What is the value of the following expression:

$\sqrt{3}{\mathrm{cot}15}^{\circ}$

Nann
2021-09-13
Answered

What is the value of the following expression:

$\sqrt{3}{\mathrm{cot}15}^{\circ}$

You can still ask an expert for help

Alannej

Answered 2021-09-14
Author has **104** answers

First, find $\mathrm{tan}15}^{\circ$

$\mathrm{tan}(A-B)=\frac{\mathrm{tan}A-\mathrm{tan}B}{1+\mathrm{tan}A\mathrm{tan}B}$

${\mathrm{tan}15}^{\circ}=\mathrm{tan}({60}^{\circ}-{45}^{\circ})$

$\mathrm{tan}15}^{\circ}=\frac{{\mathrm{tan}60}^{\circ}-{\mathrm{tan}45}^{\circ}}{1+{\mathrm{tan}60}^{\circ}{\mathrm{tan}45}^{\circ}$

$\mathrm{tan}15}^{\circ}=\frac{\sqrt{3}-1}{1+\left(\sqrt{3}\right)\left(1\right)$

$=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

Then,

$\mathrm{cot}15}^{\circ}=\frac{\sqrt{3}+1}{\sqrt{3}-1$

$\mathrm{cot}15}^{\circ}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1$

$\mathrm{cot}15}^{\circ}=\frac{3+2\sqrt{3}+1}{3-1$

$\mathrm{cot}15}^{\circ}=\frac{2\sqrt{3}+4}{2$

$c{\textstyle \phantom{\rule{1em}{0ex}}}\text{or}{\textstyle \phantom{\rule{1em}{0ex}}}{15}^{\circ}=\sqrt{3}+2$

Substitute to given expression:

$\sqrt{3}{\mathrm{cot}15}^{\circ}=\sqrt{3}(\sqrt{3}+2)$

$\sqrt{3}{\mathrm{cot}15}^{\circ}=3+2\sqrt{2}$

Then,

Substitute to given expression:

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By doing some right triangle gymnastics, we can derive things like

$\mathrm{cos}(\mathrm{arctan}x)=\frac{1}{\sqrt{1+{x}^{2}}}$, for $x>0$ $\mathrm{cos}(\mathrm{arcsin}x)=\sqrt{1-{x}^{2}}$

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What about $\mathrm{arctan}\mathrm{cos}(x)$, $\mathrm{arcsin}(\mathrm{tan}x)$, etc?

$\mathrm{cos}(\mathrm{arctan}x)=\frac{1}{\sqrt{1+{x}^{2}}}$, for $x>0$ $\mathrm{cos}(\mathrm{arcsin}x)=\sqrt{1-{x}^{2}}$

$\mathrm{cos}(\mathrm{arcsin}x)=\sqrt{1-{x}^{2}}$

$\mathrm{tan}(\mathrm{arcsin}x)=\frac{x}{\sqrt{1-{x}^{2}}}$

What about $\mathrm{arctan}\mathrm{cos}(x)$, $\mathrm{arcsin}(\mathrm{tan}x)$, etc?