Ravi Kant Kumar

Value of cos²(½ cos⁻¹(3/5)) Question Find the value of: \[ \cos^2\left(\frac{1}{2}\cos^{-1}\left(\frac{3}{5}\right)\right) \] Solution Let \[ \theta = \cos^{-1}\left(\frac{3}{5}\right) \] Then, \[ \cos \theta = \frac{3}{5} \] Using identity: \[ \cos^2\left(\frac{\theta}{2}\right) = \frac{1 + \cos \theta}{2} \] Substitute: \[ \cos^2\left(\frac{\theta}{2}\right) = \frac{1 + \frac{3}{5}}{2} \] \[ = \frac{\frac{8}{5}}{2} = \frac{8}{10} = \frac{4}{5} \] Final Answer: \[ […]

Value of cos⁻¹(cos 350°) − sin⁻¹(sin 350°) Question Find the value of: \[ \cos^{-1}(\cos 350^\circ) – \sin^{-1}(\sin 350^\circ) \] Solution We use principal value ranges: \( \cos^{-1}x \in [0^\circ, 180^\circ] \) \( \sin^{-1}x \in [-90^\circ, 90^\circ] \) First, \[ \cos^{-1}(\cos 350^\circ) \] Since \( 350^\circ = 360^\circ – 10^\circ \), \[ \cos 350^\circ = \cos

Value of cos(2sin⁻¹(1/2)) Question Find the value of: \[ \cos\left(2\sin^{-1}\left(\frac{1}{2}\right)\right) \] Solution Let \[ \theta = \sin^{-1}\left(\frac{1}{2}\right) \] Then, \[ \sin \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{6} \] Now, \[ \cos(2\theta) = \cos\left(2 \cdot \frac{\pi}{6}\right) = \cos\left(\frac{\pi}{3}\right) \] \[ = \frac{1}{2} \] Final Answer: \[ \boxed{\frac{1}{2}} \] Key Concept Use known standard values to

Value of cos⁻¹(tan 3π/4) Question Find the value of: \[ \cos^{-1}(\tan \tfrac{3\pi}{4}) \] Solution First, evaluate: \[ \tan \tfrac{3\pi}{4} = -1 \] So the expression becomes: \[ \cos^{-1}(-1) \] We know that: \[ \cos \pi = -1 \] And the principal value range of \( \cos^{-1}x \) is: \[ [0, \pi] \] Thus, \[ \cos^{-1}(-1)

Evaluate sin(tan⁻¹(3/4)) Question Evaluate: \[ \sin(\tan^{-1}\left(\frac{3}{4}\right)) \] Solution Let \[ \theta = \tan^{-1}\left(\frac{3}{4}\right) \] Then, \[ \tan \theta = \frac{3}{4} = \frac{\text{Opposite}}{\text{Adjacent}} \] Take a right triangle: Opposite = 3 Adjacent = 4 Hypotenuse: \[ \sqrt{3^2 + 4^2} = 5 \] Now, \[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3}{5} \] Final Answer: \[ \boxed{\frac{3}{5}} \]

Evaluate sin(½ cos⁻¹(4/5)) Question Evaluate: \[ \sin\left(\frac{1}{2}\cos^{-1}\left(\frac{4}{5}\right)\right) \] Solution Let \[ \theta = \cos^{-1}\left(\frac{4}{5}\right) \] Then, \[ \cos \theta = \frac{4}{5} \] Construct a right triangle: Adjacent = 4 Hypotenuse = 5 Opposite = 3 So, \[ \sin \theta = \frac{3}{5} \] Now use half-angle identity: \[ \sin\frac{\theta}{2} = \sqrt{\frac{1 – \cos \theta}{2}} \] Substitute:

Value of sin⁻¹(sin 1550°) Question Find the value of: \[ \sin^{-1}(\sin 1550^\circ) \] Solution First, reduce the angle: \[ 1550^\circ = 4 \times 360^\circ + 110^\circ \] \[ \sin 1550^\circ = \sin 110^\circ \] Now evaluate: \[ \sin^{-1}(\sin 110^\circ) \] The principal value range of \( \sin^{-1}x \) is: \[ [-90^\circ, 90^\circ] \] Since \(

Value of cos(2sin⁻¹(1/3)) Question Find the value of: \[ \cos\left(2\sin^{-1}\left(\frac{1}{3}\right)\right) \] Solution Let \[ \theta = \sin^{-1}\left(\frac{1}{3}\right) \] Then, \[ \sin \theta = \frac{1}{3} \] Using identity: \[ \cos 2\theta = 1 – 2\sin^2\theta \] Substitute: \[ \cos 2\theta = 1 – 2\left(\frac{1}{3}\right)^2 \] \[ = 1 – 2 \cdot \frac{1}{9} = 1 – \frac{2}{9}

Value of sin⁻¹(sin −600°) Question Find the value of: \[ \sin^{-1}(\sin (-600^\circ)) \] Solution First, reduce the angle: \[ -600^\circ = -600^\circ + 720^\circ = 120^\circ \] \[ \sin(-600^\circ) = \sin(120^\circ) \] Now evaluate: \[ \sin^{-1}(\sin 120^\circ) \] The principal value range of \( \sin^{-1}x \) is: \[ [-90^\circ, 90^\circ] \] Since \( 120^\circ \)

Value of cos⁻¹(cos 1540°) Question Find the value of: \[ \cos^{-1}(\cos 1540^\circ) \] Solution First, reduce the angle using periodicity: \[ 1540^\circ = 4 \times 360^\circ + 100^\circ \] \[ \cos 1540^\circ = \cos 100^\circ \] Now, we evaluate: \[ \cos^{-1}(\cos 100^\circ) \] The principal value range of \( \cos^{-1}x \) is: \[ [0^\circ, 180^\circ]