Question

Considering only the principal values of the inverse trigonometric functions, the value of $\frac{3}{2}cos^{-1}\sqrt{\frac{2}{2+\pi^2}}+\frac{1}{4}sin^{-1}\frac{2\sqrt{2}\pi}{2+\pi^2}+tan^{-1}\frac{\sqrt{2}}{\pi}$ is ___.

Answer 1

$\frac{3\pi}{4}$

Answer 2

Solution:

Let

$$A = \cos^{-1}\sqrt{\frac{2}{2+\pi^2}}.$$

Then

$$\cos A = \sqrt{\frac{2}{2+\pi^2}},\quad \sin A = \sqrt{1-\frac{2}{2+\pi^2}} = \frac{\pi}{\sqrt{2+\pi^2}},\quad \text{and} \quad \tan A = \frac{\sin A}{\cos A} = \frac{\pi}{\sqrt{2}}.$$

Thus,

$$A = \tan^{-1}\frac{\pi}{\sqrt{2}}.$$

Next, notice that

$$\sin 2A = 2\sin A \cos A = \frac{2\pi\sqrt{2}}{2+\pi^2}.$$

The second term in the expression is

$$\frac{1}{4}\sin^{-1}\frac{2\sqrt{2}\pi}{2+\pi^2}.$$

Since

$$\frac{2\sqrt{2}\pi}{2+\pi^2} = \sin 2A,$$

and because $2A$ (when not in $[-\frac{\pi}{2},\frac{\pi}{2}]$) has a principal value given by $\sin^{-1}(\sin2A)=\pi-2A$ (provided $2A$ lies between $\frac{\pi}{2}$ and $\pi$), we have

$$\sin^{-1}\frac{2\sqrt{2}\pi}{2+\pi^2} = \pi-2A.$$

Thus, the given expression becomes:

$$\frac{3}{2}A + \frac{1}{4}(\pi-2A) + \tan^{-1}\frac{\sqrt{2}}{\pi}.$$

Simplify:

$$\frac{3}{2}A - \frac{1}{2}A + \frac{\pi}{4} + \tan^{-1}\frac{\sqrt{2}}{\pi} = A + \frac{\pi}{4} + \tan^{-1}\frac{\sqrt{2}}{\pi}.$$

Now, substitute back $A=\tan^{-1}\frac{\pi}{\sqrt{2}}$:

$$\tan^{-1}\frac{\pi}{\sqrt{2}}+\tan^{-1}\frac{\sqrt{2}}{\pi}+\frac{\pi}{4}.$$

Using the sum formula for arctan, when

$$\tan^{-1}x + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2} \quad (\text{for } x>0),$$

with $x=\frac{\pi}{\sqrt{2}}$ and $\frac{1}{x}=\frac{\sqrt{2}}{\pi}$, we find:

$$\tan^{-1}\frac{\pi}{\sqrt{2}}+\tan^{-1}\frac{\sqrt{2}}{\pi} = \frac{\pi}{2}.$$

Thus, the entire expression simplifies to:

$$\frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}.$$

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Explanation (minimal):

1. Set $A=\cos^{-1}\sqrt{\frac{2}{2+\pi^2}}$ and compute $\sin A$ to get $\sin 2A = \frac{2\pi\sqrt{2}}{2+\pi^2}$.
2. Replace $\sin^{-1}\frac{2\sqrt{2}\pi}{2+\pi^2}$ by $\pi-2A$ (using the principal value).
3. Simplify the expression to $A+\frac{\pi}{4}+\tan^{-1}\frac{\sqrt{2}}{\pi}$ and write $A=\tan^{-1}\frac{\pi}{\sqrt{2}}$.
4. Use the arctan sum formula to obtain $\tan^{-1}\frac{\pi}{\sqrt{2}}+\tan^{-1}\frac{\sqrt{2}}{\pi}=\frac{\pi}{2}$.
5. Final answer: $\frac{3\pi}{4}$.