How do you find the exact value for ${sin}^{- 1} (cos (2 \frac{π}{3}))$ $sin^-1 (cos (2pi/3))$ ?

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How do you find the exact value for ${sin}^{- 1} (cos (2 \frac{π}{3}))$ $sin^-1 (cos (2pi/3))$ ?

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Answer 1 · 2018-03-29T02:43:05

${sin}^{- 1} (cos (2 \frac{π}{3})) = 7 \frac{π}{6}, 11 \frac{π}{6}$ $sin^-1(cos(2pi/3))=7pi/6, 11pi/6$
Among which the first positive solution happens to be

${sin}^{- 1} (cos (2 \frac{π}{3})) = 7 \frac{π}{6}$ $sin^-1(cos(2pi/3))=7pi/6$

Explanation:

${sin}^{- 1} (cos (2 \frac{π}{3})) = ?$ $sin^-1(cos(2pi/3))=?$

$2 \frac{π}{3} = π - \frac{π}{3}$ $2pi/3=pi-pi/3$

$cos (2 \frac{π}{3}) = cos (π - \frac{π}{3})$ $cos(2pi/3)=cos(pi-pi/3)$

$cos (A - B) = cos A cos B + sin A sin B$ $cos(A-B)=cosAcosB+sinAsinB$

$cos (π - \frac{π}{3}) = cos π cos (\frac{π}{3}) + sin π sin (\frac{π}{3})$ $cos(pi-pi/3)=cospicos(pi/3)+sinpisin(pi/3)$

$cos π = - 1$ $cospi=-1$

$cos (\frac{π}{3}) = \frac{1}{2}$ $cos(pi/3)=1/2$

$sin π = 0$ $sinpi=0$

$sin (\frac{π}{3}) = \frac{\sqrt{3}}{2}$ $sin(pi/3)=sqrt3/2$

$cos (π - \frac{π}{3}) = - 1 \times \frac{1}{2} + 0 \times \frac{\sqrt{3}}{2}$ $cos(pi-pi/3)=-1xx1/2+0xxsqrt3/2$

$cos (2 \frac{π}{3}) = - \frac{1}{2}$ $cos(2pi/3)=-1/2$

${sin}^{- 1} (cos (2 \frac{π}{3})) = {sin}^{- 1} (- \frac{1}{2})$ $sin^-1(cos(2pi/3))=sin^-1(-1/2)$

Let

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

$sin u = - \frac{1}{2}$ $sinu=-1/2$

$sin (\frac{π}{6}) = \frac{1}{2}$ $sin(pi/6)=1/2$
sine ratio is negative in 3rd and 4th quadrants

For angles u<2pi,

$sin (π + \frac{π}{6}) = - sin (\frac{π}{6})$ $sin(pi+pi/6)=-sin(pi/6)$ ->

$u = π + \frac{π}{6} = 7 \frac{π}{6}$ $u=pi+pi/6=7pi/6$

$sin (2 π - \frac{π}{6}) = - sin (\frac{π}{6})$ $sin(2pi-pi/6)=-sin(pi/6)$ ->

$u = 2 π - \frac{π}{6} = 11 \frac{π}{6}$ $u=2pi-pi/6=11pi/6$

Hence,
Exact value of

${sin}^{- 1} (cos (2 \frac{π}{3})) = 7 \frac{π}{6}, 11 \frac{π}{6}$ $sin^-1(cos(2pi/3))=7pi/6, 11pi/6$

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How do you find the exact value for ${sin}^{- 1} (cos (2 \frac{π}{3}))$ $sin^-1 (cos (2pi/3))$ ?

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