How do you find the exact value of $cos^-1(sin((7pi)/2))$ ?

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Answer 1 · 2015-08-24T19:38:51

$cos^-1(sin((7pi)/2)) = pi$

$cos^-1(sin((7pi)/2))$ is

the $t$ in $[0,pi]$ with $cost = sin((7pi)/2)$

Since $sin((7pi)/2) = -1$ , we need

the $t$ in $[0,pi]$ with $cost = -1$

So $t = pi$

How do you find the exact value of cos^-1(sin((7pi)/2))cos^-1(sin((7pi)/2))?