Question

If sin Θ = negative square root 3 over 2 and π < Θ < 3 pi over 2, what are the values of cos Θ and tan Θ?

cos Θ = negative 1 over 2; tan Θ = square root 3
cos Θ = negative 1 over 2; tan Θ = −1
cos Θ = square root 3 over 4; tan Θ = −2
cos Θ = 1 over 2; tan Θ = square root 3

Answer 1

In what quadrant is the angle located?

Answer 2

\[\sin\theta = \frac{-\sqrt{3}}{2} \implies \sin^2\theta = \frac{3}{4} \implies \sin^2\theta + \cos^2\theta = \frac{3}{4} + \cos^2\theta\]

\[\implies 1 = \frac{3}{4} + \cos^2\theta \implies 1-\frac{3}{4} = \cos^2\theta \implies \cos^2\theta = \frac{1}{4}\]
Since cosine is negative in quadrant III we take negative sqare root
\[\cos\theta = -\sqrt{ \frac{1}{4} } = -\frac{1}{2} \]

Since \(\tan\theta = \sin\theta/\cos\theta\), it is easy to see that
\[ \tan\theta = \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3}\]

Answer 3

Line 2 is from the identity \(\sin^2\theta + \cos^2\theta = 1\)