How do you evaluate $sin [2 {cos}^{- 1} (\frac{3}{5})]$ $sin[2cos^(-1)(3/5)]$ ?

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How do you evaluate $sin [2 {cos}^{- 1} (\frac{3}{5})]$ $sin[2cos^(-1)(3/5)]$ ?

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Answer 1 · 2016-05-18T16:30:37

$\pm \frac{24}{25} = \pm 0.96$ $+-24/25=+-0.96$ .

Explanation:

Let $a = {cos}^{- 1} (\frac{3}{5})$ $a = cos^(-1)(3/5)$ . Then $cos a = \frac{3}{5} > 0$ $cos a = 3/5>0$ .. So,a is in the 1st or un

the 4th quadrant. Accordingly, $sin a = \pm \frac{4}{5}$ $sin a = +-4/5$ .

Now, the given sine is $sin 2 a = 2 sin a cos a = 2 (\pm \frac{4}{5}) (\frac{3}{5}) = \pm \frac{24}{25}$ $sin 2a = 2 sin a cos a=2(+-4/5)(3/5)=+-24/25$ ..

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How do you evaluate $sin [2 {cos}^{- 1} (\frac{3}{5})]$ $sin[2cos^(-1)(3/5)]$ ?

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