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Background: I'm studying roots of complex variables (i.e. not homework!), and going through a worked problem from Schaum's Outlines on Complex Variables.

In a worked problem, the following equation is presented and assumed the reader knows trig well enough: $$\cos\frac\theta2 = \pm\sqrt{\frac{(1 + \cos \theta)}{2}}$$

Can someone show me (or just hint) why this equation is true please?

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You know $\cos 2\alpha = 2\cos^2 \alpha-1$, right? Now just put $\alpha = \theta/2$.

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  • $\begingroup$ Thanks for the answer. Even from your hint and Martin's link to the half angle formulas, it still took a bit of headscratching. I thought we might be going through the quadratic formula somewhere, but it was just some basic algebra. $\endgroup$ – PeteUK May 25 '12 at 13:36
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In the context of complex numbers, use Euler's identity $e^{ix} = \cos(x) + i \sin(x)$ to write $$\left(\cos\frac{\theta}{2}\right)^2 = \left(\frac{e^{i\theta/2} + e^{-i\theta/2}}{2}\right)^2 = \frac{e^{i\theta} + e^{-i\theta} + 2}{4} = \frac{1 + \cos \theta }{2} \Rightarrow \cos\frac\theta2 = \pm\sqrt{\frac{1 + \cos \theta}{2}}$$

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    $\begingroup$ Having only learnt about cosine being expressed that way last week, following this was very helpful (and cool!). $\endgroup$ – PeteUK May 25 '12 at 13:34
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    $\begingroup$ If you're familiar with the complex exponential definition of sine and cosine, it will usually be the easiest way to prove the various elementary trig identities. $\endgroup$ – asmeurer Oct 28 '12 at 7:08
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Despite all the proofs already mentioned, drawing a picture

enter image description here

was the method, that convinced me at the first time:

$\color{blue}{\cos(\theta)^2}$ is a squeezed (by $2$ along $x$ and $y$ axis) and shifted (by $+\frac{1}{2}$ along the $y$ axis) version of $\color{green}{\cos(\theta)}$.

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    $\begingroup$ +1. This is really the kind of "proof" those of us who lack strong geometric intuition wish we were capable of. Well done! $\endgroup$ – Mathemagician1234 May 25 '12 at 14:57
  • $\begingroup$ @Mathemagician1234 Thanks $\endgroup$ – draks ... May 25 '12 at 14:59
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\begin{equation*} \begin{split} \cos 2\alpha &= \cos^{2} \alpha-\sin^{2}\alpha\\ \Rightarrow \cos\alpha &= \cos^{2}\frac{\alpha}{2}-\sin^{2}\frac{\alpha}{2}\\ &= 2\cos^{2}\frac{\alpha}{2}-1 (\because \sin^{2}\frac{\alpha}{2}=1-\cos^{2}\frac{\alpha}{2})\\ \therefore \cos^{2}\frac{\alpha}{2} &= \frac{1+\cos \alpha}{2}\\ \Rightarrow \cos\frac{\alpha}{2} &= \pm \sqrt{\frac{1+\cos \alpha}{2}}. \end{split} \end{equation*}

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