Educational

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 \[ \frac{17\pi}{8} = 2\pi + \frac{\pi}{8} \] \[ \sin\left(\frac{17\pi}{8}\right) = \sin\left(\frac{\pi}{8}\right) \] \[ \sin\left(-\frac{17\pi}{8}\right) = -\sin\left(\frac{\pi}{8}\right) \]

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 periodic property \[ \frac{17\pi}{8} = 2\pi + \frac{\pi}{8} \] \[ \sin\left(\frac{17\pi}{8}\right) = \sin\left(2\pi + \frac{\pi}{8}\right) = \sin\left(\frac{\pi}{8}\right) \] Step 2: Apply inverse sine \[ \sin^{-1}\left(\sin \frac{\pi}{8}\right) \] The principal value range of \( \sin^{-1}x

Evaluate sin⁻¹(sin 13π/7) Evaluate \( \sin^{-1}(\sin \frac{13\pi}{7}) \) Step-by-Step Solution We need to evaluate: \[ \sin^{-1}\left(\sin \frac{13\pi}{7}\right) \] Step 1: Use periodic property \[ \frac{13\pi}{7} = 2\pi – \frac{\pi}{7} \] \[ \sin\left(\frac{13\pi}{7}\right) = \sin\left(2\pi – \frac{\pi}{7}\right) = -\sin\left(\frac{\pi}{7}\right) \] Step 2: Apply inverse sine \[ \sin^{-1}\left(-\sin\frac{\pi}{7}\right) \] The principal value range of \( \sin^{-1}x \)

Evaluate sin⁻¹(sin 5π/6) Evaluate \( \sin^{-1}(\sin \frac{5\pi}{6}) \) Step-by-Step Solution We need to evaluate: \[ \sin^{-1}\left(\sin \frac{5\pi}{6}\right) \] Step 1: Find the value of sine \[ \sin \frac{5\pi}{6} = \frac{1}{2} \] Step 2: Apply inverse sine \[ \sin^{-1}\left(\frac{1}{2}\right) \] The principal value range of \( \sin^{-1}x \) is: \[ \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \] Since \( \frac{\pi}{6}