Question

If sin theta = -(sqrt(3))/2 and pi < theta < (3pi)/2, what are the values of cos theta and tan theta?

Answer 1

Answer:

Option 1 -

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

$\tan\theta=\sqrt3$

Step-by-step explanation:

Given : $\sin\theta=-\frac{\sqrt3}{2}$ and $\pi < \theta < \frac{3\pi}{2}$

To find : What are the values of $\cos \theta$ and $\tan \theta$?

Solution :

$\sin\theta=-\frac{\sqrt3}{2}$

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

Substitute the value of $\sin\theta$,

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

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

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

$\cos\theta=\sqrt{\frac{1}{4}}$

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

Since, $\pi < \theta < \frac{3\pi}{2}$ i.e. in third quadrant

We know, $\cos \theta$ in third quadrant is negative.

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

Now, $\tan\theta=\frac{\sin\theta}{\cos\theta}$

Substitute the value of $\sin\theta$ and $\cos\theta$,

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

$\tan\theta=\frac{\sqrt3}{2}\times \frac{2}{1}}$

$\tan\theta=\sqrt3$

We know, $\tan \theta$ in third quadrant is positive.

So, $\tan\theta=\sqrt3$

Therefore, Option 1 is correct.

Answer 2

For $\pi$, we expect $\cos\theta$. Then recalling that $\sin^2\theta+\cos^2\theta=1$, we have

$\cos\theta=-\sqrt{1-\sin^2\theta}=-\dfrac12$

Then by definition of tangent,

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{-\frac{\sqrt3}2}{-\frac12}=\sqrt3$