Problem 16 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Find the greatest and least value of the function \(f(x)=\left(\sin ^{-1} x\right)^{3}+\left(\cos ^{-1} x\right)^{3}\)

Step-by-Step Solution

Answer

The maximum value of the function on its domain is \(f(1) = (\sin^{-1}(1))^3 + (\cos^{-1}(1))^3\) and the minimum value is \(f(-1) = (\sin^{-1}(-1))^3 + (\cos^{-1}(-1))^3\). These are the only extrema as the function has no critical points within its domain.

Step 1: Differentiation

Compute the derivative of \(f(x)\) using the chain rule and knowing that the derivatives of \(\sin^{-1}(x)\) and \(\cos^{-1}(x)\) are \(1/\sqrt{1-x^2}\) and \(-1/\sqrt{1-x^2}\) respectively. The derivative becomes \(3 (\sin^{-1})^2(x) (-1/\sqrt{1-x^2}) + 3 (\cos^{-1})^2(x) (1/\sqrt{1-x^2})\).

Step 2: Solving for Critical Points

Solve for \(x\) in the equation \(f'(x) = 0\) to find critical points. So, set \(3 (\sin^{-1})^2(x) (1/\sqrt{1-x^2}) + 3 (\cos^{-1})^2(x) (-1/\sqrt{1-x^2}) = 0\) and solve for \(x__). We realize, however, that the derivative will not equal to zero for any \(x\) in the domain of \(f(x)\).

Step 3: Considering the Endpoints of the Domain

Since no critical points within the domain were found, we check the endpoints of the domain, which is \([-1, 1]\) in this case. So, evaluate the given function at these endpoints: \(f(-1) = (\sin^{-1}(-1))^3 + (\cos^{-1}(-1))^3\) and \(f(1) = (\sin^{-1}(1))^3 + (\cos^{-1}(1))^3\).

Step 4: Finding Maximum and Minimum Values

Substitute \(-1\) and \(1\) from the previous step into \(f(x)\) respectively. Compare these two outputs and the larger one is the maximum value and the smaller one is the minimum value of the function on its domain.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus. It helps us understand how a function changes at any given point. For the function \( f(x) = (\sin^{-1}(x))^3 + (\cos^{-1}(x))^3 \), we need to find its derivative to locate critical points. Using the chain rule, the derivatives of \( \sin^{-1}(x) \) and \( \cos^{-1}(x) \) are \( \frac{1}{\sqrt{1-x^2}} \) and \( -\frac{1}{\sqrt{1-x^2}} \) respectively. Applying these, we differentiate \( f(x) \), obtaining

• \( 3 (\sin^{-1}(x))^2 \left(-\frac{1}{\sqrt{1-x^2}}\right) + 3 (\cos^{-1}(x))^2 \left(\frac{1}{\sqrt{1-x^2}}\right) \)

Differentiation here helps us find how \( f(x) \) changes, key to understanding its behavior throughout its domain.
It's crucial when searching for extremums, like the greatest or least values of the function.

Critical Points

Critical points are values of \( x \) where a function's derivative equals zero or is undefined. To find them for \( f(x) = (\sin^{-1}(x))^3 + (\cos^{-1}(x))^3 \), we set its derivative to zero:

• \( 3 (\sin^{-1}(x))^2 \left( -\frac{1}{\sqrt{1-x^2}} \right) + 3 (\cos^{-1}(x))^2 \left( \frac{1}{\sqrt{1-x^2}} \right) = 0 \)

Solving this equation, we notice the terms sum to zero nowhere in the domain \([-1, 1]\). This implies no internal critical points exist. This insight guides us to check the endpoints to find the function's extremums.

Domain of Functions

Understanding the domain is essential as it defines where the function exists. For inverse trigonometric functions like \( \sin^{-1}(x) \) and \( \cos^{-1}(x) \), the domain is \([-1, 1]\).
This is because:

• \( \sin^{-1}(x) \) and \( \cos^{-1}(x) \) are only defined for \( -1 \leq x \leq 1 \)
• The outputs are angles \([-rac{\pi}{2}, \frac{\pi}{2}]\) for \( \sin^{-1}(x) \) and \([0, \pi]\) for \( \cos^{-1}(x) \)

With no critical points found within the interval, checking endpoints is our next step for evaluating maximum and minimum values in the function's domain.

Trigonometric Identities

Trigonometric identities help simplify complex equations, especially when dealing with multiple trigonometric functions. In the problem with \( f(x) = (\sin^{-1}(x))^3 + (\cos^{-1}(x))^3 \), the identity \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \) is handy. This relationship is vital because it confirms the complementarity of these angles, useful when calculating values at endpoints like \(-1\) and \(1\).

• At \( x = -1 \): \( \sin^{-1}(-1) = -\frac{\pi}{2} \) and \( \cos^{-1}(-1) = \pi \)
• At \( x = 1 \): \( \sin^{-1}(1) = \frac{\pi}{2} \) and \( \cos^{-1}(1) = 0 \)

Using these, we calculate the values of \( f(x) \) at the boundaries for comparison. Recognizing these identities assists us in quickly deducing the main properties of inverse trigonometric functions in this context.