Proving trig identity with exponential form of complex numbers

**arze** · September 28th 2010, 09:03 PM

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

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

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

**Prove It** · September 28th 2010, 09:07 PM

You should know that

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

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

$= 2\cos{\theta}$ .

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

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

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

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

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

Now try to go from here...

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