Question

How do you find tan2A, given sin A = 3/5 and A is in QII?

Answer 1

Find tan 2A, given `sin A = 3/5` and A in Quadrant II.

Ans: `- 24/7`

Explanation:

Use the trig identity: `tan 2A = (2tan A)/(1 - tan^2 A) (1).`
First, find `tan A = sin A/(cos A).`
`sin A = 3/5` --> `sin^2 A = 9/25` --> `cos^2 A = 1 - 9/25 = 15/25` -
--> `cos A = +- 4/5`.
Because A is in Quadrant II, its cos is negative. `cos A = - 4/5`
`tan A = sin A/(cos A) = (3/5)(-5/4) = -3/4`
Replace value of tan A = -3/4 into identity (1) -->
`tan 2A = (-3/2)/(1 - 9/16) = (-3/2)(16/7) = -24/7`
Check by calculator.
`cos A = -4/5 = -0.8` --> `A = 143.13` --> `2A = 286.26`.-->
-> `tan 2A = -3.43`
`-24/7 = -3. 43.` OK