Question

Given `sintheta=(2sqrt2)/3` and `pi/2<theta<pi`, how do you find `tan2theta`?

Answer 1

`(4sqrt2)/7`

Explanation:

`sin t = (2sqrt2)/3`. Find cos t by identity: `cos^2 t = 1 - sin^2 t`
`cos^2 t = 1 - 8/9 = 1/9` --> `cos t = +- 1/3`
Since t is Quadrant II, then cos t is negative. --> `cos t = -1/3`
Next, find sin 2t and cos 2t by the identities:
`sin 2t = 2sin t.cos t`
`cos 2t = 2cos^2 t - 1`
`sin 2t = 2sin t.cos t = 2(-1/3)((2sqrt2)/3) = (-4sqrt2)/9`
`cos 2t = 2/9 - 1 = -7/9`
`tan 2t = (sin 2t)/(cos 2t) = ((-4sqrt2)/9)(-9/7) = (4sqrt2)/7`
Check by calculator.
`cos t = - 1/3` --> t = 109.47 --> `2t = 218^@84` --> tan 2t = 0.808
`(4sqrt2)/7 = (1.41)(4)/7 = 0.808.` OK