cos−1(cos7π/6) is equal to(A) 7π/6 (B) 5π/6 (C

1.	cos−1(cos7π/6) is equal to(A) 7π/6 (B) 5π/6 (C) π/3 (D) π/6
Answer» Answer: (b) \(\frac{5\pi}6\) Given that \(cos^{-1}(cos\frac{7\pi}6)\) We know that \(cos^{-1}(cos\,x)=x\,if\,x\in[0,\pi],\) which is the principle value branch \(\therefore cos^{-1}(cos\frac{7\pi}6)=cos^{-1}[cos(2\pi-\frac{5\pi}6)]\) \(cos(2\pi-\frac{5\pi}6)=-cos\,\frac{5\pi}6\) \(=cos^{-1}(cos(2\pi-\frac{5\pi}6))=cos^{-1}(-cos\frac{5\pi}6)\) \(=\pi-cos^{-1}(cos\frac{5\pi}6)\) \(=\pi-cos^{-1}(cos(\pi-\frac{\pi}6))\) \(=\pi-cos^{-1}(-cos\frac{5\pi}6)\) \(=\pi-(\pi-cos^{-1}cos\frac{5\pi}6)\) \(=cos^{-1}(cos(\frac{5\pi}6))\) \(=\frac{5\pi}6\)

cos−1(cos7π/6) is equal to(A) 7π/6 (B) 5π/6 (C) π/3 (D) π/6

Answer»

Answer: (b) \(\frac{5\pi}6\)

Given that \(cos^{-1}(cos\frac{7\pi}6)\)

We know that \(cos^{-1}(cos\,x)=x\,if\,x\in[0,\pi],\) which is the principle value branch

\(\therefore cos^{-1}(cos\frac{7\pi}6)=cos^{-1}[cos(2\pi-\frac{5\pi}6)]\)

\(cos(2\pi-\frac{5\pi}6)=-cos\,\frac{5\pi}6\)

\(=cos^{-1}(cos(2\pi-\frac{5\pi}6))=cos^{-1}(-cos\frac{5\pi}6)\)

\(=\pi-cos^{-1}(cos\frac{5\pi}6)\)

\(=\pi-cos^{-1}(cos(\pi-\frac{\pi}6))\)

\(=\pi-cos^{-1}(-cos\frac{5\pi}6)\)

\(=\pi-(\pi-cos^{-1}cos\frac{5\pi}6)\)

\(=cos^{-1}(cos(\frac{5\pi}6))\)

\(=\frac{5\pi}6\)