How do you find two exact values of sin((cos^-1)(sqrt5/6))?

Dec 12, 2015

The two exact values of $\sin \left({\cos}^{- 1} \left(\frac{\sqrt{5}}{6}\right)\right)$ are
$\frac{\sqrt{31}}{6}$ and $- \frac{\sqrt{31}}{6}$

Explanation:

Let's consider the right triangle with angle $x$ where $\cos \left(x\right) = \frac{\sqrt{5}}{6}$

Then ${\cos}^{- 1} \left(\frac{\sqrt{5}}{6}\right) = x$ and so
$\sin \left({\cos}^{- 1} \left(\frac{\sqrt{5}}{6}\right)\right) = \sin \left(x\right) = \frac{y}{6}$

${\left(\sqrt{5}\right)}^{2} + {y}^{2} = {6}^{2} \implies y = \sqrt{36 - 5} = \sqrt{31}$

So we have one value of $\sin \left({\cos}^{- 1} \left(\frac{\sqrt{5}}{6}\right)\right)$ as $\frac{\sqrt{31}}{6}$

But assuming we are looking at ${\cos}^{- 1}$ as multivalued, looking at it geometrically loses one solution, as we only look at one possibility for ${\cos}^{- 1} \left(\frac{\sqrt{5}}{6}\right)$. To get the other, note that as as the cosine function is even,

$\cos \left(- x\right) = \cos \left(x\right) = \frac{\sqrt{5}}{6}$

So we need to consider $- x$ as our other value. Then, as the sine function is odd,

$\sin \left(- x\right) = - \sin \left(x\right) = - \frac{\sqrt{31}}{6}$

Thus the two exact values of $\sin \left({\cos}^{- 1} \left(\frac{\sqrt{5}}{6}\right)\right)$ are $\frac{\sqrt{31}}{6}$ and $- \frac{\sqrt{31}}{6}$