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Mathematics Question on Trigonometry

If a=sin⁡−1(sin⁡(5))a = \sin^{-1} (\sin(5)) and b=cos⁡−1(cos⁡(5))b = \cos^{-1} (\cos(5)), then a2+b2a^2 + b^2 is equal to

A

4π2+254\pi^2 + 25

B

8π2−40π+508\pi^2 - 40\pi + 50

C

4π2−20π+504\pi^2 - 20\pi + 50

D

2525

Answer

8π2−40π+508\pi^2 - 40\pi + 50

Explanation

Solution

Calculate a=sin⁡−1(sin⁡(5))a = \sin^{-1}(\sin(5)). To find aa, note that sin⁡−1(sin⁡(x))\sin^{-1}(\sin(x)) gives a result in the range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Since 5 is outside this range, we need to adjust it. We have:

a=sin⁡−1(sin⁡(5))=5−2π.a = \sin^{-1}(\sin(5)) = 5 - 2\pi.

Thus,

a=5−2π.a = 5 - 2\pi.

Calculate b=cos⁡−1(cos⁡(5))b = \cos^{-1}(\cos(5)). To find bb, note that cos⁡−1(cos⁡(x))\cos^{-1}(\cos(x)) gives a result in the range [0,π][0, \pi].

Since 5 is within this range, we can write:

b=cos⁡−1(cos⁡(5))=2π−5.b = \cos^{-1}(\cos(5)) = 2\pi - 5.

Calculate a2+b2a^2 + b^2. Now, substitute a=5−2πa = 5 - 2\pi and b=2π−5b = 2\pi - 5:

a2+b2=(5−2π)2+(2π−5)2.a^2 + b^2 = (5 - 2\pi)^2 + (2\pi - 5)^2.

Expanding both terms:

=(5−2π)2+(2π−5)2=(25−20π+4π2)+(4π2−20π+25).= (5 - 2\pi)^2 + (2\pi - 5)^2 = (25 - 20\pi + 4\pi^2) + (4\pi^2 - 20\pi + 25).

Combine like terms:

=8π2−40π+50.= 8\pi^2 - 40\pi + 50.

Thus, the answer is:

8π2−40π+508\pi^2 - 40\pi + 50