The value of cos−1 12+2sin−1 12 is

Let cos−1 12=x⇒ cos x =12=cos π3⇒ x=π3∈[0,π]∵ principal value branch of cos−1 is [0,π] Again, let sin−1 12=y Again, let sin−1 12=y⇒ sin y=12=sin π6⇒ y=π6∈−π2,π2 ∵ principal value branch of sin−1 is −π2,π2∴cos−1 12+2sin−1 12=π3+2⋅π6=π3+π3=2π3

The value of cos−1 12+2sin−1 12 is

Let cos−1 12=x⇒ cos x =12=cos π3⇒ x=π3∈[0,π]∵ principal value branch of cos−1 is [0,π] Again, let sin−1 12=y Again, let sin−1 12=y⇒ sin y=12=sin π6⇒ y=π6∈−π2,π2 ∵ principal value branch of sin−1 is −π2,π2∴cos−1 12+2sin−1 12=π3+2⋅π6=π3+π3=2π3

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The value of $\cos^{- 1} (\frac{1}{2}) + 2 \sin^{- 1} (\frac{1}{2})$ is

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$\frac{π}{2}$

$\frac{π}{3}$

$\frac{2 π}{3}$

answer is D.

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Detailed Solution

Let $\cos^{- 1} (\frac{1}{2}) = x$
$\begin{array}{lrl} \Rightarrow \cos x = \frac{1}{2} = \cos \frac{π}{3} \\ \Rightarrow x = \frac{π}{3} \in [0, π] \end{array}$
$[∵ principal value branch of \cos^{- 1} is [0, π]]$
$Again, let \sin^{- 1} (\frac{1}{2}) = y$
$\begin{aligned} Again, let \sin^{- 1} (\frac{1}{2}) = y \\ \Rightarrow \sin y = \frac{1}{2} = \sin \frac{π}{6} \\ \Rightarrow y = \frac{π}{6} \in [- \frac{π}{2}, \frac{π}{2}] \end{aligned}$
$\begin{array}{l} [∵ principal value branch of \sin^{- 1} is [- \frac{π}{2}, \frac{π}{2}]] \\ ∴ \cos^{- 1} (\frac{1}{2}) + 2 \sin^{- 1} (\frac{1}{2}) = \frac{π}{3} + 2 \cdot \frac{π}{6} \\ = \frac{π}{3} + \frac{π}{3} = \frac{2 π}{3} \end{array}$