Simplify Inverse Trigonometry Expressions

Question

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

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

Answer 1

HINT...draw a right-angled triangle with adjacent side $x$ and hypotenuse $4$

Answer 2

Let $\displaystyle y=\tan\left(\cos^{-1}\left(\frac{x}{4}\right)\right)\;,$ Now Put $\displaystyle \cos^{-1}\left(\frac{x}{4}\right)=\phi\;,$

Then $\displaystyle \frac{x}{4} = \cos \phi\;,$ Then $\displaystyle \sec \phi = \frac{4}{x},$ So using $\displaystyle \tan \phi = \pm \sqrt{\sec^2 \phi -1} = \pm \frac{\sqrt{16-x^2}}{x}$

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

Answer 3

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

This implies $$\frac{x}{4} = \cos \theta $$

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,

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