If Cos(t)=-7/8 with #pi<=t<=3pi/2#, how do you find the value of sin(2t)?

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If Cos(t)=-7/8 with #pi<=t<=3pi/2#, how do you find the value of sin(2t)?

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Answer 1 · 2016-04-07T02:30:36

#(7sqrt15)/32 = 0.85#

Explanation:

sin 2t = 2sin t.cos t
Knowing #cos t = -7/8#, find sin t.
Trig identity: #sin^2 a = 1 - cos^2 a# -->
#sin^2 t = 1 - cos^2 t = 1 - 49/64 = 15/64#
sin t = +- sqrt15/8.
Arc t is in Quadrant III, then, sin t is negative.
#sin 2t = 2(-7/8)(- sqrt15/8) = (7sqrt15)/32= 0.85#

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If Cos(t)=-7/8 with #pi<=t<=3pi/2#, how do you find the value of sin(2t)?

1 Answer

Explanation:

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