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

The values of cos Θ and tan Θ are
`-(1)/(2)` and √3 respectively.

Explanation:

Given,

`sin\theta = -(√(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 √(1 - sin^2 \theta)`

We have,

`\pi < \theta < (3\pi)/(2)`

⇒
`\theta` lies in third quadrant,

⇒
`cos\theta` is negative.

`\implies cos \theta = - √(1 - sin^2 \theta)=-\sqrt{1 -(-(√(3) )/(2))^2}=-\sqrt{1 -(3)/(4)}=-\sqrt{(1)/(4)}=-(1)/(2)`

Now,

`tan \theta = (sin \theta)/(cos \theta)=(-√(3)/2)/(-1/2)=√(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