If $sin x = (\frac{4}{5})$ $sinx=(4/5)$ , how do you find $sin 2 x$ $sin2x$ ?

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If $sin x = (\frac{4}{5})$ $sinx=(4/5)$ , how do you find $sin 2 x$ $sin2x$ ?

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Answer 1 · 2016-04-11T05:53:11

$sin 2 x = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25}$ $sin2x = 2*4/5*3/5=24/25$

Explanation:

Since $sin x = \frac{4}{5}$ $sinx=4/5$ we have a (3, 4, 5) triangle and we can totally define all sines and cosines, $cos x = \frac{3}{5}$ $cosx=3/5$

Now we use the double angle identities:
$sin 2 x = 2 sin x \cdot cos x$ $sin2x=2sinx*cosx$
$sin 2 x = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25}$ $sin2x = 2*4/5*3/5=24/25$

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If $sin x = (\frac{4}{5})$ $sinx=(4/5)$ , how do you find $sin 2 x$ $sin2x$ ?

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If $sin x = (\frac{4}{5})$ $sinx=(4/5)$ , how do you find $sin 2 x$ $sin2x$ ?