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

Answer:

The values of cos Θ and tan Θ are $-\frac{1}{2}$ and √3 respectively.

Step-by-step explanation:

Given,

$sin\theta = -\frac{\sqrt{3} }{2}$

Since, by the trigonometric identity,

$sin^2 \theta + cos^2\theta = 1$

$\implies cos^2 \theta = 1 - sin^2 \theta$

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

We have,

$\pi < \theta < \frac{3\pi}{2}$

⇒ $\theta$ lies in third quadrant,

⇒ $cos\theta$ is negative.

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

Now,

$tan \theta = \frac{sin \theta}{cos \theta}=\frac{-\sqrt{3}/2}{-1/2}=\sqrt{3}$

Answer 2

Sin(teta) = -√3/2

pi<teta<3pi/2

because of range of pi we can conclude that teta is in third quadrant of x-y coordinate system which will determine us what is the sign of cos(teta) and tan(teta)
cos(teta) = √(1-sin^2(teta)) = 1/2
but because teta is in third quadrant, cos(teta) has to be negative which means:
cos(teta) = -1/2

tan(teta) = sin(teta)/cos(teta) = √3

tan of an angle that is in third quadrant is positive which is what we got.

Answer is:
cos(teta) = -1/2
tan(teta) = √3