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

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

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

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

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

$cos 2x=1-18/25$

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

$cos 2x=7/25$

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