Inverse trigonometric equation

**Hardwork** · October 17th 2010, 12:46 PM

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

**Also sprach Zarathustra** · October 17th 2010, 12:49 PM

Just one.

**Hardwork** · October 17th 2010, 12:50 PM

Originally Posted by Also sprach Zarathustra

Just one.

How do you show that?

**General** · October 17th 2010, 01:17 PM

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.

**Hardwork** · October 17th 2010, 01:39 PM

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 $x$ . How do we go about that?

**skeeter** · October 17th 2010, 02:22 PM

Originally Posted by Hardwork

Ok. I know that much. But let's say that we want to find $x$ . How do we go about that?

it's just a matter of being familiar with both functions.

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

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

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

if you insist on doing it algebraically ...

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

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

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

$x^2 = 1 - x^2$

$2x^2 - 1 = 0$

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

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

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

and

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

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

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