How do you calculate $cos (arcsin - \frac{3}{5})$ $cos (Arcsin -3/5)$ ?

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How do you calculate $cos (arcsin - \frac{3}{5})$ $cos (Arcsin -3/5)$ ?

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Answer 1 · 2015-06-11T05:11:46

$cos ({sin}^{- 1} (- \frac{3}{5})) = \frac{4}{5}$ $cos(sin^(-1)(-3/5))=4/5$

Explanation:

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

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

Which is solution.

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How do you calculate $cos (arcsin - \frac{3}{5})$ $cos (Arcsin -3/5)$ ?

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