Question

Find sin θ if θ is in Quadrant III and tan θ = . 0.958

Answer 1

Use the following identities:
$sec^2 = 1 + tan^2 \\ \\ sec = \frac{1}{cos} \\ \\ sin^2 = 1 - cos^2$
Also because the angle is in quadrant 3, sin must be negative.
Therefore
$sin = - \sqrt{1 - \frac{1}{1 + tan^2}}$
Subbing in tan = 0.958
$sin \theta = -0.69178$