Ethen Wong

2022-01-28

Given $\mathrm{cos}\left(t\right)=\left(\frac{1}{2}\right)\left({e}^{it}+{e}^{-it}\right)$ solve for ${\mathrm{cos}}^{-1}\left(x\right)$
The hint says to let $x=\mathrm{cos}\left(t\right)$ and $z={e}^{it}$. So I started first by substituting:
$x=\frac{12}{z+\frac{1}{z}}$ --> multiply both sides by 2z
$2xz={z}^{2}+1$
${z}^{2}-2zx+1=0$ --> quadratic formula, solve for z
$z=x±\sqrt{{x}^{2}-1}$
Im

Ydaxq

Expert

You've done the hard part. Choosing the positive square root, you have $it=\mathrm{log}\left(x+\sqrt{{x}^{2}-1}\right)$ (where $\mathrm{log}=\mathrm{ln}$). It's easy enough to solve for t by multiplying by -i.
Note that $x-\sqrt{{x}^{2}-1}=\frac{1}{x+\sqrt{{x}^{2}-1}}$,so indeed,
$-it=-\mathrm{log}\left(x+\sqrt{{x}^{2}-1}\right)=\mathrm{log}\left(x-\sqrt{{x}^{2}-1}\right)$

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