Ravi Kant Kumar

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) \[ -1 \leq x \leq 1 \]

Find x²+y²+z² from squared inverse sine sum Question: If \[ (\sin^{-1}x)^2 + (\sin^{-1}y)^2 + (\sin^{-1}z)^2 = \frac{3\pi^2}{4} \] Find \(x^2 + y^2 + z^2\). Concept: The principal range: \[ -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2} \] Maximum square: \[ (\sin^{-1}x)^2 \leq \left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{4} \] — Solution: Given: \[ (\sin^{-1}x)^2 + (\sin^{-1}y)^2 + (\sin^{-1}z)^2 = \frac{3\pi^2}{4}

Domain of cos⁻¹(x² − 4) Find the Domain of Definition of f(x) = cos-1(x² − 4) Solution: Given function: \[ f(x) = \cos^{-1}(x^2 – 4) \] For the inverse cosine function, the input must lie in the interval: \[ -1 \leq x^2 – 4 \leq 1 \] Add 4 to all parts: \[ 3 \leq

Find x²+y²+z²+t² from sum of sin⁻¹ Question: If \[ \sin^{-1}x+\sin^{-1}y+\sin^{-1}z+\sin^{-1}t = 2\pi \] Find \(x^2 + y^2 + z^2 + t^2\). Concept: The principal value range of \( \sin^{-1}x \) is: \[ -\frac{\pi}{2} \leq \sin^{-1}x \leq \frac{\pi}{2} \] So each term is at most \( \frac{\pi}{2} \). — Solution: Maximum possible sum of four terms:

Domain of sin⁻¹(x) + sin⁻¹(2x) Question: Find the domain of: \[ f(x) = \sin^{-1}(x) + \sin^{-1}(2x) \] Concept: For \( \sin^{-1}(t) \) to be defined: \[ -1 \leq t \leq 1 \] — Solution: Step 1: Apply conditions From \( \sin^{-1}(x) \): \[ -1 \leq x \leq 1 \] From \( \sin^{-1}(2x) \): \[ -1