If sinx = -3/5, x is in quadrant III, then how do you find tan2x?

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If sinx = -3/5, x is in quadrant III, then how do you find tan2x?

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Answer 1 · 2016-07-10T22:37:49

#24/7#

Explanation:

#sin x = - 3/5# . Find cos x by identity: #cos^2 x = 1 - sin^2 x#
#cos^2 x = 1 - 9/25 = 16/25#
#cos x = +- 4/5.#
Since x is in Quadrant III, then cos x is negative.
#tan x = sin/(cos) = (-3/5)(-5/4) = 3/4#
Use trig identity:
#tan 2x = (2tan x)/(1 - tan^2 x)#
#tan 2x = (6/4)/(1 - 9/16) = (6/4)(16/7) = 24/7#

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If sinx = -3/5, x is in quadrant III, then how do you find tan2x?

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Explanation:

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