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

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asked Aug 3, 2019 42.2k views

1 Answer

Chunk · Answer 1 · 2019-08-09T21:34:22+0000

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

$sin \theta = -0.69178$

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

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