How do you evaluate $cos [2 {sin}^{- 1} (\frac{5}{13})]$ $cos[ 2sin^-1(5/13)]$ ?

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How do you evaluate $cos [2 {sin}^{- 1} (\frac{5}{13})]$ $cos[ 2sin^-1(5/13)]$ ?

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Answer 1 · 2015-06-11T09:11:06

The solution is $\frac{119}{169}$ $119/169$ or 0.7041

Explanation:

Assume ${sin}^{- 1} (\frac{5}{13}) = x$ $sin^(-1)(5/13)=x$ _(1)
$\Rightarrow sin x = \frac{5}{13}$ $=>sinx=5/13$ _____(2)
Therefore
$cos (2 {sin}^{- 1} (\frac{5}{13}))$ $cos(2sin^(-1)(5/13))$ = $cos 2 x$ $cos2x$ [since freom equation (1)]
$= 1 - 2 {sin}^{2} x$ $=1-2sin^2x$
$= 1 - 2 {(\frac{5}{13})}^{2}$ $=1-2(5/13)^2$
$= \frac{119}{169}$ $=119/169$

which is required solution.

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How do you evaluate $cos [2 {sin}^{- 1} (\frac{5}{13})]$ $cos[ 2sin^-1(5/13)]$ ?

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