aligass2004yi

2022-06-25

If $\mathrm{tan}\left(22{\frac{1}{2}}^{\circ }\right)=\sqrt{2}-1$, then prove that$\mathrm{tan}\left(11{\frac{1}{4}}^{\circ }\right)=\sqrt{4+2\sqrt{2}}-\sqrt{2}-1$

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Eli Shaffer

Expert

$\mathrm{tan}\left(2x\right)=\frac{\mathrm{sin}\left(2x\right)}{\mathrm{cos}\left(2x\right)}=\frac{2\mathrm{sin}x\mathrm{cos}x}{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x}=\frac{2\mathrm{tan}x}{1-{\mathrm{tan}}^{2}x}$

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Kyla Ayers

Expert

Hint:
$\mathrm{tan}2\alpha =\frac{2\mathrm{tan}\alpha }{1-{\mathrm{tan}}^{2}\alpha }$
Let $\alpha =11\frac{1}{4}$ and $\mathrm{tan}\alpha =x$, then
$\sqrt{2}-1=\frac{2x}{1-{x}^{2}}$
$x=...$

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