# How do you evaluate cos^-1(sin(7pi/2))?

Jul 29, 2016

$- 3 \pi$

#### Explanation:

For any a, $\sin a = \cos \left(\frac{\pi}{2} - a\right)$.

Use bijective 1-1 inversion ${f}^{- 1} \left(f \left(x\right)\right) = x$ that returns x.

Here,

${\cos}^{- 1} \left(\sin \left(\frac{7}{2} \pi\right)\right)$

$= {\cos}^{- 1} \left(\cos \left(\frac{\pi}{2} - \frac{7}{2} \pi\right)\right)$.

$= {\cos}^{- 1} \left(\cos \left(- 3 \pi\right)\right)$.

$= - 3 \pi$

Despite that both $\pm 3 \pi$ point to the same direction, the sense of

rotation is opposite, for the negative sign. So, I make this self-

correction to my previous answer from $3 \pi$ to $- 3 \pi$ .

See how it works, in the reverse process.

${\sin}^{- 1} \left(\cos \left(- 3 \pi\right)\right)$

=sin^(-1)sin(pi/2-(-3pi)))

$- {\sin}^{- 1} \left(\sin \left(\frac{7}{2} \pi\right)\right)$

$= \frac{7}{2} \pi$, using ${f}^{- 1} f \left(x\right) = x$

Interestingly, owing to the principal value convention, your

calculator gives the answer as ${180}^{o} = \pi$. For the reversed

calculation, it gives $- {90}^{\odot} = - \frac{\pi}{2}$, and not $\frac{7}{2} \pi$..