Find the principal value of \(cos^{-1} [cos (– 680°)]\).

1.	Find the principal value of \(cos^{-1} [cos (– 680°)]\).
Answer» \(cos^{-1}[cos(-680^o)]\)\(=cos^{-1}[cos(680^o)]\) \([\because cos(-\theta)=cos\,\theta]\) \(=cos^{-1}[cos\,(720^o-40^o)]\) \(=cos^{-1}[cos\,(4\pi-40^o)]\)\(=cos^{-1}(cos\,40^o)\) \(=40^o\) or \(\frac{2\pi}9\) \([\because40^o=\frac{2\pi}9\in/0,\pi/]\)

Answer»

\(cos^{-1}[cos(-680^o)]\)\(=cos^{-1}[cos(680^o)]\) \([\because cos(-\theta)=cos\,\theta]\)

\(=cos^{-1}[cos\,(720^o-40^o)]\)

\(=cos^{-1}[cos\,(4\pi-40^o)]\)\(=cos^{-1}(cos\,40^o)\)

\(=40^o\) or \(\frac{2\pi}9\) \([\because40^o=\frac{2\pi}9\in/0,\pi/]\)

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