How do you calculate $cos (Arcsin -3/5)$ ?

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Answer 1 · 2015-06-11T05:11:46

$cos(sin^(-1)(-3/5))=4/5$

$cos(sin^(-1)(-3/5))=cos(-sin^(-1)(3/5))$
$=cos(sin^(-1)(3/5))$ ____(1)
Assume $sin^-1(3/5)=x$ _____(2)

Substitute equation(2) in equation(1)
$cos(sin^(-1)(-3/5))=cosx$
$=sqrt(1-sin^2x)$
$=sqrt(1-(3/5)^2)$ [since from equation 3]
$=4/5$

Which is solution.

How do you calculate cos (Arcsin -3/5)cos (Arcsin -3/5)?