Problem 35 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\alpha=\sin ^{-1} \frac{\sqrt{3}}{2}+\sin ^{-1} \frac{1}{3}\) and \(\beta=\cos ^{-1} \frac{\sqrt{3}}{2}+\cos ^{-1} \frac{1}{3}\), then (A) \(\alpha>\beta\) (B) \(\alpha=\beta\) (C) \(\alpha<\beta\) (D) \(\alpha+\beta=2 \pi\)

Step-by-Step Solution

Answer

(A) \(\alpha > \beta\)

Step 1: Understand Inverse Functions

First, we must understand the inverse sinusoidal functions involved in the problem. We have \[ \alpha = \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) + \sin^{-1} \left( \frac{1}{3} \right) \] and \[ \beta = \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) + \cos^{-1} \left( \frac{1}{3} \right) \]. The values returned by \(\sin^{-1}\) (or arcsine) range from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), and the values from \(\cos^{-1}\) (or arccosine) range from \(0\) to \(\pi\).

Step 2: Calculate Individual Inverse Values

Calculate each of the inverse functions:- \( \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{3} \) because \( \sin \left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2} \).- \( \sin^{-1} \left( \frac{1}{3} \right) \rightarrow \text{Find in radians using a calculator as arcsine value} \).- \( \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) = \frac{\pi}{6} \) since \( \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} \).- \( \cos^{-1} \left( \frac{1}{3} \right) \rightarrow \text{Find in radians using a calculator as arccosine value} \).

Step 3: Evaluate α and β in Expression

Now, substitute each value back into \(\alpha\) and \(\beta\) expressions.Since we do not immediately compute \( \sin^{-1} \left(\frac{1}{3}\right) \) or \( \cos^{-1} \left(\frac{1}{3}\right) \) without a calculator, consider:- \( \alpha = \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) + \sin^{-1} \left( \frac{1}{3} \right) = \frac{\pi}{3} + \sin^{-1} \left( \frac{1}{3} \right)\)- \( \beta = \cos^{-1} \left( \frac{\sqrt{3}}{2} \right) + \cos^{-1} \left( \frac{1}{3} \right) = \frac{\pi}{6} + \cos^{-1} \left( \frac{1}{3} \right)\).

Step 4: Use Identity for Inverses

Recall the identity:\(\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}\)for all \(x\) between \([-1,1]\).Therefore,\[ \sin^{-1} \left( \frac{1}{3} \right) + \cos^{-1} \left( \frac{1}{3} \right) = \frac{\pi}{2}\].

Step 5: Put Together to Compare α and β

Using the identity from Step 4 and plugging back into expressions:\[\alpha = \frac{\pi}{3} + \frac{\pi}{2} = \frac{\pi}{3} + \frac{\pi}{2} = \frac{5\pi}{6}\]\[\beta = \frac{\pi}{6} + \frac{\pi}{2} = \frac{\pi}{2} + \frac{\pi}{6} = \frac{2\pi}{3}\]Compare \(\alpha\) and \(\beta\): since \(\frac{5\pi}{6} > \frac{2\pi}{3}\), we have \( \alpha > \beta \).

Step 6: Final Answer and Conclusion

The comparison shows \( \alpha > \beta \). Therefore, the correct answer is (A) \(\alpha > \beta\).

Key Concepts

Arcsine and Arccosine

Inverse trigonometric functions serve to reverse the effect of the basic trigonometric functions, allowing us to find angles when given the function values. The arcsine function, denoted as \( \sin^{-1}(x) \), gives us the angle whose sine is \( x \), and its range spans from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\). Conversely, the arccosine function, \( \cos^{-1}(x) \), finds the angle whose cosine is \( x \), with a range from \(0\) to \(\pi\). This is important to remember because it affects how the results are interpreted in different quadrants.

• Arcsine works effectively on the first and fourth quadrants.
• Arccosine is applicable predominantly in the first and second quadrants.

When dealing with these functions in calculations, especially without a calculator, it's useful to remember key angles and their respective sine and cosine values, like \( \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2} \) and \( \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \). This knowledge allows quick evaluations when exact computation is required.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for any value of the included variables. One of the most powerful identities utilized for inverse trigonometric functions is \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \) for any \( x \) in the interval \([-1, 1]\). This is derived from the Pythagorean identity and the fact that sine and cosine are cofunctions.

• This identity facilitates solving problems involving sums of inverse trigonometric functions.
• Knowing this identity allows for a quick calculation of unknown angles when one inverse function is solved.

In this example, it was used to determine that \( \sin^{-1}(\frac{1}{3}) + \cos^{-1}(\frac{1}{3}) = \frac{\pi}{2} \). Thus, identities help simplify expressions and conduct angle arithmetic effortlessly.

Angle Comparison

When comparing angles, it is crucial to ensure they are expressed similarly, usually in radians or degrees, and within the relevant ranges laid by their inverse functions.
Calculating \( \alpha \) and \( \beta \) from the problems: \( \alpha = \frac{\pi}{3} + \sin^{-1}(\frac{1}{3}) \) and \( \beta = \frac{\pi}{6} + \cos^{-1}(\frac{1}{3}) \). By substituting from our identity \( \sin^{-1}(\frac{1}{3}) + \cos^{-1}(\frac{1}{3}) = \frac{\pi}{2} \), it rearranges to give us \( \sin^{-1}(\frac{1}{3}) = \frac{\pi}{2} - \cos^{-1}(\frac{1}{3}) \). Substituting back, we find \( \alpha = \frac{5\pi}{6} \) and \( \beta = \frac{2\pi}{3} \). Since \( \frac{5\pi}{6} > \frac{2\pi}{3} \), we determine \( \alpha > \beta \).
This structured procedure helps avoid errors when working through complex trigonometric problems, giving clear criteria for direct angle comparison.