# How do you evaluate cos(sin^-1((sqrt3/2)) without a calculator?

Feb 4, 2017

$\cos \left({\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right)\right) = \frac{1}{2}$

#### Explanation:

Consider an equilateral triangle with sides of length $2$, bisected to form two right angled triangles:

Remembering that $\sin = \text{opposite"/"hypotenuse}$, we can see that:

$\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

Since $\frac{\sqrt{3}}{2} > 0$ and $\frac{\pi}{3}$ is in Q1, we can deduce:

${\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$

From the same diagram, remembering $\cos = \text{adjacent"/"hypotenuse}$, we can see that:

$\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$

So:

$\cos \left({\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right)\right) = \cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$

Feb 4, 2017

$\cos \left({\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right)\right) = \frac{1}{2}$

#### Explanation:

Starting from:

${\cos}^{2} \theta + {\sin}^{2} \theta = 1$

Subtract ${\sin}^{2} \theta$ from both sides to get:

${\cos}^{2} \theta = 1 - {\sin}^{2} \theta$

Take the square root to find:

$\cos \theta = \pm \sqrt{1 - {\sin}^{2} \theta}$

If $\theta = {\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right)$ then $\sin \left(\theta\right) = \frac{\sqrt{3}}{2}$ and we find:

$\cos \theta = \pm \sqrt{1 - {\left(\frac{\sqrt{3}}{2}\right)}^{2}} = \pm \sqrt{1 - \frac{3}{4}} = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2}$

Further note that $\frac{\sqrt{3}}{2} > 0$, so ${\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right)$ must be in Q1. Hence $\cos \theta > 0$ and we need to choose the positive square root.

So:

$\cos \left({\sin}^{- 1} \left(\frac{\sqrt{3}}{2}\right)\right) = \frac{1}{2}$