Question

Given `costheta=-(2sqrt10)/11` and `pi<theta<(3pi)/2`, how do you find `tan2theta`?

Answer 1

`(8sqrt10)/41`

Explanation:

Use the 2 trig identities:
`sin^2 a = 1 - cos^2 a`
`tan2a = (2tan a)/(1 - tan^2 a)`
First fin sin t.
`sin^2 t = 1 - cos^2 t = 1 - 40/121 = 81/121`
`sin t = -9/11` (since t is in Quadrant III)
`tan t = (-9/11)(-11/2sqrt10) = 9/(2sqrt10) = (9sqrt10)/20`
`tan 2t = ((18sqrt10)/20)/(1 - 810/400) =`
`= (2sqrt10/10)(-400/410) = (8sqrt10)/41`