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Express $e^{-\frac{1}{2}i\theta} - e^{\frac{1}{2}i\theta}$ in trigonometric form, and show that $(1-e^{i\theta})^2 = -4e^{i\theta}\sin^2\left(\frac{1}{2}\theta\right)$

By inspection, I noticed that the former expression is identical to $-2i\sin\left(\frac{1}{2}\theta\right)$. I cannot for the life of me figure out the second part. My first attempt involved expanding the L.H.S to achieve $$(1-e^{i\theta})^2 = e^{2i\theta} - 2e^{i\theta} + 1$$ and attempted to multiply the R.H.S of the expanded form by $e^{-\frac{1}{2}\theta}$ to try and use the identity that was achieved in the former part of the question, but could not immediately notice anything of use.

Any help or hints would be greatly appreciated.

lulu
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  • Hint: Compare the square of what you got from the first expression to the RHS you are aiming for. What factor is missing? – DanLewis3264 Oct 13 '21 at 18:43

1 Answers1

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Just to clarify, using the fact that $$e^{ix} = \cos(x) + i\sin(x)$$ and expanding everything out gives us the first identity:

\begin{align*} e^{-\frac{1}{2}i\theta} - e^{\frac{1}{2}i\theta} &= \cos\left(-\frac{\theta}{2}\right) + i\sin\left(-\frac{\theta}{2}\right) - \cos\left(\frac{\theta}{2}\right) - i\sin\left(\frac{\theta}{2}\right) = -2i\sin\left(\frac{\theta}{2}\right). \end{align*}

Next, notice that $$-4\sin^{2}\left(-\frac{\theta}{2}\right) = \left(e^{-\frac{1}{2}i\theta} - e^{\frac{1}{2}i\theta}\right)^{2} = e^{i\theta} - 2 + e^{-i\theta}.$$

Can you take it from there?

DMcMor
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