How do you solve the inverse trig function #arcsin (sin 5pi/6)#?

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Answer 1 · 2015-05-20T21:44:39

When #-1 <= x <= 1#, #arcsin(x)# is defined as the value #theta#

in the range #-pi/2 <= theta <= pi/2# such that #sin theta = x#.

The value #(5pi)/6# does not fall into this range.

Instead we find:

#pi/2 < (5pi)/6 < pi#

Use the equality #sin(pi-theta) = sin theta# and note that

#(5pi)/6 = pi - pi/6#

So #sin (pi/6) = sin (pi - pi/6) = sin ((5pi)/6)#

So #arcsin(sin((5pi)/6)) = pi/6#