Answer
$sin~B = -\frac{\sqrt{7}}{4}$
$cos~B = \frac{3}{4}$
Work Step by Step
If $540^{\circ} \lt 2B \lt 720^{\circ},~~$ then $~~270^{\circ} \lt B \lt 360^{\circ},~~$ so the angle $B$ is in quadrant IV.
Since $B$ is in quadrant IV, $sin~B$ is negative.
We can find the value of $sin~B$:
$cos~2B = 1-2~sin^2~B$
$sin^2~B = \frac{1-cos~2B}{2}$
$sin ~B = -\sqrt{\frac{1-cos~2B}{2}}$
$sin~B = -\sqrt{\frac{1-(\frac{1}{8})}{2}}$
$sin~B = -\sqrt{\frac{(\frac{7}{8})}{2}}$
$sin~B = -\sqrt{\frac{7}{16}}$
$sin~B = -\frac{\sqrt{7}}{4}$
Since $B$ is in quadrant IV, $cos~B$ is positive.
We can find the value of $cos~B$:
$cos~2B = 2~cos^2~B-1$
$cos^2~B = \frac{1+cos~2B}{2}$
$cos~B = \sqrt{\frac{1+cos~2B}{2}}$
$cos~B = \sqrt{\frac{1+(\frac{1}{8})}{2}}$
$cos~B = \sqrt{\frac{(\frac{9}{8})}{2}}$
$cos~B = \sqrt{\frac{9}{16}}$
$cos~B = \frac{3}{4}$
