Problem
Considering only the principal values of the inverse trigonometric functions, find the value of:
\( \frac{3}{2}\cos^{-1}\sqrt{\frac{2}{2+\pi^2}} + \frac{1}{4}\sin^{-1}\frac{2\sqrt{2}\pi}{2+\pi^2}+\tan^{-1}\frac{\sqrt{2}}{\pi} \)
Solution
Scary looking terms! But that means things will get simplified when we try to combine them – there’s no other way. Otherwise, there would have been simpler / familar terms, e.g. tan-11, cos-11/2, etc.
So, we’ll proceed by assuming \( \tan^{-1}\frac{\sqrt{2}}{\pi} \) = θ. Then, we’ll try to convert the entire expression in terms of θ. Finally, we’ll hope that the terms involving θ cancel out, and we’ll be left with something that’s easy to calculate.
Now, if tan-1√2/π = θ, then √2/π = tanθ. Since 0 < √2/π < 1, we have 0 < θ < π/4.
Using this relation, the expression \( \sqrt{\frac{2}{2+\pi^2}} \) simplifies to |sinθ|.
And, the expression \( \frac{2\sqrt{2}\pi}{2+\pi^2} \) simplifies to sin2θ.
So, the given expression becomes (3/2)cos-1|sinθ| + (1/4)sin-1(sin2θ) + θ.
Now, since sinθ is positive, we can write |sinθ| as sinθ, and using the identity sin-1x + cos-1x = π/2, the first term changes to (3/2)(π/2 – sin-1sinθ).
Since 0 < θ < π/4, sin-1sinθ equals θ. So the first term simplifies to (3/2)(π/2 – θ) or 3π/4 – 3θ/2.
Coming to the second term – since 0 < θ < π/4, we have 0 < 2θ < π/2, and therefore sin-1(sin2θ) = 2θ.
So, the second term becomes (1/4)(2θ) or θ/2.
Finally, if we add the three terms, we’ll get the sum as 3π/4 – 3θ/2 + θ/2 + θ, which simplies to 3π/4. Nice!
The question paper asked to express the answer as a decimal, after rounding off to two decimal places.
For that, we can approximate π as 3.14, and we’ll get the value as 2.36.
Comments
A nice opening / warm-up problem – though nothing too difficult. Things would’ve gotten interesting had θ been in the range (π/4, π/2). What do you think would’ve changed in the steps above?
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