If sinx = -3/5, x is in quadrant III, then how do you find cos2x?

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Answer 1 · 2016-06-19T08:02:57

#cos 2x=7/25#

Given #sin x=-3/5#

Using half-angle formulas

#sin x=+-sqrt((1-cos 2x)/2)#

#-3/5=-sqrt((1-cos 2x)/2)#

square both sides of the equation

#(-3/5)^2=(-sqrt((1-cos 2x)/2))^2#

#9/25=(1-cos 2x)/2#

#9/25=1/2-(cos 2x)/2#

#(cos 2x)/2=1/2-9/25#

#cos 2x=1-18/25#

#cos 2x=(25-18)/25#

#cos 2x=7/25#

God bless....I hope the explanation is useful.