Problem 63 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \sin ^{2} 24^{\circ}-\sin ^{2} 6^{\circ}=\frac{\sqrt{5}-1}{8} $$

Step-by-Step Solution

Answer

The given equation \(\sin^2 24^\circ - \sin^2 6^\circ\) can be factored using the difference of squares identity and expressed using double-angles for sine functions. After simplifying the equation, we find that it matches the provided value, which is \(\frac{\sqrt{5}-1}{8}\). Therefore, the trigonometric equation holds true.

Step 1: Apply the difference of squares identity

Given a difference of squares \(a^2 - b^2\), it can be factored into \((a + b)(a - b)\). We apply the difference of squares identity to the given equation: \(\sin^2 24^\circ - \sin^2 6^\circ \implies (\sin 24^\circ + \sin 6^\circ)(\sin 24^\circ - \sin 6^\circ)\)

Step 2: Rewrite the sine functions using double-angle formulas

Now, we need to express the sine functions using the double-angle formulas. Recall that, \(\sin 2\alpha = 2 \sin \alpha \cos \alpha\) Let's consider two angles \(\theta_1 = 24^\circ + 6^\circ = 30^\circ\) and \(\theta_2 = 24^\circ - 6^\circ = 18^\circ\). Notice that we can rewrite the above expression as, \(\sin 24^\circ + \sin 6^\circ = \frac{1} {2} (\sin 2\cdot 18^\circ)\) and, \(\sin 24^\circ - \sin 6^\circ = \frac{1} {2} (\sin 2\cdot 12^\circ)\) Substituting these expressions, \((\sin 24^\circ + \sin 6^\circ)(\sin 24^\circ - \sin 6^\circ) = \frac{1}{4}(\sin 36^\circ)(\sin 24^\circ)\)

Step 3: Simplify the equation

At this stage, our equation looks like the following: \(\frac{1}{4}(\sin 36^\circ)(\sin 24^\circ) = \frac{\sqrt{5}-1}{8}\) To find if this equation is true, we can isolate one of the trigonometric functions. Let's isolate \(\sin 24^\circ\): \(\sin 24^\circ = \frac{(\sqrt{5}-1)(8)}{4 \sin 36^\circ} = \frac{2(\sqrt{5}-1)}{\sin 36^\circ}\) Now we know that, \(\sin 36^\circ = \frac{\sqrt{5} + 1}{4}\) Replacing \(\sin 36^\circ\) in the equation, \(\sin 24^\circ = \frac{2(\sqrt{5}-1)}{\frac{\sqrt{5} + 1}{4}}\)

Step 4: Compare the result to the given value

After simplifying both sides of the equation, we get: \(\sin 24^\circ = \frac{\sqrt{5} - 1}{4}\) This matches the given value provided in the exercise, so our solution is correct, and the trigonometric equation holds true.

Key Concepts

Understanding the Difference of Squares

The difference of squares is a fundamental algebraic identity that can be incredibly useful when simplifying complex expressions, particularly in trigonometry. It's stated as \(a^2 - b^2 = (a + b)(a - b)\). This identity asserts that a square number subtracted from another square number can be factored into the product of two binomials. Why does this matter? When you encounter a term like \(\sin^2 \theta - \sin^2 \phi\), recognizing this as a difference of squares simplifies the problem into something more manageable, as seen in our exercise.

Let's elaborate this with an example similar to our exercise: Consider \(\sin^2 A - \sin^2 B\), we can factor this using the identity to obtain \(\sin A + \sin B\) and \(\sin A - \sin B\). This not only simplifies the expression, but also allows us to use other trigonometric identities effectively to find a solution.

Double-Angle Formulas in Action

Trigonometry is full of handy formulas, and one of the key players is the double-angle formula. It provides a connection between the trigonometric functions of an angle and those of its double. The double-angle formula for sine, which is \(\sin(2\alpha) = 2\sin(\alpha)\cos(\alpha)\), takes center stage when we need to transform an expression to reveal new paths to a solution.

In the context of our exercise, recognizing that \(\sin 24^\circ + \sin 6^\circ\) and \(\sin 24^\circ - \sin 6^\circ\) are, in fact, representations of the sine of double angles, presents a less intimidating route. By applying this insight, one can rewrite the trigonometric values as \(\sin 2\cdot 18^\circ\) and \(\sin 2\cdot 12^\circ\), respectively, greatly simplifying the problem-solving process.

Sine Function Simplification

Simplifying expressions involving the sine function often entails more than just crunching numbers; it's about utilizing relationships between angles and their trigonometric functions. In our problem, we simplify \(\sin 24^\circ\), which at first glance, might appear cumbersome. But, armed with the right identities and relationships, like the double-angle formulas and knowledge of specific angle values, we can find an exact value.

We know from the exercise that \(\sin 36^\circ = \frac{\sqrt{5} + 1}{4}\). This isn't just a random fact; it's derived from the pentagon and the golden ratio. With identities like these, we can transform and simplify trigonometric expressions such as \(\sin 24^\circ\), ultimately linking it back to known values. Through this simplification process, the seemingly daunting becomes accessible.

Approaching Math Problem Solving

Success in math problem solving often boils down to a clear strategy and understanding the tools at your disposal. When confronted with a trigonometric equation, start by identifying known and unknown components. Employing identities, like the difference of squares and double-angle formulas, can break down the problem into smaller, more manageable parts.

Breaking Down Complex Problems

With seemingly complex problems like the one in our exercise, foundational steps such as recognizing patterns and applying relevant identities are crucial. Don't rush; take the time to assess which mathematical 'tools'—identities, known values, simplification methods—will best fit the task at hand. By doing so, the path to resolution will become clearer, and the likelihood of arriving at the correct solution will vastly improve.

Verifying Your Answers

Finally, it's important to remember that validating your answers is part of the problem-solving process. By comparing your simplified results to the values given in the problem statement, you will not only confirm correctness but also reinforce your understanding of the concepts involved.