# Thread: inverse trig function

1. ## inverse trig function

i'm trying to study for a test and i'm having trouble understanding how to solve these, my textbook doesn't really cover this very well. could someone please help me?

solve for x:

$\arcsin(\frac{x}{2})+\arccos(\frac{-x \sqrt{3}}{2})=\pi$

$\arcsin(\frac{2}{x})-\arccos(\frac{\sqrt{3}}{x})=\arccos(\frac{1}{x})$

2. Hello yaykittyeee
Originally Posted by yaykittyeee
i'm trying to study for a test and i'm having trouble understanding how to solve these, my textbook doesn't really cover this very well. could someone please help me?

solve for x:

$\arcsin(\frac{x}{2})+\arccos(\frac{-x \sqrt{3}}{2})=\pi$

$\arcsin(\frac{2}{x})-\arccos(\frac{\sqrt{3}}{x})=\arccos(\frac{1}{x})$
Inverse trig functions are always a pain. Get rid of them as soon as you can. Here's the first one:

Let $y = \arcsin\left(\frac{x}{2}\right)$. So $\sin y = \frac{x}{2}$

Let $z = \arccos\left(\frac{-x\sqrt3}{2}\right)$. So $\cos z = \frac{-x\sqrt3}{2}$

And the original equation becomes:
$y+z=\pi$

$\Rightarrow z = \pi -y$

$\Rightarrow \cos z = -\frac{x\sqrt3}{2}= \cos(\pi-y) = -\cos y$
But $\cos y = \sqrt{1 - \sin^2 y}=\sqrt{1-\frac{x^2}{4}}$

So $\frac{x\sqrt3}{2}=\sqrt{1-\frac{x^2}{4}}$

Solve this for $x$ by squaring both sides. You get a quadratic, one of whose roots is not valid. The other is the answer to the question (which is $x = 1$).

Number 2 is a bit trickier, but start it in the same way. Let:
$y = \arcsin\left(\frac{2}{x}\right)$ i.e. $\sin y = ...$

$z = \arccos\left(\frac{\sqrt3}{x}\right)$ i.e. $\cos z = ...$

$w = \arccos\left(\frac{1}{x}\right)$ i.e. $\cos w = ...$
Then say:
$y-z=w$
and:
$\frac{1}{x}=\cos w = \cos(y-z)$
$=\cos y \cos z + \sin y \sin z$

$= ...$ (substitute to get everything in terms of $x$)
Then solve the equation in $x$.

Do you want to have another go now?