1. ## Inverse trigonometric equation

How many $\displaystyle \displaystyle x\in\mathbb{R}$ exist such that $\displaystyle \displaystyle \arcsin{x} = \arccos{x}$?

2. Just one.

3. Originally Posted by Also sprach Zarathustra
Just one.
How do you show that?

4. There is a simple way:

Draw the graphs for the arcsin(x) & arccos(x),
The x-coordinate of the point of intersection is the solution.

5. Originally Posted by General
Draw the graphs for the arcsin(x) & arccos(x),
The x-coordinate of the point of intersection is the solution.
Ok. I know that much. But let's say that we want to find $\displaystyle x$. How do we go about that?

6. Originally Posted by Hardwork
Ok. I know that much. But let's say that we want to find $\displaystyle x$. How do we go about that?
it's just a matter of being familiar with both functions.

$\displaystyle y = \arcsin(x)$ has range $\displaystyle -\frac{\pi}{2} \le y \le \frac{\pi}{2}$

$\displaystyle y = \arccos(x)$ has range $\displaystyle 0 \le y \le \pi$

the ranges intersect in quad I ... the angle in quad I that has the same sine and cosine value is $\displaystyle y = \frac{\pi}{4}$

if you insist on doing it algebraically ...

$\displaystyle \arcsin(x) = \arccos(x)$

$\displaystyle \sin[\arcsin(x)] = \sin[\arccos(x)]$

$\displaystyle x = \sqrt{1 - x^2}$

$\displaystyle x^2 = 1 - x^2$

$\displaystyle 2x^2 - 1 = 0$

$\displaystyle (\sqrt{2} x - 1)(\sqrt{2} x + 1) = 0$

$\displaystyle x = \frac{1}{\sqrt{2}}$ ... the negative solution is invalid since

$\displaystyle y = \arcsin\left(-\frac{1}{\sqrt{2}}\right) = -\frac{\pi}{4}$

and

$\displaystyle y = \arccos\left(-\frac{1}{\sqrt{2}}\right) = \frac{3\pi}{4}$

$\displaystyle \arcsin\left(\frac{1}{\sqrt{2}}\right) = \arccos\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$