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

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

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Answer 1 · 2016-12-13T14:38:24

#pi/6 + 2kpi#
#(5pi)/6 + 2kpi#

Explanation:

#sin^-1 (1/2) --> arcsin (1/2)#
Trig table of special arcs gives -->
#sin x = 1/2# --> arc #x = pi/6#
Unit circle gives another arc that has the same sin value (1/2) -->
arc #x = pi - pi/6 = (5pi)/6#
General answers:
#x = pi/6 + 2kpi#
#x = (5pi)/6 + 2kpi#

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

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