# What is the value of sin -1 (cos x)?

Oct 2, 2015

$\arcsin \left(\cos \left(x\right)\right) = x + \frac{\pi}{2}$

#### Explanation:

Assuming you mistyped and meant ${\sin}^{- 1} \left(\cos \left(x\right)\right)$ or simply $\arcsin \left(\cos \left(x\right)\right)$, we can easily solve this by putting it on terms of the sine function.

We know that the cosine function, is nothing more than the sine $\frac{\pi}{2}$ radians out of phase, as proved below:

$\cos \left(\theta - \frac{\pi}{2}\right) = \cos \left(\theta\right) \cos \left(- \frac{\pi}{2}\right) - \sin \left(\theta\right) \sin \left(- \frac{\pi}{2}\right)$
$\cos \left(\theta - \frac{\pi}{2}\right) = \cos \left(\theta\right) \cdot 0 - \left(- \sin \left(\theta\right) \sin \left(\frac{\pi}{2}\right)\right)$
$\cos \left(\theta - \frac{\pi}{2}\right) = \sin \left(\theta\right) \cdot 1 = \sin \left(\theta\right)$

So we can say that the sine function, 90 degrees ahead, is the cosine function.

$\arcsin \left(\cos \left(x\right)\right) = \arcsin \left(\sin \left(x + \frac{\pi}{2}\right)\right)$

Using the property of inverse functions that ${f}^{- 1} \left(f \left(x\right)\right) = x$, we have

$\arcsin \left(\cos \left(x\right)\right) = x + \frac{\pi}{2}$

If you must use degrees, just convert those $\frac{\pi}{2}$ radians to 90º degrees.