Question

If sin theta=3/5 and theta has its terminal side in quadrant II, how do you find the exact value of tan 2 theta?

Answer 1

`- 24/7`

Explanation:

`sin t = 3/5`. Find cos t.
`cos^2 t = 1 - sin^2 t = 1 - 9/25 = 16/25`
`cos t = +- 4/5.`
Since t is located in Quadrant II, cos t is negative.
`cos t = - 4/5` --> `tan t = (sin t)/(cos t) = (3/5)(-4/5) = -3/4`
`tan 2t = (2tan t)/(1 - tan^2 t) = (-3/2)/(1 - 9/16) =`
`= -(3/2)/(7/16) = - (3/2)(16/7) = - 24/7`
Check by calculator
`tan t = -3/4` -> `t = - 36^@87` --> `tan 2t = tan (-73.74) = -3.43`
tan 2t = - 24/7 = 3.43. OK