How do you evaluate ${sin}^{- 1} (sin (\frac{3 π}{10}))$ $sin^-1(sin((3pi)/10))$ ?

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How do you evaluate ${sin}^{- 1} (sin (\frac{3 π}{10}))$ $sin^-1(sin((3pi)/10))$ ?

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Answer 1 · 2016-09-23T14:17:20

$\frac{3 π}{10}$ $(3pi)/10$ .

Explanation:

Let us recall the Defn. of ${sin}^{- 1}$ $sin^-1$ fun. :

${sin}^{- 1} x = θ, | x | \leq 1 \Leftrightarrow sin θ = x, θ \in [- \frac{π}{2}, \frac{π}{2}]$ $sin^-1x=theta, |x| le1 iff sintheta=x, theta in [-pi/2,pi/2]$

$:. sin^-1(sintheta)=theta, AAtheta in [-pi/2,pi/2].$

In our case. since, $3pi/10 in [-pi/2,pi/2], sin^-1(sin((3pi)/10)=(3pi)/10$ .

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How do you evaluate ${sin}^{- 1} (sin (\frac{3 π}{10}))$ $sin^-1(sin((3pi)/10))$ ?

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Explanation:

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How do you evaluate ${sin}^{- 1} (sin (\frac{3 π}{10}))$ $sin^-1(sin((3pi)/10))$ ?