Answer
- $\frac{\sqrt {15}}{4} $
Work Step by Step
Given-
$\sin A$ = $\frac{1}{4}$, $A$ terminates in QII
From first Pythagorean identity-
$\cos A$ = ± $\sqrt {1 - \sin^{2} A}$
As $A$ terminates in QII, therefore
$\cos A$ = - $\sqrt {1 - \sin^{2} A}$
= - $\sqrt {1 - (\frac{1}{4})^{2} }$
= - $\sqrt {1 - \frac{1}{16} }$
= - $\sqrt {\frac{16 - 1}{16} }$
= - $\sqrt {\frac{15}{16} }$
= - $\frac{\sqrt {15}}{4} $
