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!!!

  • $\begingroup$ I went further and got $e^(i2\theta + i\pi/2) = e^(2i\theta).e^(i\pi/2) = (\cos(2\theta)+i\sin(2\theta))(\cos(\pi/2)+i\sin(\pi/2))=(\cos(2\theta)+i\sin(2\theta))(i) = i\cos(2\theta)-sin(2\theta) $ $\endgroup$
    – Leo
    Nov 11, 2015 at 2:10
  • 1
    $\begingroup$ Hint: $ e^{i(2\theta+\pi/2)} = i(e^{i\theta})^2 $ $\endgroup$ Nov 11, 2015 at 2:11
  • $\begingroup$ Ah i see now :) De Moivre's theorm to remove $2\theta$ from the inside and then use that 2 as a binomial expansion! Thanks - I won't forget that small step ever again. $\endgroup$
    – Leo
    Nov 11, 2015 at 2:29

1 Answer 1


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, $\cos(2\theta+\pi/2) =-2\cos\theta\sin\theta$.

  • $\begingroup$ Ah I see!!! So you're able to take a $2\theta$ out by literally bringing it out the brackets i.e. De Moivre's theorem; and then like in my comment above evaluate $e^(i\pi/2)$ and multiply it by the binomial expansion of the polar form! I was running into difficulty because I couldn't figure out how to remove the $2\theta$ but obviously I was forgetting I could use De Moivre. Thank you!!! $\endgroup$
    – Leo
    Nov 11, 2015 at 2:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.