Question

How do I solve for tan (theta) and theta?

if `sin(theta)=0.5` and `cos(theta)=-sqrt (3/2)`, evaluate:
a) `tan(theta)`
b) `theta`

Answer 1

`tan(theta)=-sqrt(6)`

`theta=ilog((sqrt(2))*(1+sqrt(3))/2)`

Explanation:

First,Solve for `tan(theta)` By using the fact That `tan(theta)=sin(theta)/cos(theta)`

and Since You said that `sin(theta)=0.5=1/2`

and `cos(theta)=-sqrt(3)/2`

That means `tan(theta)=sin(theta)/cos(theta)=(1/2)/(-sqrt(3)/2)=` `-1/sqrt(3)=-sqrt(3)/3`

Lets Use the `cos(theta)` equation to find `theta`

`theta=cos^-1(-sqrt(6)/2)=ilog(u)`

when `u=(sqrt(6)+sqrt(2))/2=(sqrt(2))(1+sqrt(3))/2`

Answer 2

`tan(theta)=1/-sqrt(3)~~-0.5773...`

`theta=150º`

Explanation:

I think you meant to write:

`sin(theta)=0.5`
`cos(theta)=color(red)(-sqrt(3)/2)`

For part (a), we can use the definition that `tan` is just `sin/cos`:

`tan(theta)=sin(theta)/cos(theta)`

`=>0.5/(-sqrt(3)/2)=1/-sqrt(3)`

For part (b), think about where on the unit circle the `y`-coordinate of a point is `0.5`; it's at `30º` and `150º`. Here are those points:

![desmos.com/calculator]()

Similarly, also think about where on the unit circle the `x`-coordinate of a point is `-sqrt(3)/2`; it's at `150º` and `210º`

![desmos.com/calculator]()

The point that both of these criteria share is `(-sqrt(3)/2,0.5)`, or the `150º` rotation.

Therefore, `theta` is be `150º`.