How do you evaluate ${cos}^{- 1} (cos (\frac{7 π}{6}))$ $cos^-1(cos((7pi)/6))$ ?

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Answer 1 · 2016-08-07T17:49:38

$= \frac{7 π}{6}$ $=(7pi)/6$

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

Answer 2 · 2016-08-07T19:08:02

$5 \frac{π}{6}$ $5pi/6$ .

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

But, $(cos (π + θ) = - cos θ$ $(cos(pi+theta)=-costheta$

$:. Reqd. Value=cos^-1(-cos(pi/6))=cos^-1(-sqrt3/2)$

Since, $cos^-1(-x)=pi-cos^-1x, AA |x|<=1$ .

The Reqd. value $=pi-cos^-1(sqrt3/2)=pi-pi/6$

$=5pi/6$ .

How do you evaluate cos^-1(cos((7pi)/6))cos−1(cos(7π6))cos^-1(cos((7pi)/6))?