + isin(\theta))^n = cos(n \theta) +isin(n \theta ) ) $$

watchmath · Answer

$\cos (2\theta)_1\sin(2\theta)=(\cos \theta +i \sin \theta)^2=(\cos^2\theta-sin^2\theta)+i(2\sin\theta\cos\theta)$

anonymous · Answer

^ that yeh :|

watchmath · Answer

The left most expression should be $\cos(2\theta)+i\sin(2\theta)$

anonymous · Answer

there is actually something to do here yes?  besides using laws of exponents i mean.

anonymous · Answer

oops my apologies  whatchmath wrote it. should should also mumble something about setting real part = real part and imaginary part = imaginary part.

anonymous · Answer

watchmath i mean. yikes it is early

watchmath · Answer

Are you in US too satellite?

anonymous · Answer

yes US

anonymous · Answer

trying to learn latex and practicing here as much as i can