Problem 48 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Assertion: The general solution of the equation \(2^{\sin } \theta+\) \(2^{\cos } \theta=2^{1-\frac{1}{\sqrt{2}}}\) is \(\theta=n \pi+\frac{\pi}{4}\) Reason: For any two numbers \(a\) and \(b\), A.M. \(\geq\) G.M.

Step-by-Step Solution

Answer

The solution to the equation is \(\theta = n\pi + \frac{\pi}{4}\).

Step 1: Understand the Equation

We are given the equation \(2^{\sin \theta} + 2^{\cos \theta} = 2^{1-\frac{1}{\sqrt{2}}}\). We need to find the general solution of this equation for \(\theta\).

Step 2: Apply A.M. ≥ G.M.

For two numbers \(a = 2^{\sin \theta}\) and \(b = 2^{\cos \theta}\), the Arithmetic Mean and Geometric Mean inequality states \(\frac{a+b}{2} \geq \sqrt{ab}\). In this context, \(a + b = 2^{1-\frac{1}{\sqrt{2}}}\), so \(\frac{2^{\sin \theta} + 2^{\cos \theta}}{2} \geq \sqrt{2^{\sin \theta + \cos \theta}}\).

Step 3: Equality in A.M. = G.M.

Equality in the A.M. ≥ G.M. inequality occurs when \(a = b\). For the equation \(2^{\sin \theta} = 2^{\cos \theta}\), it implies \(\sin \theta = \cos \theta\). This happens when \(\theta = n\pi + \frac{\pi}{4}\) for integer \(n\).

Step 4: Verify Solution

Substitute \(\theta = n\pi + \frac{\pi}{4}\) back into the original equation to verify. For \(\theta = n\pi + \frac{\pi}{4}\), both \(\sin \theta\) and \(\cos \theta\) equal \(\frac{1}{\sqrt{2}}\). So, the equation simplifies as follows: \(2^{1/\sqrt{2}} + 2^{1/\sqrt{2}} = 2^{1-\frac{1}{\sqrt{2}}}\), which satisfies the given equation.

Step 5:Conclusion

Therefore, the general solution to the equation is \(\theta = n\pi + \frac{\pi}{4}\), matching the assertion in the problem statement.

Key Concepts

Arithmetic Mean-Geometric Mean Inequality

The Arithmetic Mean-Geometric Mean (AM-GM) Inequality is a fundamental concept in mathematics, particularly useful in analyzing trigonometric equations like the one in our exercise. It provides a relationship between the arithmetic and geometric means of two non-negative numbers.
For two numbers, say \(a\) and \(b\), the AM-GM Inequality states:

• The arithmetic mean: \(\frac{a+b}{2}\).
• The geometric mean: \(\sqrt{ab}\).

According to the inequality, \(\frac{a+b}{2} \geq \sqrt{ab}\), meaning that the arithmetic mean will always be greater than or equal to the geometric mean of those numbers.
In the context of our trigonometric equation problem, we applied this inequality by setting \(a = 2^{\sin \theta}\) and \(b = 2^{\cos \theta}\). When the inequality \(\frac{2^{\sin \theta} + 2^{\cos \theta}}{2} \geq \sqrt{2^{\sin \theta + \cos \theta}}\) holds true, and when equality \(a = b\) is achieved, it implies that \(\sin \theta = \cos \theta\), leading to a key part of the equation's solution.

Solution Verification

Solution verification involves testing the general solution of an equation to ensure it satisfies the original equation. This step is crucial because it confirms the validity of the solution.
For the given equation \(2^{\sin \theta} + 2^{\cos \theta} = 2^{1-\frac{1}{\sqrt{2}}}\), we proposed a general solution \(\theta = n\pi + \frac{\pi}{4}\). To verify this, we substitute back into the equation.
When \(\theta = n\pi + \frac{\pi}{4}\), both \(\sin \theta\) and \(\cos \theta\) become \(\frac{1}{\sqrt{2}}\). Substituting these values, the equation becomes \(2^{1/\sqrt{2}} + 2^{1/\sqrt{2}} = 2^{1-\frac{1}{\sqrt{2}}}\).
By this calculation, the values on both sides of the equation match, confirming that our proposed \(\theta\) is indeed a valid solution.

General Solution

The concept of a general solution in trigonometric equations is to find an expression for \(\theta\) that satisfies the equation for all valid integers \(n\). It addresses repetitive aspects of trigonometric functions.
In this problem, the solution gives \(\theta = n\pi + \frac{\pi}{4}\), indicating a periodic nature inherent to \sin and \cos functions, where \(n\) can be any integer.
Why \(\theta = n\pi + \frac{\pi}{4}\) works:

• Trigonometric functions have a period of \(\pi\). This means that adding any multiple of \(\pi\) to \(\theta\) results in the same \sin and \cos values.
• \(\frac{\pi}{4}\) ensures each \sin \theta and \cos \theta is equal.

Thus, the general solution captures all possible angles that satisfy the original equation, providing a comprehensive answer to the problem statement.