Educational

Solve inverse trig equation Question Solve: \[ 3\sin^{-1}\left(\frac{2x}{1+x^2}\right) – 4\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) + 2\tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{3} \] Solution Let \[ x = \tan\theta \] Then use standard identities: \[ \frac{2x}{1+x^2} = \sin 2\theta \] \[ \frac{1-x^2}{1+x^2} = \cos 2\theta \] \[ \frac{2x}{1-x^2} = \tan 2\theta \] So equation becomes: \[ 3\sin^{-1}(\sin 2\theta) – 4\cos^{-1}(\cos 2\theta) + 2\tan^{-1}(\tan […]

Value of θ = sin⁻¹(sin(−600°)) Question If \[ \theta = \sin^{-1}(\sin(-600^\circ)) \] Find one possible value of \( \theta \). Solution First reduce the angle: \[ -600^\circ = -600^\circ + 720^\circ = 120^\circ \] \[ \sin(-600^\circ) = \sin(120^\circ) \] \[ = \sin(180^\circ – 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} \] Now, \[ \theta = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right)

Value of sin{2cos⁻¹(−3/5)} Question Evaluate: \[ \sin\left(2\cos^{-1}\left(-\frac{3}{5}\right)\right) \] Solution Let \[ \theta = \cos^{-1}\left(-\frac{3}{5}\right) \Rightarrow \cos\theta = -\frac{3}{5} \] Since \( \theta \in [0,\pi] \), angle lies in second quadrant ⇒ sinθ > 0 \[ \sin\theta = \sqrt{1 – \cos^2\theta} = \sqrt{1 – \frac{9}{25}} = \frac{4}{5} \] Now use identity: \[ \sin 2\theta = 2\sin\theta

Value of cos⁻¹(cos 5π/3) + sin⁻¹(sin 5π/3) Question Evaluate: \[ \cos^{-1}(\cos \tfrac{5\pi}{3}) + \sin^{-1}(\sin \tfrac{5\pi}{3}) \] Solution Step 1: Evaluate \( \cos^{-1}(\cos \tfrac{5\pi}{3}) \) Principal range of \( \cos^{-1}x \) is: \[ [0, \pi] \] \[ \frac{5\pi}{3} = 2\pi – \frac{\pi}{3} \Rightarrow \cos \tfrac{5\pi}{3} = \cos \tfrac{\pi}{3} \] \[ \cos^{-1}(\cos \tfrac{\pi}{3}) = \frac{\pi}{3} \] Step

If tan⁻¹3 + tan⁻¹x = tan⁻¹8, find x Question If \[ \tan^{-1}3 + \tan^{-1}x = \tan^{-1}8 \] Find \( x \). Solution Use identity: \[ \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \] So, \[ \tan^{-1}\left(\frac{3 + x}{1 – 3x}\right) = \tan^{-1}(8) \] Thus, \[ \frac{3 + x}{1 – 3x} = 8 \] \[ 3 + x

If tan⁻¹3 + tan⁻¹x = tan⁻¹8, find x Question If \[ \tan^{-1}3 + \tan^{-1}x = \tan^{-1}8 \] Find \( x \). Solution Use identity: \[ \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \] So, \[ \tan^{-1}\left(\frac{3 + x}{1 – 3x}\right) = \tan^{-1}(8) \] Thus, \[ \frac{3 + x}{1 – 3x} = 8 \] \[ 3 + x

Find expression from cos⁻¹(x/2) + cos⁻¹(y/3) = θ Question If \[ \cos^{-1}\left(\frac{x}{2}\right) + \cos^{-1}\left(\frac{y}{3}\right) = \theta \] Find: \[ 9x^2 – 12xy\cos\theta + 4y^2 \] Solution Let \[ \cos^{-1}\left(\frac{x}{2}\right) = A,\quad \cos^{-1}\left(\frac{y}{3}\right) = B \] Then, \[ A + B = \theta \] So, \[ \cos A = \frac{x}{2}, \quad \cos B = \frac{y}{3} \]

Value of tan⁻¹(1/11) + tan⁻¹(2/11) Question Evaluate: \[ \tan^{-1}\left(\frac{1}{11}\right) + \tan^{-1}\left(\frac{2}{11}\right) \] Solution Use identity: \[ \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \] Substitute \( a = \frac{1}{11}, b = \frac{2}{11} \): \[ = \tan^{-1}\left(\frac{\frac{1}{11} + \frac{2}{11}}{1 – \frac{2}{121}}\right) \] \[ = \tan^{-1}\left(\frac{3/11}{119/121}\right) \] \[ = \tan^{-1}\left(\frac{3}{11} \cdot \frac{121}{119}\right) = \tan^{-1}\left(\frac{33}{119}\right) \] \[ = \tan^{-1}\left(\frac{3}{\frac{119}{11}}\right) =

Find f(8π/9) Question Let \[ f(x) = e^{\cos^{-1}(\sin(x + \tfrac{\pi}{3}))} \] Find \( f\left(\tfrac{8\pi}{9}\right) \). Solution Substitute: \[ x + \frac{\pi}{3} = \frac{8\pi}{9} + \frac{\pi}{3} = \frac{8\pi}{9} + \frac{3\pi}{9} = \frac{11\pi}{9} \] Now, \[ \sin\left(\frac{11\pi}{9}\right) = \sin\left(\pi + \frac{2\pi}{9}\right) = -\sin\left(\frac{2\pi}{9}\right) \] So expression becomes: \[ \cos^{-1}\left(-\sin\left(\frac{2\pi}{9}\right)\right) \] Use identity: \[ \sin\theta = \cos\left(\frac{\pi}{2} –

Find α − β from inverse tangent expressions Question If \[ \alpha = \tan^{-1}\left(\frac{\sqrt{3}x}{2y – x}\right), \quad \beta = \tan^{-1}\left(\frac{2x – y}{\sqrt{3}y}\right) \] Find \( \alpha – \beta \). Solution Use identity: \[ \tan(\alpha – \beta) = \frac{\tan\alpha – \tan\beta}{1 + \tan\alpha \tan\beta} \] Substitute: \[ \tan(\alpha – \beta) = \frac{\frac{\sqrt{3}x}{2y – x} – \frac{2x