How do you find ${sin}^{- 1} (sin (\frac{5 π}{6}))$ $sin^-1(sin ((5pi)/6))$ ?

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How do you find ${sin}^{- 1} (sin (\frac{5 π}{6}))$ $sin^-1(sin ((5pi)/6))$ ?

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Answer 1 · 2018-03-27T05:06:47

${sin}^{- 1} (\frac{sin (5 π)}{6}) = \frac{π}{6}$ $sin^-1(sin(5pi)/6)=pi/6$

Explanation:

$sin (\frac{5 π}{6}) = \frac{1}{2}$ $sin((5pi)/6)=1/2$ , so we're really being asked for ${sin}^{- 1} (\frac{1}{2}) .$ $sin^-1(1/2).$

Let

$y = {sin}^{- 1} (\frac{1}{2})$ $y=sin^-1(1/2)$

Then, from the definition of arcsine,

$\frac{1}{2} = sin (y)$ $1/2=sin(y)$

( $y = {sin}^{- 1} (a) \Leftrightarrow a = sin (y)$ $y=sin^-1(a) hArr a=sin(y)$ )

In other words, we want to know where the sine function returns $\frac{1}{2} .$ $1/2.$

Sine is equal to $\frac{1}{2}$ $1/2$ for $\frac{π}{6} + 2 n π, \frac{5 π}{6} + 2 n π,$ $pi/6+2npi, (5pi)/6+2npi,$ where $n$ $n$ is any integer. But really, since the domain of arcsine is $[- \frac{π}{2}, \frac{π}{2}], \frac{π}{6}$ $[-pi/2, pi/2], pi/6$ is the only valid answer.

Answer 2 · 2018-03-27T06:44:09

$\frac{π}{6}$ $pi/6$

Explanation:

Hint:
We know that ,
${sin}^{- 1} (sin θ) = θ,$ $color(red)(sin^-1(sintheta)=theta,$ where, $θ \in [- \frac{π}{2}, \frac{π}{2}]$ $color(red)(theta in [-pi/2,pi/2]$
Even though,
${sin}^{- 1} (sin (\frac{5 π}{6})) \neq (\frac{5 π}{6}),$ $color(blue)(sin^-1(sin((5pi)/6))!=((5pi)/6),$ because, $\frac{5 π}{6} \notin [- \frac{π}{2}, \frac{π}{2}]$ $color(blue)((5pi)/6 !in [-pi/2,pi/2]$
So, we reduce $θ = \frac{5 π}{6} .$ $theta=(5pi)/6.$ But how ? By using Addition and Subtraction Formulas for sine and cosine .Now enjoy the answer.

****Answer:****

Here,

$\frac{5 π}{6} = π - \frac{π}{6}$ $(5pi)/6=pi-pi/6$

$({sin}^{- 1} (sin (\frac{5 π}{6})) = {sin}^{- 1} (sin (π - \frac{π}{6}))$ $(sin^-1(sin((5pi)/6))=sin^-1(sin(pi-pi/6))$

$=sin^-1(sin(pi/6))....to$ , Apply $color(red)([sin(pi-theta)=sintheta ]$

$=pi/6......to$ , where, $pi/6in[-pi/2,pi/2]$

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How do you find ${sin}^{- 1} (sin (\frac{5 π}{6}))$ $sin^-1(sin ((5pi)/6))$ ?

2 Answers

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