Ethen Wong

Answered

2022-01-28

Given $\mathrm{cos}\left(t\right)=\left(\frac{1}{2}\right)({e}^{it}+{e}^{-it})$ 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

$z=x\pm \sqrt{{x}^{2}-1}$

Im

The hint says to let

Im

Answer & Explanation

Ydaxq

Expert

2022-01-29Added 12 answers

You've done the hard part. Choosing the positive square root, you have $it=\mathrm{log}(x+\sqrt{{x}^{2}-1})$ (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,

$-it=-\mathrm{log}(x+\sqrt{{x}^{2}-1})=\mathrm{log}(x-\sqrt{{x}^{2}-1})$

Note that

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