Evaluating inverse sine and inverse cosine at angle 5

Key concept: This question tests principal values of inverse trigonometric functions. The key is to rewrite $\sin^{-1}(\sin 5)$ and $\cos^{-1}(\cos 5)$ using the correct principal-value intervals, then square and add the results.

Step by step

Identify the principal value ranges

For inverse trigonometric functions, the output is not the original angle in general. Instead, it is the angle in the principal range:

• $\sin^{-1}(x)$ lies in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$
• $\cos^{-1}(x)$ lies in $[0,\pi]$

We need

$a=\sin^{-1}(\sin 5), \qquad b=\cos^{-1}(\cos 5).$

Since $5$ is in radians, compare it with $\pi$ and $2\pi$.

Compute a

Because $5$ lies between $\frac{3\pi}{2}$ and $2\pi$, the equivalent angle in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ is

$a = 5-2\pi.$

Compute b

For cosine, the principal value must be in $[0,\pi]$. The angle with the same cosine as 5 in that interval is

$b = 2\pi-5,$

because cosine is even and periodic.

Square and add

Now compute:

$a^2+b^2=(5-2\pi)^2+(2\pi-5)^2.$

These are equal, so

$a^2+b^2=2(5-2\pi)^2.$

Expand:

$2(25-20\pi+4\pi^2)=50-40\pi+8\pi^2.$

Thus,

$a^2+b^2=8\pi^2-40\pi+50.$

Final answer

$\boxed{8\pi^2-40\pi+50}$

Common idea behind the problem

Inverse trig functions do not simply undo every angle. They return a unique principal value. Once you place the given angle into the correct principal interval, the rest is direct substitution and algebra.

Pitfall alert

A very common mistake is to say $\sin^{-1}(\sin 5)=5$ and $\cos^{-1}(\cos 5)=5$. That ignores principal values. Another mistake is mixing degrees and radians, which completely changes the interval comparison. Since the question uses JEE-style notation, the angle is in radians unless stated otherwise. It is also easy to forget that cosine has a principal range starting at 0, so the inverse cosine answer cannot be negative. Always map the angle back into the correct output interval before squaring.

Try different conditions

If the problem were changed to $a=\sin^{-1}(\sin 4)$ and $b=\cos^{-1}(\cos 4)$, the same principal-value logic would apply. Since $4$ lies between $\pi$ and $\frac{3\pi}{2}$, the inverse sine result would be the angle in $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ with the same sine, and the inverse cosine result would be the angle in $[0,\pi]$ with the same cosine. The exact expressions would change, but the method remains: locate the angle, convert it to the principal range, then compute the required combination.

Further reading

principal value range, inverse trigonometric function, radian measure