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?

Answer 1

In the attachments I have answered this problem.

Since the angle lies in third quadrant,
Cos theta is negative and tan theta is positive.

See the attachments for detailed solution.

I hope this answer helps you

Answer 2

Answer:

$\cos\theta=-\frac{1}{2}$

$\tan\theta=\sqrt3$

Step-by-step explanation:

Given that $\sin\theta=-\frac{\sqrt3}{2},\pi<\theta<\frac{3\pi}{2}$

θ is in quadrant III, hence only tan θ and cot θ is positive. Hence, cosθ will be negative.

Now, we have the formula

$\cos\theta=\pm\sqrt{1-\sin^2\theta}$

Since, cosθ is positive, hence we have

$\cos\theta=-\sqrt{1-\sin^2\theta}$

$\cos\theta=-\sqrt{1-(-\frac{\sqrt3}{2})^2}$

$\cos\theta=-\sqrt{1-\frac{3}{4}}\\\\\cos\theta=-\sqrt{\frac{1}{4}}\\\\\cos\theta=-\frac{1}{2}$

And the value of tanθ is

$\tan\theta=\frac{\sin\theta}{\cos\theta}\\\\\tan\theta=\frac{-\sqrt3/2}{-1/2}\\\\\tan\theta=\sqrt3$