How do you solve and find the value of #sin(2sin^-1(1/2))#?

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How do you solve and find the value of #sin(2sin^-1(1/2))#?

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Answer 1 · 2016-12-21T04:33:17

#+- sqrt3/2#

Explanation:

#sin (2sin^-1 (1/2)) --> sin (2arcsin (1/2))#
Trig table and unit circle give:
#sin x= 1/2# --> arc #x = pi/6# and arc #x = (5pi)/6#

a. #sin 2x = sin (2(pi)/6) = sin (pi/3) = sqrt3/2#
b. #sin 2x = sin (2(5pi)/6) = sin ((5pi)/3) = - sin (pi/3) = - sqrt3/2#
Finally:
#sin (2sin^-1 (1/2)) = +- sqrt3/2#

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How do you solve and find the value of #sin(2sin^-1(1/2))#?

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