Educational

Solve cos(2sin⁻¹(−x)) = 0 Problem Solve: \( \cos\left(2\sin^{-1}(-x)\right) = 0 \) Solution Use identity: \[ \cos(2\theta) = 1 – 2\sin^2\theta \] Let \( \theta = \sin^{-1}(-x) \) Then: \[ \sin \theta = -x \] Now, \[ \cos\left(2\sin^{-1}(-x)\right) = 1 – 2(-x)^2 \] \[ = 1 – 2x^2 \] Given: \[ 1 – 2x^2 = 0 […]

Solve cos(sin⁻¹x) = 1/6 Problem Solve: \( \cos(\sin^{-1}x) = \frac{1}{6} \) Solution Let \( \theta = \sin^{-1}x \) Then: \[ \sin \theta = x = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \] So take: Perpendicular = x Hypotenuse = 1 Base: \[ \sqrt{1 – x^2} \] Now, \[ \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} = \sqrt{1 – x^2} \] Given: \[ \sqrt{1

Prove sin(cos⁻¹(3/5) + sin⁻¹(5/13)) = 63/65 Problem Prove: \( \sin\left(\cos^{-1}\left(\frac{3}{5}\right) + \sin^{-1}\left(\frac{5}{13}\right)\right) = \frac{63}{65} \) Solution Let: \[ A = \cos^{-1}\left(\frac{3}{5}\right), \quad B = \sin^{-1}\left(\frac{5}{13}\right) \] Step 1: Find sin A and cos A \[ \cos A = \frac{3}{5} = \frac{\text{Base}}{\text{Hypotenuse}} \] Base = 3 Hypotenuse = 5 Perpendicular: \[ \sqrt{5^2 – 3^2} = \sqrt{25

Prove tan(sin⁻¹(5/13) + cos⁻¹(3/5)) = 63/16 Problem Prove: \( \tan\left(\sin^{-1}\left(\frac{5}{13}\right) + \cos^{-1}\left(\frac{3}{5}\right)\right) = \frac{63}{16} \) Solution Let: \[ A = \sin^{-1}\left(\frac{5}{13}\right), \quad B = \cos^{-1}\left(\frac{3}{5}\right) \] Step 1: Find tan A \[ \sin A = \frac{5}{13} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \] Perpendicular = 5 Hypotenuse = 13 Base: \[ \sqrt{13^2 – 5^2} = \sqrt{169 – 25} =

Prove cos(sin⁻¹(3/5) + cot⁻¹(3/2)) = 6/(5√13) Problem Prove: \( \cos\left(\sin^{-1}\left(\frac{3}{5}\right) + \cot^{-1}\left(\frac{3}{2}\right)\right) = \frac{6}{5\sqrt{13}} \) Solution Let: \[ A = \sin^{-1}\left(\frac{3}{5}\right), \quad B = \cot^{-1}\left(\frac{3}{2}\right) \] Step 1: Find sin A and cos A \[ \sin A = \frac{3}{5} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \] Perpendicular = 3 Hypotenuse = 5 Base: \[ \sqrt{5^2 – 3^2} = \sqrt{25

Prove tan(cos⁻¹(4/5) + tan⁻¹(2/3)) = 17/6 Problem Prove: \( \tan\left(\cos^{-1}\left(\frac{4}{5}\right) + \tan^{-1}\left(\frac{2}{3}\right)\right) = \frac{17}{6} \) Solution Let: \[ A = \cos^{-1}\left(\frac{4}{5}\right), \quad B = \tan^{-1}\left(\frac{2}{3}\right) \] Step 1: Find tan A \[ \cos A = \frac{4}{5} = \frac{\text{Base}}{\text{Hypotenuse}} \] Base = 4 Hypotenuse = 5 Perpendicular: \[ \sqrt{5^2 – 4^2} = \sqrt{25 – 16} =

Evaluate cos(tan⁻¹(24/7)) Problem Evaluate: \( \cos\left(\tan^{-1}\left(\frac{24}{7}\right)\right) \) Solution Let \( \theta = \tan^{-1}\left(\frac{24}{7}\right) \) Then: \[ \tan \theta = \frac{24}{7} = \frac{\text{Perpendicular}}{\text{Base}} \] Perpendicular = 24 Base = 7 Hypotenuse: \[ \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \] Now, using: \[ \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} \] \[ \cos \theta =

Evaluate cot(cos⁻¹(3/5)) Problem Evaluate: \( \cot\left(\cos^{-1}\left(\frac{3}{5}\right)\right) \) Solution Let \( \theta = \cos^{-1}\left(\frac{3}{5}\right) \) Then: \[ \cos \theta = \frac{3}{5} = \frac{\text{Base}}{\text{Hypotenuse}} \] Base = 3 Hypotenuse = 5 Perpendicular: \[ \sqrt{5^2 – 3^2} = \sqrt{25 – 9} = \sqrt{16} = 4 \] Now, using: \[ \cot \theta = \frac{\text{Base}}{\text{Perpendicular}} \] \[ \cot \theta =

Evaluate tan(cos⁻¹(8/17)) Problem Evaluate: \( \tan\left(\cos^{-1}\left(\frac{8}{17}\right)\right) \) Solution Let \( \theta = \cos^{-1}\left(\frac{8}{17}\right) \) Then: \[ \cos \theta = \frac{8}{17} = \frac{\text{Base}}{\text{Hypotenuse}} \] So, Base = 8 Hypotenuse = 17 Perpendicular: \[ \sqrt{17^2 – 8^2} = \sqrt{289 – 64} = \sqrt{225} = 15 \] Now, using: \[ \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \] \[ \tan \theta

Evaluate tan(cos⁻¹(8/17)) Problem Evaluate: \( \tan\left(\cos^{-1}\left(\frac{8}{17}\right)\right) \) Solution Let \( \theta = \cos^{-1}\left(\frac{8}{17}\right) \) Then: \[ \cos \theta = \frac{8}{17} \] Construct a right triangle: Adjacent = 8 Hypotenuse = 17 Opposite: \[ \sqrt{17^2 – 8^2} = \sqrt{289 – 64} = \sqrt{225} = 15 \] Now, \[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{15}{8} \] Therefore: