Animated Solution for Mathematics - Inverse Trigonometric Functions: If $a={\sin }^{-1}(\sin (5))$ and $b={\cos }^{-1}(\cos (5))$, then $a^2+b^2$ is equal to

Welcome, fellow traveler of the mathematical landscape. Today, we confront a problem that looks deceptively simple.

You see $a={\sin }^{-1}(\sin 5)$ and $b={\cos }^{-1}(\cos 5)$, and your intuition screams, 'The answer is just $5^2+5^2=50$!' But pause.

In the world of JEE Advanced, intuition is a powerful tool, but it must be tempered by the rigor of definitions. We are dealing with inverse trigonometric functions, and these functions are not simple identity mappings. They are restricted, guarded by principal value branches.

Let us peel back the layers of this problem together.

Mapping the Territory

First, we must locate our input, $5$ radians, on the number line. We know that $\pi \approx 3.14$.

Therefore, $\frac{3\pi }{2}\approx 4.71$ and $2\pi \approx 6.28$. Our value, $5$, sits comfortably between $4.71$ and $6.28$.

This places $5$ in the fourth quadrant of the unit circle. We are not in the principal range of $[-\frac{\pi }{2},\frac{\pi }{2}]$ for sine, nor $[0,\pi ]$ for cosine. We must find the equivalent angles within those principal ranges.

The Sine Inverse Branch

Let us tackle $a={\sin }^{-1}(\sin 5)$. The graph of $y={\sin }^{-1}(\sin x)$ is a beautiful, repeating sawtooth wave.

In the interval $[\frac{3\pi }{2},\frac{5\pi }{2}]$, the graph is a straight line defined by the equation:

Substituting our value $x=5$, we get $a=5-2\pi$. Notice that since $5<6.28$, $a$ is a negative value. This makes sense; the sine of $5$ radians is negative, so its inverse must return a negative angle.

The Cosine Inverse Branch

Now, consider $b={\cos }^{-1}(\cos 5)$. The cosine inverse function is even more elegant.

In the interval $[\pi ,2\pi ]$, the graph is a line with a negative slope, defined by:

Substituting $x=5$, we find $b=2\pi -5$. Observe the beauty here: $b$ is positive. This is consistent with the definition of ${\cos }^{-1}$, which always outputs values in $[0,\pi ]$.

The Symmetry of Squares

Here is where the problem rewards our patience. We have $a=5-2\pi$ and $b=2\pi -5$.

If you look closely, you will see that $b=-(5-2\pi )$, which means $a=-b$. They are additive inverses!

When we calculate $a^2+b^2$, the negative sign vanishes. We are essentially calculating $2(5-2\pi {)}^2$. This symmetry is the key that unlocks the final calculation.

We expand $(5-2\pi {)}^2$ using the identity $(x-y{)}^2=x^2-2xy+y^2$. This gives us:

Multiplying this entire expression by $2$, we arrive at $50-40\pi +8{\pi }^2$. Rearranging to the standard form, we get the final result:

We have navigated the trap, respected the domain, and arrived at the truth. Keep this rigor in your toolkit, and no inverse trig problem will ever intimidate you again.