Principal Value of sin⁻¹(−√3/2) + cos⁻¹(√3/2) Evaluate: sin-1(−√3/2) + cos-1(√3/2) Solution: Using principal values: \[ \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3} \] (Since range of sin-1(x) is \([- \pi/2, \pi/2]\)) \[ \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \] (Since range of cos-1(x) is \([0, \pi]\)) Substitute values: \[ -\frac{\pi}{3} + \frac{\pi}{6} = -\frac{2\pi}{6} + \frac{\pi}{6} = -\frac{\pi}{6} \] Final Answer: Value […]

Principal Value of sin⁻¹(−1/2) + 2cos⁻¹(−√3/2) Evaluate: sin-1(−1/2) + 2cos-1(−√3/2) Solution: Using principal values: \[ \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \] (Since range of sin-1(x) is \([- \pi/2, \pi/2]\)) \[ \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6} \] (Since range of cos-1(x) is \([0, \pi]\)) Substitute values: \[ \sin^{-1}\left(-\frac{1}{2}\right) + 2\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{6} + 2 \times \frac{5\pi}{6} \] \[ = -\frac{\pi}{6}

Principal Value of cos⁻¹(1/2) − 2sin⁻¹(−1/2) Evaluate: cos-1(1/2) − 2sin-1(−1/2) Solution: We know the principal values: \[ \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \] \[ \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \] (Since the range of sin-1(x) is \([- \pi/2, \pi/2]\)) Substitute values: \[ \cos^{-1}\left(\frac{1}{2}\right) – 2\sin^{-1}\left(-\frac{1}{2}\right) = \frac{\pi}{3} – 2\left(-\frac{\pi}{6}\right) \] \[ = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3} \] Final

Principal Value of cos⁻¹(1/2) + 2sin⁻¹(1/2) Evaluate: cos-1(1/2) + 2sin-1(1/2) Solution: We know: \[ \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} \] \[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \] Substitute values: \[ \cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3} + 2 \times \frac{\pi}{6} \] \[ = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3} \] Final Answer: Value = \[ \frac{2\pi}{3} \] Next Question / Full

Principal Value of cos⁻¹(tan 3π/4) Find the Principal Value of cos-1(tan 3π/4) Solution: Given: \[ y = \cos^{-1}\left(\tan \frac{3\pi}{4}\right) \] Step 1: Evaluate tan(3π/4) \[ \tan \frac{3\pi}{4} = -1 \] So, \[ y = \cos^{-1}(-1) \] Step 2: Use principal value range The principal value range of cos-1(x) is: \[ [0, \pi] \] We know:

Principal Value of cos⁻¹(sin 4π/3) Find the Principal Value of cos-1(sin 4π/3) Solution: Given: \[ y = \cos^{-1}(\sin \tfrac{4\pi}{3}) \] First evaluate: \[ \sin \tfrac{4\pi}{3} = -\frac{\sqrt{3}}{2} \] So, \[ y = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) \] We know: \[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \] Thus, \[ \cos y = -\frac{\sqrt{3}}{2} = \cos\left(\frac{5\pi}{6}\right) \] Since principal value range of

Principal Value of cos⁻¹(−1/√2) Find the Principal Value of cos-1(−1/√2) Solution: Let \[ y = \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) \] Then, \[ \cos y = -\frac{1}{\sqrt{2}} \] We know that: \[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \] So, \[ \cos y = \cos\left(\pi – \frac{\pi}{4}\right) = \cos\left(\frac{3\pi}{4}\right) \] Since the principal value range of cos-1(x) is: \[ [0, \pi] \]

Principal Value of cos⁻¹(−√3/2) Find the Principal Value of cos-1(−√3/2) Solution: Let \[ y = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) \] Then, \[ \cos y = -\frac{\sqrt{3}}{2} \] We know that: \[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \] So, \[ \cos y = \cos\left(\pi – \frac{\pi}{6}\right) = \cos\left(\frac{5\pi}{6}\right) \] Since the principal value range of cos-1(x) is: \[ [0, \pi] \]

Domain of cos⁻¹(x) + cos x Find the Domain of f(x) = cos-1(x) + cos(x) Solution: Given function: \[ f(x) = \cos^{-1}(x) + \cos(x) \] Step 1: Domain of cos-1(x) \[ -1 \leq x \leq 1 \] Step 2: Domain of cos(x) \[ \cos(x) \text{ is defined for all real numbers} \] Step 3: Intersection

Domain of 2cos⁻¹(2x) + sin⁻¹(x) Find the Domain of f(x) = 2cos-1(2x) + sin-1(x) Solution: Given function: \[ f(x) = 2\cos^{-1}(2x) + \sin^{-1}(x) \] Step 1: Domain of cos-1(2x) \[ -1 \leq 2x \leq 1 \] \[ -\frac{1}{2} \leq x \leq \frac{1}{2} \] Step 2: Domain of sin-1(x) \[ -1 \leq x \leq 1 \]