Question

PLEASE HELP ME!! LOTS OF POINTS!!
If sine theta equals negative square root of three over two and the range of theta is pi less than theta less than three times pi over two, what are the values of cos Θ and tan Θ? Please help!!, I don't understnad how to figure out which quadrant it's in to help me solve the question!!

Answer 1

Answer:

Just find sin^2theta and change them with the help of known identities.

You can also use Pythagoras theorm. It will work..

Hope it helps

Answer 2

Answer:

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

$tan\theta=\sqrt3$

Step-by-step explanation:

We are given that

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

Where $\theta$ lies between $\pi\;and\;\frac{3\pi}{2}$

We have to find the value of $cos\theta$ and $tan\theta$

$\theta$ lies in third quadrant

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

$cos\theta=\sqrt{1-(\frac{\sqrt3}{2})^2}=\sqrt{1-\frac{3}{4}}=\sqrt{\frac{1}{4}}=\pm\frac{1}{2}$

$cos\theta$ is negative in third quadrant

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

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

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

$tan\theta=\sqrt3$

$tan\theta$ is positive in third quadrant

Hence, $tan\theta=\sqrt3$