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 = (1)/(cos) \\ \\ sin^2 = 1 - cos^2`
Also because the angle is in quadrant 3, sin must be negative.
Therefore

`sin = - \sqrt{1 - (1)/(1 + tan^2)}`
Subbing in tan = 0.958

`sin \theta = -0.69178`