# How do you find the exact value of Sin^-1(cos(pi/4))?

May 11, 2018

$\textrm{A r c} \textrm{\sin} \left(\cos \left(\frac{\pi}{4}\right)\right) = \frac{\pi}{4}$

$\arcsin \left(\cos \left(\frac{\pi}{4}\right)\right) = \frac{\pi}{4} + \pi k , \quad$ integer $k$

#### Explanation:

Oh my, $\frac{\pi}{4}$ again. What a surprise.

Question writers, questions like this are your big chance to get away from the two cliche triangles of trig.

I prefer the notation $\textrm{A r c} \textrm{\sin} \left(x\right)$ for the principal value and $\arcsin \left(x\right)$ for all values. Let's run the problem both ways.

Of course,

$\sin \left(\frac{\pi}{4}\right) = \cos \left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

text{Arc}text{sin}(cos (pi/4) ) = text{Arc}text{sin}( 1/sqrt{2} ) = pi/4

$x = \arcsin \left(\cos \left(\frac{\pi}{4}\right)\right)$ is equivalent to

$\sin \left(x\right) = \cos \left(\frac{\pi}{4}\right)$

I always remember $\cos x = \cos a$ has solutions $x = \pm a + 2 \pi k \quad$ integer $k$.

$\cos \left(\frac{\pi}{2} - x\right) = \cos \left(\frac{\pi}{4}\right)$

$\frac{\pi}{2} - x = \pm \frac{\pi}{4} + 2 \pi k$

$x = \frac{\pi}{2} \pm \frac{\pi}{4} + 2 \pi k$

$x = \left\{\frac{\pi}{4} , \frac{3 \pi}{4}\right\} + 2 \pi k$

We can rewrite that as

$x = \frac{\pi}{4} + \pi k$

$\arcsin \left(\cos \left(\frac{\pi}{4}\right)\right) = \frac{\pi}{4} + \pi k , \quad$ integer $k$