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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!!!

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  • $\begingroup$ 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 '15 at 2:10
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    $\begingroup$ Hint: $ e^{i(2\theta+\pi/2)} = i(e^{i\theta})^2 $ $\endgroup$ – stochasticboy321 Nov 11 '15 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 '15 at 2:29
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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, $\cos(2\theta+\pi/2) =-2\cos\theta\sin\theta$.

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  • $\begingroup$ 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 '15 at 2:26

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