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 $arcsin(sin((5pi)/6)) = pi/6$

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