Evaluate.
Q.2. Using principal values, evaluate the following.
$\cos^{- 1} (\cos \frac{2 π}{3}) + \sin^{- 1} (\sin \frac{2 π}{3})$ .

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Solution

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$\cos^{- 1} (\cos \frac{2 π}{3}) + \sin^{- 1} (\sin \frac{2 π}{3}) C o n s i d e r, \cos^{- 1} (\cos \frac{2 π}{3}) = \cos^{- 1} (\frac{- 1}{2}) I f a = \cos^{- 1} b \Rightarrow \cos (a) = b f o r 0 \leq a \leq π x = \cos^{- 1} (\frac{- 1}{2}) \Rightarrow cosx = \frac{- 1}{2}, 0 \leq x \leq π Now, cosx = \frac{- 1}{2}, 0 \leq x \leq π General solution for cosx = \frac{- 1}{2} x = \frac{2 π}{3} + 2 nπ, x = \frac{4 π}{3} + 2 nπ Solutions for the range 0 \leq x \leq π x = \frac{2 π}{3} Consider, \sin^{- 1} (\sin \frac{2 π}{3}) = \sin^{- 1} (\frac{\sqrt{3}}{2}) If a = \sin^{- 1} b \Rightarrow \sin (a) = b for \frac{- π}{2} \leq a \leq \frac{π}{2} x = \sin^{- 1} (\frac{\sqrt{3}}{2}) \Rightarrow sinx = \frac{\sqrt{3}}{2}, \frac{- π}{2} \leq x \leq \frac{π}{2} General solution for sinx = \frac{\sqrt{3}}{2} Now, sinx = \frac{\sqrt{3}}{2}, \frac{- π}{2} \leq x \leq \frac{π}{2} x = \frac{π}{3} + 2 nπ, x = \frac{2 π}{3} + 2 nπ Solutions for the range \frac{- π}{2} \leq x \leq \frac{π}{2} x = \frac{π}{3} So, \cos^{- 1} (\cos \frac{2 π}{3}) + \sin^{- 1} (\sin \frac{2 π}{3}) = \frac{2 π}{3} + \frac{π}{3} = π$
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