JEE Challenger
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MathMedAdv 2022 • Paper 1

Value of Inverse Trigonometric Expression involving Square Roots and Pi

Considering only the principal values of the inverse trigonometric functions, the value of 32cos⁡−122+π2+14sin⁡−122π2+π2+tan⁡−12π\frac{3}{2} \cos^{-1} \sqrt{\frac{2}{2 + \pi^2}} + \frac{1}{4} \sin^{-1} \frac{2\sqrt{2} \pi}{2 + \pi^2} + \tan^{-1} \frac{\sqrt{2}}{\pi} is __________ .

Official Numerical Answer2.36

Topics & Concepts

Inverse Trigonometric FunctionsInverse Trigonometric Functions

Step-by-Step Solution

To find the value of the given expression considering only the principal values of the inverse trigonometric functions:

Let E=32cos⁡−122+π2+14sin⁡−122π2+π2+tan⁡−12π\text{Let } E = \frac{3}{2} \cos^{-1} \sqrt{\frac{2}{2 + \pi^2}} + \frac{1}{4} \sin^{-1} \frac{2\sqrt{2} \pi}{2 + \pi^2} + \tan^{-1} \frac{\sqrt{2}}{\pi}

Let θ=tan⁡−1(2π)\theta = \tan^{-1} \left(\frac{\sqrt{2}}{\pi}\right).

Since 2π>0\frac{\sqrt{2}}{\pi} > 0, we have θ∈(0,π2)\theta \in \left(0, \frac{\pi}{2}\right) with: tan⁡θ=2π\tan \theta = \frac{\sqrt{2}}{\pi}

Using a right-angled triangle with perpendicular 2\sqrt{2} and base π\pi, the hypotenuse is: Hypotenuse=(2)2+π2=2+π2\text{Hypotenuse} = \sqrt{(\sqrt{2})^2 + \pi^2} = \sqrt{2 + \pi^2}

From this, we get: sin⁡θ=22+π2=22+π2\sin \theta = \frac{\sqrt{2}}{\sqrt{2 + \pi^2}} = \sqrt{\frac{2}{2 + \pi^2}} cos⁡θ=π2+π2\cos \theta = \frac{\pi}{\sqrt{2 + \pi^2}}

Step 1: Simplify the first term

Since sin⁡θ=22+π2\sin \theta = \sqrt{\frac{2}{2 + \pi^2}} and θ∈(0,π4)\theta \in \left(0, \frac{\pi}{4}\right), we can write: cos⁡−122+π2=cos⁡−1(sin⁡θ)=cos⁡−1(cos⁡(π2−θ))\cos^{-1} \sqrt{\frac{2}{2 + \pi^2}} = \cos^{-1} (\sin \theta) = \cos^{-1} \left(\cos\left(\frac{\pi}{2} - \theta\right)\right)

Since (π2−θ)∈(π4,π2)⊂[0,π]\left(\frac{\pi}{2} - \theta\right) \in \left(\frac{\pi}{4}, \frac{\pi}{2}\right) \subset [0, \pi], we have: cos⁡−122+π2=π2−θ\cos^{-1} \sqrt{\frac{2}{2 + \pi^2}} = \frac{\pi}{2} - \theta

Step 2: Simplify the second term

Using the double-angle identity for sine: sin⁡(2θ)=2sin⁡θcos⁡θ=2⋅22+π2⋅π2+π2=22π2+π2\sin(2\theta) = 2 \sin \theta \cos \theta = 2 \cdot \frac{\sqrt{2}}{\sqrt{2 + \pi^2}} \cdot \frac{\pi}{\sqrt{2 + \pi^2}} = \frac{2\sqrt{2}\pi}{2 + \pi^2}

Since tan⁡θ=2π<1\tan \theta = \frac{\sqrt{2}}{\pi} < 1, we have 0<θ<π40 < \theta < \frac{\pi}{4}, which implies 0<2θ<π20 < 2\theta < \frac{\pi}{2}. Since 2θ∈[−π2,π2]2\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], the principal value gives: sin⁡−1(22π2+π2)=sin⁡−1(sin⁡2θ)=2θ\sin^{-1} \left(\frac{2\sqrt{2}\pi}{2 + \pi^2}\right) = \sin^{-1}(\sin 2\theta) = 2\theta

Step 3: Substitute back into the expression

Substitute the simplified forms into the expression EE: E=32(π2−θ)+14(2θ)+θE = \frac{3}{2} \left(\frac{\pi}{2} - \theta\right) + \frac{1}{4} (2\theta) + \theta

E=3π4−32θ+12θ+θE = \frac{3\pi}{4} - \frac{3}{2}\theta + \frac{1}{2}\theta + \theta

E=3π4+(−32+12+1)θE = \frac{3\pi}{4} + \left(-\frac{3}{2} + \frac{1}{2} + 1\right)\theta

E=3π4E = \frac{3\pi}{4}

Numerical Value:

E=3×3.14159265...4≈2.35619...E = \frac{3 \times 3.14159265...}{4} \approx 2.35619...

Rounding off to two decimal places gives 2.362.36 (or 2.352.35 upon truncating).

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