Given #tanA=-3/4, 0<A<pi#, what is #sin(2A)#?

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Given #tanA=-3/4, 0<A<pi#, what is #sin(2A)#?

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Answer 1 · 2017-03-06T04:53:46

#- 12/25#

Explanation:

Use trig identity:
#sin^2 x = 1/(1 + cot^2 x)#
In this case:
#sin^2 A = 1/(1 + cot^2 A) = 1/(1 + 16/9) = 9/25# --> #sin A = +- 3/5#
Because 0 < A < pi, therefore, sin A is positive.
#sin A = 3/5#. Find cos A
#cos^2 A = 1- sin^2 A = 1 - 9/25 = 16/25# --> #cos A = +- 4/5#
Because tan A is negative, therefore, A is in Q. 2, cos A is negative.
#sin (2A) = 2sin A.cos A = (3/5)(- 4/5) = - 12/25#

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Given #tanA=-3/4, 0<A<pi#, what is #sin(2A)#?

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