Problem 18 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve the following trigonometric equations: Find the number of solution of the equation \(\sin 5 x \cdot \cos 3 x=\sin 6 x \cdot \cos 2 x\) in \([0, \pi]\).

Step-by-Step Solution

Answer

The equation will have infinite solutions in the interval \([0, \pi]\) as the \(\sin function\) repeats itself after every cycle of \(2\pi\).

Step 1: Using Trigonometric Identities

To simplify the equation we can use identity of \(\sin(a-b)=\sin a*\cos b-\cos a*\sin b\). Here, apply the identity to both sides so the equation becomes: \(\sin(5x+3x)=\sin(6x+2x)\)Simplifying, we get a simpler equation:\(\sin 8x=\sin 8x\)

Step 2: Identifying the solutions

Since after the simplification we got the same terms on both sides of the equation, it is essentially proving the condition of equality, which is always true. This implies the equality will hold for any value of x.

Step 3: Counting the solutions

In order to find the number of solutions, we need to know the interval which in this case is \([0, \pi]\). Since the period of the \(\sin function\) is \(2\pi\), it completes its one cycle in the interval \([0, \pi]\). As the value of the \(\sin function\) repeats itself after a period, there will be as many solutions as the number of times the \(\sin function\) completes its cycle in the interval.

Key Concepts

Trigonometric Identities

Understanding trigonometric identities is crucial for simplifying complex trigonometric equations. In our original exercise, the identity \( \sin(a-b) = \sin a \cdot \cos b - \cos a \cdot \sin b \) was utilized. This particular identity helps in expressing the difference of angles using the sine function, transforming products of sine and cosine into a more manageable form.

• Applying identities can simplify equations, making it easier to solve them.
• Recognizing the appropriate identity helps convert complex expressions into simpler equivalent forms.

To tackle the given problem, we used identities to express the products on either side of the equation, eventually simplifying it to \( \sin(8x) = \sin(8x) \). This demonstrates how using the right trigonometric identities can reduce complex terms to simpler equations.

Solution of Equations

Solving trigonometric equations requires specific strategies to isolate the variable, which in our case, is the angle \(x\). After simplifying the original equation, we found that both sides were identical: \( \sin(8x) = \sin(8x) \). This means the equation holds true for all values of \(x\), not just specific ones. In general, solving equations involves:

• Simplifying both sides of the equation.
• Applying trigonometric identities wherever necessary.
• Ensuring all transformations are reversible.

In our solved exercise, the procedure showed that after using identities and simplifying, the equation proves to be an equality valid for all permissible values of \(x\). Thus, interpreting such simplified results helps determine the solution's nature.

Sine Function

The sine function is one of the fundamental trigonometric functions and is periodic with a period of \(2\pi\). This periodic nature is significant, especially in the context of solving trigonometric equations over a given interval such as \([0, \pi]\).

• The sine function oscillates between -1 and 1.
• It completes one full wave, or cycle, over \(2\pi\), meaning every interval of \(2\pi\) units along the horizontal axis, its values repeat.
• Within the interval \([0, \pi]\), the sine function completes half of its cycle.

Understanding this periodicity helps in analyzing how many times certain values (like the angle \(x\) in \(\sin(8x)\)) satisfy an equation within a specified range.

Interval Analysis

Interval analysis involves examining the behavior of functions over specific intervals, an essential aspect when solving trigonometric equations. Given our interval \([0, \pi]\), understanding how often functions like the sine function cycle through their values helps in counting solutions.

• The interval \([0, \pi]\) captures half a cycle of the sine function.
• Because \(\sin(8x)\) was shown to equal itself, every value of \(x\) within this interval is a valid solution.
• Analyzing such intervals ensures that all potential solutions are considered and counted accurately.

By interpreting the results in the context of the specified interval, one can determine the number of valid solutions efficiently, ensuring a correct and comprehensive understanding of the problem.