Compute inverse trig functions

**Alice96** · September 20th 2009, 03:06 PM

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!

**redsoxfan325** · September 20th 2009, 03:17 PM

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.

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

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

So we know that:

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

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

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

**Alice96** · September 20th 2009, 04:20 PM

I got arccos(x) = arcsin(sqrt(1-x^2)).
I'm pretty sure this is right as I've checked it on my calculator.
Thanks for your help!

**redsoxfan325** · September 20th 2009, 04:24 PM

Originally Posted by Alice96

I got arccos(x) = arcsin(sqrt(1-x^2)).
I'm pretty sure this is right as I've checked it on my calculator.
Thanks for your help!

Yes it is, though if your professor doesn't want it in terms of $\arcsin$ then it would be:

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

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