Ravi Kant Kumar

Evaluate cos⁻¹(cos 3) Evaluate \( \cos^{-1}(\cos 3) \) Step-by-Step Solution We need to evaluate: \[ \cos^{-1}(\cos 3) \] Step 1: Principal value range The principal value range of \( \cos^{-1}x \) is: \[ [0, \pi] \] Step 2: Check the angle Since \( 3 \in (0, \pi) \), it already lies in the principal range. […]

Evaluate cos⁻¹(cos 13π/6) Evaluate \( \cos^{-1}(\cos \frac{13\pi}{6}) \) Step-by-Step Solution We need to evaluate: \[ \cos^{-1}\left(\cos \frac{13\pi}{6}\right) \] Step 1: Reduce the angle \[ \frac{13\pi}{6} = 2\pi + \frac{\pi}{6} \] \[ \cos\left(\frac{13\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) \] Step 2: Apply inverse cosine \[ \cos^{-1}\left(\cos \frac{\pi}{6}\right) \] The principal value range of \( \cos^{-1}x \) is: \[ [0,

Evaluate cos⁻¹(cos 4π/3) Evaluate \( \cos^{-1}(\cos \frac{4\pi}{3}) \) Step-by-Step Solution We need to evaluate: \[ \cos^{-1}\left(\cos \frac{4\pi}{3}\right) \] Step 1: Principal value range The principal value range of \( \cos^{-1}x \) is: \[ [0, \pi] \] Step 2: Adjust the angle Since \( \frac{4\pi}{3} > \pi \), we use identity: \[ \cos(2\pi – x) =

Evaluate cos⁻¹(cos 5π/4) Evaluate \( \cos^{-1}(\cos \frac{5\pi}{4}) \) Step-by-Step Solution We need to evaluate: \[ \cos^{-1}\left(\cos \frac{5\pi}{4}\right) \] Step 1: Principal value range The principal value range of \( \cos^{-1}x \) is: \[ [0, \pi] \] Step 2: Adjust the angle Since \( \frac{5\pi}{4} > \pi \), we use identity: \[ \cos(2\pi – x) =

Evaluate cos⁻¹(cos −π/4) Evaluate \( \cos^{-1}(\cos -\frac{\pi}{4}) \) Step-by-Step Solution We need to evaluate: \[ \cos^{-1}\left(\cos -\frac{\pi}{4}\right) \] Step 1: Use identity \[ \cos(-x) = \cos x \] \[ \cos\left(-\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) \] Step 2: Apply inverse cosine \[ \cos^{-1}\left(\cos \frac{\pi}{4}\right) \] The principal value range of \( \cos^{-1}x \) is: \[ [0, \pi] \]

Evaluate sin⁻¹(sin 2) Evaluate \( \sin^{-1}(\sin 2) \) Step-by-Step Solution We need to evaluate: \[ \sin^{-1}(\sin 2) \] Step 1: Principal value range The principal value range of \( \sin^{-1}x \) is: \[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] Step 2: Check the angle Since \( 2 \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), we convert it using identity: \[ \sin(\pi

Evaluate sin⁻¹(sin 12) Evaluate \( \sin^{-1}(\sin 12) \) Step-by-Step Solution We need to evaluate: \[ \sin^{-1}(\sin 12) \] Step 1: Principal value range The principal value range of \( \sin^{-1}x \) is: \[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] Step 2: Reduce the angle Since 12 is not in the principal range, we find an equivalent angle: \[

Evaluate sin⁻¹(sin 4) Evaluate \( \sin^{-1}(\sin 4) \) Step-by-Step Solution We need to evaluate: \[ \sin^{-1}(\sin 4) \] Step 1: Principal value range The principal value range of \( \sin^{-1}x \) is: \[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] Step 2: Identify interval of 4 Since \( 4 \in (\pi, \frac{3\pi}{2}) \), we write: \[ 4 = \pi

Evaluate sin⁻¹(sin 3) Evaluate \( \sin^{-1}(\sin 3) \) Step-by-Step Solution We need to evaluate: \[ \sin^{-1}(\sin 3) \] Step 1: Principal value range The principal value range of \( \sin^{-1}x \) is: \[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] Step 2: Check the angle Since \( 3 \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), we cannot directly write: \[ \sin^{-1}(\sin 3)

Evaluate sin⁻¹(sin −17π/8) Evaluate \( \sin^{-1}(\sin -\frac{17\pi}{8}) \) Step-by-Step Solution We need to evaluate: \[ \sin^{-1}\left(\sin -\frac{17\pi}{8}\right) \] Step 1: Use identity \[ \sin(-x) = -\sin x \] \[ \sin\left(-\frac{17\pi}{8}\right) = -\sin\left(\frac{17\pi}{8}\right) \] Step 2: Reduce the angle \[ \frac{17\pi}{8} = 2\pi + \frac{\pi}{8} \] \[ \sin\left(\frac{17\pi}{8}\right) = \sin\left(\frac{\pi}{8}\right) \] \[ \sin\left(-\frac{17\pi}{8}\right) = -\sin\left(\frac{\pi}{8}\right) \]