Use the trig identity: tan 2A = (2tan A)/(1 - tan^2 A) (1).tan2A=2tanA1−tan2A(1).
First, find tan A = sin A/(cos A).tanA=sinAcosA.
sin A = 3/5sinA=35 --> sin^2 A = 9/25sin2A=925 --> cos^2 A = 1 - 9/25 = 15/25cos2A=1−925=1525 -
--> cos A = +- 4/5cosA=±45.
Because A is in Quadrant II, its cos is negative. cos A = - 4/5cosA=−45
tan A = sin A/(cos A) = (3/5)(-5/4) = -3/4tanA=sinAcosA=(35)(−54)=−34
Replace value of tan A = -3/4 into identity (1) -->
tan 2A = (-3/2)/(1 - 9/16) = (-3/2)(16/7) = -24/7tan2A=−321−916=(−32)(167)=−247
Check by calculator.
cos A = -4/5 = -0.8cosA=−45=−0.8 --> A = 143.13A=143.13 --> 2A = 286.262A=286.26.-->
-> tan 2A = -3.43tan2A=−3.43
-24/7 = -3. 43.−247=−3.43. OK
