Trig identities with complex numbers

[Darya]

Recently I've watched a video by 3Blue1Brown, where he (really intuitively) explains complex numbers as rotations.
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He then explained how not to memorize the formula cos(a+b) and visualize it as rotating in 2 angles instead. cos(a+b)=(cos(a)+isin(a))(cos(b)+isin(b)). The real part is cos(a)cos(b)-sin(a)sin(b).
Naturally, I then wondered if I can get all the other trig identities this way.
However, as sin(a+b), it's not as trivial. I tried sin(a+b)=(sin(a)+icos(a))(sin(b)+icos(b)) but it's not correct at all. What's wrong with my logic?

Edit: the imaginary part in cos(a+b) turns out to be i(sinacosb+cosasinb) - exactly what am solving for. Why's that?

 

[topsquark]

[math]e^{ in \theta } = ( cos( \theta ) + i ~ sin( \theta ) )^n = cos( n \theta ) + i ~ sin( n \theta )[/math]so the real part will deal with cosine function and the imaginary part will deal with the sine function.

-Dan

 

[Steven G]

Using the equation that Dan gave above you can find for example sin(3θ) and get the cos(3θ) for free!