# Thread: Compute inverse trig functions

1. ## Compute inverse trig functions

This is in the review section of my calculus books so I think it should be solvable without calc but I'm not sure. I dont even know how to start this problem.

Using only arctangent and other commonly available functions compute the arcsine and the arcosine.

Any and all help is greatly appreciated!

2. Originally Posted by Alice96
This is in the review section of my calculus books so I think it should be solvable without calc but I'm not sure. I dont even know how to start this problem.

Using only arctangent and other commonly available functions compute the arcsine and the arcosine.

Any and all help is greatly appreciated!
You want to draw out the triangle - it will help.

$\displaystyle \arcsin(x)$ is the angle $\displaystyle \theta$ such that $\displaystyle \sin(\theta)=\frac{x}{1}$. So if the opposite side of $\displaystyle \theta$ is $\displaystyle x$ and the hypotenuse is $\displaystyle 1$, what is the adjacent side to $\displaystyle \theta$?

Well, using the Pythagorean Theorem, we know that it equals $\displaystyle \sqrt{1-x^2}$. Now that we know the opposite and adjacent sides to $\displaystyle \theta$, we can find $\displaystyle \tan(\theta)$, which equals $\displaystyle \frac{x}{\sqrt{1-x^2}}$.

So we know that:

$\displaystyle \arcsin(x)=\theta$
$\displaystyle \tan(\theta)=\frac{x}{\sqrt{1-x^2}}\implies\theta=\arctan\left(\frac{x}{\sqrt{1-x^2}}\right)$

Thus $\displaystyle \arcsin(x)=\arctan\left(\frac{x}{\sqrt{1-x^2}}\right)$

Using an analogous method to this one, see if you can compute $\displaystyle \arccos(x)$.

3. I got arccos(x) = arcsin(sqrt(1-x^2)).
I'm pretty sure this is right as I've checked it on my calculator.
Yes it is, though if your professor doesn't want it in terms of $\displaystyle \arcsin$ then it would be:
$\displaystyle \arccos(x)=\arctan\left(\frac{\sqrt{1-x^2}}{x}\right)$