# Thread: Proving trig identity with exponential form of complex numbers

1. ## Proving trig identity with exponential form of complex numbers

Express $\displaystyle \cos\theta$ and $\displaystyle \sin\theta$ in terms of $\displaystyle e^{i\theta}$ and $\displaystyle e^{-i\theta}$.
Use your results to prove that:
$\displaystyle 16\cos^3\theta\sin^2\theta\equiv 2\cos\theta-\cos 3\theta-\cos 5\theta$

I have done:
$\displaystyle \cos\theta=Re(e^{i\theta})$ and $\displaystyle Re(e^{-i\theta})$
$\displaystyle \sin\theta=Im(e^{i\theta})$ and $\displaystyle -Im(e^{-i\theta})$

The proving part:
$\displaystyle LHS\equiv 16[Re(e^{i\theta})]^3[Im(e^{i\theta})]^2$
I'm stuck with the $\displaystyle [Im(e^{i\theta})]^2$ part. The power of 2 indicates its real, but I haven't figured out how to express it in terms of $\displaystyle Re(e^{i\theta})$.
Thanks!

2. You should know that

$\displaystyle e^{i\theta} = \cos{\theta} + i\sin{\theta}$ and $\displaystyle e^{-i\theta} = \cos{\theta} - i\sin{\theta}$.

Then $\displaystyle e^{i\theta} + e^{-i\theta} = \cos{\theta} + i\sin{\theta} + \cos{\theta} - i\sin{\theta}$

$\displaystyle = 2\cos{\theta}$.

Thus $\displaystyle \cos{\theta} = \frac{1}{2}(e^{i\theta} + e^{-i\theta})$.

$\displaystyle e^{i\theta} - e^{-i\theta} = \cos{\theta} + i\sin{\theta} - (\cos{\theta} - i\sin{\theta})$

$\displaystyle = 2i\sin{\theta}$.

Thus $\displaystyle \sin{\theta} = \frac{1}{2i}(e^{i\theta} - e^{-i\theta})$

$\displaystyle = -\frac{i}{2}(e^{i\theta} - e^{-i\theta})$.

Now try to go from here...