# If cos(x) = (1/4), where x lies in quadrant 4, how do you find sin(x - pi/6)?

Jun 17, 2016

$- \frac{3 \sqrt{5} + 1}{8}$

#### Explanation:

so you can find
$\sin x = - \sqrt{1 - {\cos}^{2} x}$

$= - \sqrt{1 - \frac{1}{16}}$

$= - \sqrt{\frac{15}{16}}$

$= - \frac{\sqrt{15}}{4}$

Then
$\sin \left(x - \frac{\pi}{6}\right) = \sin x \cos \left(\frac{\pi}{6}\right) - \cos x \sin \left(\frac{\pi}{6}\right)$

$\sin x \cdot \frac{\sqrt{3}}{2} - \cos x \cdot \frac{1}{2}$

and, by substituting sinx and cosx values, you have:

$- \frac{\sqrt{15}}{4} \cdot \frac{\sqrt{3}}{2} - \frac{1}{4} \cdot \frac{1}{2}$

$- \frac{\sqrt{45}}{8} - \frac{1}{8}$

$- \frac{3 \sqrt{5} + 1}{8}$