What does $arcsin (sin (\frac{- π}{2}))$ $arcsin(sin ((-pi)/2))$ equal?

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Answer 1 · 2016-01-20T10:53:42

$- \frac{π}{2}$ $-pi/2$

If you want to know how please follow.

$sin (- \frac{π}{2}) = - sin (\frac{π}{2})$ $sin(-pi/2) = -sin(pi/2)$ since $sin (x)$ $sin(x)$ is a odd function.

$sin (- \frac{π}{2}) = - 1$ $sin(-pi/2) = -1$ since $sin (\frac{π}{2}) = 1$ $sin(pi/2) = 1$

$arcsin (sin (- \frac{π}{2})) = arcsin (- 1)$ $arcsin(sin(-pi/2)) = arcsin(-1)$

Now comes the range of $arcsin (x)$ $arcsin(x)$
The range of $arcsin (x)$ $arcsin(x)$ is $[- \frac{π}{2}, \frac{π}{2}]$ $[-pi/2,pi/2]$

So $arcsin (- 1)$ $arcsin(-1)$ would give $- \frac{π}{2}$ $-pi/2$

Therefore,

$arcsin (sin (- \frac{π}{2})) = - \frac{π}{2}$ $arcsin(sin(-pi/2)) = -pi/2$

What does arcsin(sin ((-pi)/2)) arcsin(sin(−π2))arcsin(sin ((-pi)/2)) equal?