How do you calculate the $cos[2 sin^ -1 (1/2)]$ ?

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Answer 1 · 2016-02-26T02:44:39

$1/2$

First, focus on just $sin^-1(1/2)$ . This is another way of asking, for what angle is sine equal to $"1/2"$ ? Recall that

$sin(pi/6)=1/2$

Thus, since $sin$ and $arcsin$ are inverse functions,

$sin^-1(1/2)=pi/6$

Substitute this into the original expression:

$cos[2sin^-1(1/2)]=cos(2*pi/6)=cos(pi/3)=1/2$

How do you calculate the cos[2 sin^ -1 (1/2)]cos[2 sin^ -1 (1/2)]?