How do you find the value of ${sin}^{- 1} (sin ({cos}^{- 1} (sin (\frac{π}{12}))))$ $sin^(-1)(sin (cos^(-1) (sin (pi/12))))$ ?

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How do you find the value of ${sin}^{- 1} (sin ({cos}^{- 1} (sin (\frac{π}{12}))))$ $sin^(-1)(sin (cos^(-1) (sin (pi/12))))$ ?

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Answer 1 · 2016-07-17T07:36:31

${sin}^{- 1} (sin ({cos}^{- 1} (sin (\frac{π}{12})))) = 5 \frac{π}{12}$ $sin^-1(sin(cos^-1(sin(pi/12))))=5pi/12$ .

Explanation:

We will use the following Rules :

$(R 1) : sin θ = cos (\frac{π}{2} - θ)$ $(R1) : sintheta=cos(pi/2-theta)$

$(R 2) : {cos}^{- 1} (cos θ) = θ, θ \in [o, π]$ $(R2) : cos^-1(costheta)=theta, theta in [o,pi]$

$(R 3) : {sin}^{- 1} (sin θ) = θ, θ \in [- \frac{π}{2}, \frac{π}{2}]$ $(R3) : sin^-1(sintheta)=theta, theta in [-pi/2,pi/2]$

Let us note that, by

$(R 1), sin (\frac{π}{12}) = cos (\frac{π}{2} - \frac{π}{12}) = cos (5 \frac{π}{12})$ $(R1), sin(pi/12)=cos(pi/2-pi/12)=cos(5pi/12)$

So, ${cos}^{- 1} (sin (\frac{π}{12})) = {cos}^{- 1} (cos (5 \frac{π}{12}))$ $cos^-1(sin(pi/12))=cos^-1(cos(5pi/12))$ , where,

$5 \frac{π}{12} \in [0, π]$ $5pi/12 in [0,pi]$ , so, using $(R 2)$ $(R2)$ , we get,

${cos}^{- 1} (cos (5 \frac{π}{12})) = 5 \frac{π}{12}$ $cos^-1(cos(5pi/12))=5pi/12$

Again, as $5 \frac{π}{12} \in [- \frac{π}{2}, \frac{π}{2}]$ $5pi/12 in [-pi/2,pi/2]$ , by $(R 3)$ $(R3)$ , we have,

${sin}^{- 1} (sin (5 \frac{π}{12})) = 5 \frac{π}{12}$ $sin^-1(sin(5pi/12))=5pi/12$

Finally, ${sin}^{- 1} (sin ({cos}^{- 1} (sin (\frac{π}{12})))) = 5 \frac{π}{12}$ $sin^-1(sin(cos^-1(sin(pi/12))))=5pi/12$ .

Hope, this will be of Help! Enjoy Maths.!

Answer 2 · 2016-07-17T07:45:15

$\frac{5 π}{12} = 75^{o}$ $(5pi)/12=75^o$

Explanation:

Use $sin a = cos (\frac{π}{2} - a) and,$ $sin a = cos (pi/2-a) and,$

if $y = f (x), x = f^{- 1} y, f f^{- 1} y = y and f^{- 1} y f (x) = x$ $y=f(x) , x = f^(-1)y, f f^(-1)y=y and f^(-1)yf(x)=x$ .

The given expression is

${sin}^{- 1} {sin cos}^{- 1} sin (\frac{π}{12})$ $sin^(-1)sin cos^(-1)sin(pi/12)$

$= {sin}^{- 1} sin ({cos}^{- 1} cos (\frac{π}{2} - \frac{π}{12}))$ $=sin^(-1)sin (cos^(-1)cos(pi/2-pi/12))$

$= {sin}^{- 1} sin (\frac{5 π}{12})$ $=sin^(-1)sin((5pi)/12)$

$= \frac{5 π}{12}$ $=(5pi)/12$

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