How do you solve the inverse trig function $sin(sin^-1 (1/3))$ ?

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Answer 1 · 2015-05-25T16:58:58

$sin^-1(1/3)$ or $arcsin(1/3)$ is the angle $theta$ for which $sin(theta) = 1/3$

Therefore $sin(sin^-1(1/3))$
$= sin(theta)$ for the value of $theta$ for which $sin(theta) = 1/3$

That is
$sin(sin^-1(1/3)) = 1/3$

How do you solve the inverse trig function sin(sin^-1 (1/3))sin(sin^-1 (1/3))?