How do you solve and find the value of $sin (2 {sin}^{- 1} (\frac{1}{2}))$ $sin(2sin^-1(1/2))$ ?

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How do you solve and find the value of $sin (2 {sin}^{- 1} (\frac{1}{2}))$ $sin(2sin^-1(1/2))$ ?

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Answer 1 · 2016-12-21T04:33:17

$\pm \frac{\sqrt{3}}{2}$ $+- sqrt3/2$

Explanation:

$sin (2 {sin}^{- 1} (\frac{1}{2})) - \to sin (2 arcsin (\frac{1}{2}))$ $sin (2sin^-1 (1/2)) --> sin (2arcsin (1/2))$
Trig table and unit circle give:
$sin x = \frac{1}{2}$ $sin x= 1/2$ --> arc $x = \frac{π}{6}$ $x = pi/6$ and arc $x = \frac{5 π}{6}$ $x = (5pi)/6$

a. $sin 2 x = sin (2 \frac{π}{6}) = sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$ $sin 2x = sin (2(pi)/6) = sin (pi/3) = sqrt3/2$
b. $sin 2 x = sin (2 \frac{5 π}{6}) = sin (\frac{5 π}{3}) = - sin (\frac{π}{3}) = - \frac{\sqrt{3}}{2}$ $sin 2x = sin (2(5pi)/6) = sin ((5pi)/3) = - sin (pi/3) = - sqrt3/2$
Finally:
$sin (2 {sin}^{- 1} (\frac{1}{2})) = \pm \frac{\sqrt{3}}{2}$ $sin (2sin^-1 (1/2)) = +- sqrt3/2$

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How do you solve and find the value of $sin (2 {sin}^{- 1} (\frac{1}{2}))$ $sin(2sin^-1(1/2))$ ?

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How do you solve and find the value of $sin (2 {sin}^{- 1} (\frac{1}{2}))$ $sin(2sin^-1(1/2))$ ?