Q.2. Using principal values, evaluate the following.
cos - 1 cos 2 π 3 + sin - 1 sin 2 π 3 .

Dear student
cos-1cos2π3+sin-1sin2π3Consider, cos-1cos2π3=cos-1-12If a=cos-1bcosa=b for  0aπx=cos-1-12cosx=-12, 0xπNow, cosx=-12  , 0xπGeneral solution for cosx=-12x=2π3+2, x=4π3+2Solutions for the range 0xπx=2π3Consider, sin-1sin2π3=sin-132If a=sin-1bsina=b for  -π2aπ2x=sin-132sinx=32, -π2xπ2General solution for sinx=32Now, sinx=32  ,-π2xπ2x=π3+2, x=2π3+2Solutions for the range -π2xπ2x=π3So, cos-1cos2π3+sin-1sin2π3=2π3+π3=π

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I too got the same answer but in back its written as "pi" is the required answer. :/ Thats why I got confused. Is it possible that the back page answer is wrong? I think so. Thanks :D

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cos^-1(cos(2π/3)) + sin^-1(sin(2π/3) = 2π/3 + sin^-1(sin(π-π/3)) (This is because principal value branch of cos^-1 lies in [0,π] while that of sin^-1 lies in [-π/2,π/2]) = 2π/3 + sin^-1(sin(π/3)) = 2π/3 + π/3 = 3π/3 = π P.S. It coincides with the answer at the back of your textbook! Sorry if the answer is not formatted. That's the fault of meritnation website.
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