Question

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

Answers:
cos theta = -(1)/2; tan theta = sqrt(3)
cos theta = -(1)/2; tan theta = -1
cos theta = (sqrt(3))/4; tan theta = -2
cos theta = 1/2; tan theta = sqrt(3)

Answer 1

For
`\pi<\theta<\frac{3\pi}2`, we expect
`\cos\theta<0`. Then recalling that
`\sin^2\theta+\cos^2\theta=1`, we have

`\cos\theta=-√(1-\sin^2\theta)=-\frac12`

Then by definition of tangent,

`\tan\theta=(\sin\theta)/(\cos\theta)=\frac{-\frac{\sqrt3}2}{-\frac12}=\sqrt3`

Answer 2

Option 1 -

`\cos\theta=-(1)/(2)`

`\tan\theta=\sqrt3`

Explanation:

Given :
`\sin\theta=-(\sqrt3)/(2)` and
`\pi < \theta < (3\pi)/(2)`

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

Solution :

`\sin\theta=-(\sqrt3)/(2)`

We know,
`\cos\theta=√(1-sin^2\theta)`

Substitute the value of
`\sin\theta`,

`\cos\theta=\sqrt{1-(-(\sqrt3)/(2))^2}`

`\cos\theta=\sqrt{1-(3)/(4)}`

`\cos\theta=\sqrt{(4-3)/(4)}`

`\cos\theta=\sqrt{(1)/(4)}`

`\cos\theta=(1)/(2)`

Since,
`\pi < \theta < (3\pi)/(2)` i.e. in third quadrant

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

So,
`\cos\theta=-(1)/(2)`

Now,
`\tan\theta=(\sin\theta)/(\cos\theta)`

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

`\tan\theta=(-(\sqrt3)/(2))/(-(1)/(2))`

`\tan\theta=(\sqrt3)/(2)* (2)/(1)}`

`\tan\theta=\sqrt3`

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

So,
`\tan\theta=\sqrt3`

Therefore, Option 1 is correct.