Educational

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

Domain of sin⁻¹(√(x² − 1)) Question: Find the domain of: \[ f(x) = \sin^{-1}\!\left(\sqrt{x^2 – 1}\right) \] Concept: For \( \sin^{-1}(t) \) to be defined: \[ -1 \leq t \leq 1 \] Also, the square root requires: \[ x^2 – 1 \geq 0 \] — Solution: Step 1: Square root condition \[ x^2 – 1

Domain of sin⁻¹(x) + sin(x) Question: Find the domain of the function: \[ f(x) = \sin^{-1}(x) + \sin(x) \] Concept: – \( \sin(x) \) is defined for all real numbers – \( \sin^{-1}(x) \) is defined only when: \[ -1 \leq x \leq 1 \] — Solution: Since \( \sin(x) \) has no restriction, the

Domain of sin⁻¹(x²) Question: Find the domain of the function: \[ f(x) = \sin^{-1}(x^2) \] Concept: For \( \sin^{-1}(t) \) to be defined: \[ -1 \leq t \leq 1 \] — Solution: Step 1: Apply condition \[ -1 \leq x^2 \leq 1 \] Step 2: Solve inequality Since \( x^2 \geq 0 \), the condition