Question #122b2

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Question #122b2

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Answer 1 · 2016-11-28T14:53:04

$\frac{24}{25}$ $24/25$

Explanation:

Use the trig identity:
sin (a + b) = sin a.cos b + sin b.cos a
In this formula, we know sin a and sin b.
We must find cos a and cos b.
$sin a = \frac{3}{5}$ $sin a = 3/5$ --> ${cos}^{2} a = 1 - \frac{9}{25} = \frac{16}{25}$ $cos^2 a = 1 - 9/25 = 16/25$ --> $cos a = \frac{4}{5}$ $cos a = 4/5$ (because a is in Quadrant I)
$sin b = \frac{4}{5}$ $sin b = 4/5$ --> ${cos}^{2} b = 1 - \frac{16}{25} = \frac{9}{25}$ $cos^2 b = 1 - 16/25 = 9/25$ --> $cos b = \frac{3}{5}$ $cos b = 3/5$ (Quadrant 1)
Finally,
$sin (a + b) = (\frac{3}{5}) (\frac{4}{5}) + (\frac{3}{5}) (\frac{4}{5}) = 2 (\frac{3}{5}) (\frac{4}{5}) = \frac{24}{25}$ $sin (a + b) = (3/5)(4/5) + (3/5)(4/5) = 2(3/5)(4/5) = 24/25$

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Question #122b2

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