Question

If sin theta = -4/5 and theta is in quadrant 3/ find tan (theta/2)?

Answer 1

`tantheta=4/3`

Explanation:

If we're in the third quadrant, both sine and cosine are negative.

Since `tantheta=sintheta/costheta`, the tangent in the third quadrant will be positive due to the negatives on the sine and cosine cancelling each other out. So, keep this in mind.

Recall the identity

`sin^2theta+cos^2theta=1`

Since `sintheta=-4/5, sin^2theta=(-4/5)^2=16/25`

Plug this into the identity and solve for `costheta:`

`16/25+cos^2theta=25/25`

`cos^2theta=(25-16)/25`

`cos^2theta=9/25`

`costheta=+-sqrt(9/25)`

We're in the third quadrant, so we want the negative cosine.

`costheta=-3/5`

So, since `tantheta=sintheta/costheta,` in this case,

`tantheta=(-4/5)/(-3/5)=-4/5*-5/3=4/3`

Answer 2

`tan (t/2) = - 2`

Explanation:

`sin t = -4/5`
`cos^2 t = 1 - sin^2 t = 1 - 16/25 = 9/25`
`cos t = +- 3/5`.
Since t is in Quadrant 3, then, cos t is negative
`cos t = - 3/5`
Use half angle identity;
`tan (t/2) = (1 - cos t)/(sin t)`
In this case:
`tan (t/2) = ( 1 + 3/5)/(-4/5) = (8/5)(-5/4) = - 2`