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

1 Answer
Dec 2, 2015

Answer:

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