I've been suck on a question for around 2 hours now that reads: Using the exponential form of a complex number and De Moivre's theorem, show that:
$$ cos(2\theta + \pi/2) = -2cos\theta sin\theta $$
I've never tried De Moiver's theorem using exponential form as polar form was all that was ever needed i.e:
$$ \cos(n\theta)+i\sin(n\theta)=(\cos\theta+i\sin\theta)^n $$
and for the opposite direction using:
$$ (2cos\theta)^n = 2^n\cos^n\theta =(z+1/z)^n $$
and $(z-1/z)$ for the $2i\sin\theta$ expansion.
Needless to say, I haven't learn this with an additional $+ \pi/2 $
Here is my attempt with the LHS:
$ \cos(2\theta+\pi/2)+ i\sin(2\theta+\pi/2) = e^i(2\theta+\pi/2)$ (The (2\theta+\pi/2) should all be multiplied by i as the exponent.
After I got that result I didn't know what to do after… I'm not asking for the answer but any help would be appreciated!
In the end, I tried to evaluate the RHS and got: $ -2\cos\theta\sin\theta = -sin(2\theta) $, but still was not able to evaluate the LHS to give this result.
I am easily able to evaluate the LHS using trig identity: $\cos(2\theta+\pi/2)=\cos(2\theta)\cos(\pi/2)-\sin(2\theta)\sin(\pi/2)=0-\sin(2\theta)=-2\cos\theta\sin\theta$
Thanks in advance!!!
Best Answer
As you noted, $\cos(2\theta+\pi/2) = \operatorname{Re}(e^{i(2\theta+\pi/2)})$.
$$e^{i(2\theta+\pi/2)}=(e^{i\theta})^2e^{i\pi/2}=(\cos\theta+i\sin\theta)^2(i)$$ $$=i(\cos^2\theta+2i\cos\theta\sin\theta-\sin^2\theta)$$ $$=-2\cos\theta\sin\theta+i\cos^2\theta\sin^2\theta$$
Taking the real part, $\cos(2\theta+\pi/2) =-2\cos\theta\sin\theta$.