Given $tan A = - \frac{3}{4}, 0 < A < π$ $tanA=-3/4, 0<A<pi$ , what is $sin (2 A)$ $sin(2A)$ ?

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Given $tan A = - \frac{3}{4}, 0 < A < π$ $tanA=-3/4, 0<A<pi$ , what is $sin (2 A)$ $sin(2A)$ ?

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Answer 1 · 2017-03-06T04:53:46

$- \frac{12}{25}$ $- 12/25$

Explanation:

Use trig identity:
${sin}^{2} x = \frac{1}{1 + {cot}^{2} x}$ $sin^2 x = 1/(1 + cot^2 x)$
In this case:
${sin}^{2} A = \frac{1}{1 + {cot}^{2} A} = \frac{1}{1 + \frac{16}{9}} = \frac{9}{25}$ $sin^2 A = 1/(1 + cot^2 A) = 1/(1 + 16/9) = 9/25$ --> $sin A = \pm \frac{3}{5}$ $sin A = +- 3/5$
Because 0 < A < pi, therefore, sin A is positive.
$sin A = \frac{3}{5}$ $sin A = 3/5$ . Find cos A
${cos}^{2} A = 1 - {sin}^{2} A = 1 - \frac{9}{25} = \frac{16}{25}$ $cos^2 A = 1- sin^2 A = 1 - 9/25 = 16/25$ --> $cos A = \pm \frac{4}{5}$ $cos A = +- 4/5$
Because tan A is negative, therefore, A is in Q. 2, cos A is negative.
$sin (2 A) = 2 sin A . cos A = (\frac{3}{5}) (- \frac{4}{5}) = - \frac{12}{25}$ $sin (2A) = 2sin A.cos A = (3/5)(- 4/5) = - 12/25$

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Given $tan A = - \frac{3}{4}, 0 < A < π$ $tanA=-3/4, 0<A<pi$ , what is $sin (2 A)$ $sin(2A)$ ?

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Explanation:

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