# How do you evaluate cos(arcsin (1/4))?

Sep 10, 2015

$\frac{\sqrt{15}}{4} \approx 0.968$

#### Explanation:

By the fact that ${\cos}^{2} \left(\theta\right) + {\sin}^{2} \left(\theta\right) = 1$ for all $\theta$, when we know the value of $\sin \left(\theta\right)$, there are two possible corresponding values for $\cos \left(\theta\right)$, namely $\cos \left(\theta\right) = \pm \sqrt{1 - {\sin}^{2} \left(\theta\right)}$.

Next note that $\arcsin \left(x\right)$ always gives an answer between $- \frac{\pi}{2}$ and $\frac{\pi}{2}$ radians, where the cosine function is positive. Hence, $\cos \left(\arcsin \left(\frac{1}{4}\right)\right) \ge q 0$.

Thus, $\cos \left(\arcsin \left(\frac{1}{4}\right)\right) = \sqrt{1 - {\sin}^{2} \left(\arcsin \left(\frac{1}{4}\right)\right)}$

$= \sqrt{1 - {\left(\frac{1}{4}\right)}^{2}} = \sqrt{1 - \frac{1}{16}} = \sqrt{\frac{15}{16}} = \frac{\sqrt{15}}{4} \approx 0.968 .$

It's also possible to solve this problem by drawing a right triangle, labeling one of the non-right angles as $\arcsin \left(\frac{1}{4}\right)$, using the Pythagorean Theorem and SOH, CAH, TOA to label and find possible side lengths and ultimately the final answer.

Jul 19, 2016

Less formal style of solution

$\frac{\sqrt{15}}{4} \approx 0.9682$ to 4 decimal places

#### Explanation:

arcsin of some value gives you the angle that was used derive the that value of the sine

$\textcolor{b r o w n}{\text{sin, cos and tangent are just another way of defining ratios}}$

The value given can be used in conjunction with the properties of sine to determine a related triangle.

From this and using Pythagoras we can determine the length of the adjacent.

${x}^{2} + {1}^{2} = {4}^{2}$

$\implies x = \sqrt{15}$

so $\cos \left(\theta\right) = \frac{x}{4} = \frac{\sqrt{15}}{4} \approx 0.9682$ to 4 decimal places