I need to simplify this expression. I'm not even sure where to start.

$\mathrm{tan}(\mathrm{arccos}(\frac{x}{4}))$

$\mathrm{tan}(\mathrm{arccos}(\frac{x}{4}))$

Leland Morrow
2022-06-29
Answered

I need to simplify this expression. I'm not even sure where to start.

$\mathrm{tan}(\mathrm{arccos}(\frac{x}{4}))$

$\mathrm{tan}(\mathrm{arccos}(\frac{x}{4}))$

You can still ask an expert for help

Jovan Wong

Answered 2022-06-30
Author has **23** answers

Let $y=\mathrm{tan}({\mathrm{cos}}^{-1}\left(\frac{x}{4}\right))\phantom{\rule{thickmathspace}{0ex}},$, Now Put ${\mathrm{cos}}^{-1}\left(\frac{x}{4}\right)=\varphi \phantom{\rule{thickmathspace}{0ex}},$

Then $\frac{x}{4}=\mathrm{cos}\varphi \phantom{\rule{thickmathspace}{0ex}},$, Then $\mathrm{sec}\varphi =\frac{4}{x},$, So using $\mathrm{tan}\varphi =\pm \sqrt{{\mathrm{sec}}^{2}\varphi -1}=\pm \frac{\sqrt{16-{x}^{2}}}{x}$

So We get $y=\mathrm{tan}\varphi =\pm \frac{\sqrt{16-{x}^{2}}}{x}$

Then $\frac{x}{4}=\mathrm{cos}\varphi \phantom{\rule{thickmathspace}{0ex}},$, Then $\mathrm{sec}\varphi =\frac{4}{x},$, So using $\mathrm{tan}\varphi =\pm \sqrt{{\mathrm{sec}}^{2}\varphi -1}=\pm \frac{\sqrt{16-{x}^{2}}}{x}$

So We get $y=\mathrm{tan}\varphi =\pm \frac{\sqrt{16-{x}^{2}}}{x}$

varitero5w

Answered 2022-07-01
Author has **6** answers

Let $\mathrm{arccos}\frac{x}{4}=\theta $

This implies

$\frac{x}{4}=\mathrm{cos}\theta $

Forming a right angled triangle with the above and using pythagoras' theorem, opposite side to angle $\theta $ will be: $\pm \sqrt{16-{x}^{2}}$

Therefore,

$\mathrm{tan}\theta =\mathrm{tan}(\mathrm{arccos}\frac{x}{4})=\frac{\sqrt{16-{x}^{2}}}{x}$

This implies

$\frac{x}{4}=\mathrm{cos}\theta $

Forming a right angled triangle with the above and using pythagoras' theorem, opposite side to angle $\theta $ will be: $\pm \sqrt{16-{x}^{2}}$

Therefore,

$\mathrm{tan}\theta =\mathrm{tan}(\mathrm{arccos}\frac{x}{4})=\frac{\sqrt{16-{x}^{2}}}{x}$

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I know that if I substitute

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