If sinx = -3/5, x is in quadrant III, then how do you find sin2x?

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If sinx = -3/5, x is in quadrant III, then how do you find sin2x?

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Answer 1 · 2016-06-05T06:19:09

$sin 2 x = \frac{24}{25}$ $sin2x=24/25$

Explanation:

$sin 2 x = 2 sin x cos x$ $sin2x=2sinxcosx$
$cos x = - \sqrt{1 - {sin}^{2} x}$ $cosx=-sqrt(1-sin^2x$ (cosx <0 in quadrant III)
so $cos x = - \sqrt{1 - \frac{9}{25}}$ $cosx=-sqrt(1-9/25)$
that's $cos x = - \frac{4}{5}$ $cosx=-4/5$
and $sin 2 x = 2 (- \frac{3}{5}) (- \frac{4}{5})$ $sin2x=2(-3/5)(-4/5)$
that's $\frac{24}{25}$ $24/25$

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If sinx = -3/5, x is in quadrant III, then how do you find sin2x?

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