If sin theta = 1/4, theta in quadrant 2, how do you find the exact value of sin(theta - pi/3)?

Jun 18, 2018

$\sin \left(t - \frac{\pi}{3}\right) = \frac{1 + 3 \sqrt{5}}{8}$

Explanation:

$\sin t = \frac{1}{4}$. Find cos t
${\cos}^{2} t = 1 - {\sin}^{2} t = 1 - \frac{1}{16} = \frac{15}{16}$
$\cos t = - \frac{\sqrt{15}}{4}$ (because t lies in Quadrant 2)
From trig identity, we get:
$\sin \left(t - \frac{\pi}{3}\right) = \sin t . \cos \left(\frac{\pi}{3}\right) - \sin \left(\frac{\pi}{3}\right) . \cos t$ (1)
Note.
$\sin t = \frac{1}{4}$ --> $\cos t = - \frac{\sqrt{15}}{4}$ -->
$\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$ --> $\sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$
Equation (1) becomes:
$\sin \left(t - \frac{\pi}{3}\right) = \frac{1}{8} + \frac{\sqrt{3.} \sqrt{15}}{8} = \frac{1 + 3 \sqrt{5}}{8}$
Check by calculator.
$\sin t = 0.25$ --> $t = {162}^{\circ} 52$ --> $\left(t - 60\right) = {105}^{\circ} 22$
$\sin \left(t - 60\right) = 0.96$
$\frac{1 + 3 \sqrt{5}}{8} = 0.96$. Proved.