Problem 20 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve the following equations and tick the correct one. The sum of all solution of the equation \(\cos \theta \cdot \cos \left(\frac{\pi}{3}+\theta\right) \cdot \cos \left(\frac{\pi}{3}-\theta\right)=\frac{1}{4}\) is (a) \(15 \pi\) (b) \(30 \pi\) (c) \(\frac{100 \pi}{3}\) (d) None

Step-by-Step Solution

Answer

(d) None

Step 1: Simplification

Using the product-to-sum identity, we can simplify the equation by converting the product of cosines in the equation to the sum of cosines. The product-to-sum identities are: \( cos(A)cos(B) = \frac{1}{2}cos(A-B) + \frac{1}{2}cos(A+B) \) and \( cos(A)cos(B) = \frac{1}{2}cos(A-B) - \frac{1}{2}cos(A+B) \). In this case, your equation becomes \( \frac{1}{2}\cos(2\frac{\pi}{3})+\frac{1}{2}\cos(2\theta)=\frac{1}{4} \).

Step 2: Solve the Simplified Trigonometric Equation

We simplify this equation to: \( \cos(2\theta) = \frac{1}{2}-cos(2\frac{\pi}{3}) = \frac{1}{2}-(-\frac{1}{2})=1 \).\n Now we need to find all solutions of \( \cos(2\theta) = 1 \). We know that the cosine function equals 1 at \( 2\theta = 2n\pi \) where n is any integer. Therefore, the solution for \( \theta \) is \( n\pi \) where n, an integer.

Step 3: Sum of all Solutions

We know that theta, a solution, is \( n\pi \) and the sum of all solutions is therefore \( \sum_{n=-\infty}^{\infty}n\pi \), but this is an arithmetic sequence where the number of terms is infinite. This series does not converge (there's no finite sum), so the sum of all solutions is undefined.

Key Concepts

Product-to-Sum Identities

Understanding product-to-sum identities is crucial when solving complex trigonometric equations. These identities allow us to transform the product of two trigonometric functions into a sum or difference of two separate functions. This simplification can make an otherwise challenging problem much more manageable.

For the cosine function, which is relevant to our exercise, the identity states:\[\begin{equation}\cos(A)\cos(B) = \frac{1}{2}[\cos(A - B) + \cos(A + B)].\end{equation}\]In the example provided, this identity is applied to break down the product into a more solvable form, highlighting the transformative power of these identities in trigonometric equations.

Trigonometric Functions

The concept of trigonometric functions plays a fundamental role in various areas of mathematics, including geometry, calculus, and algebra. These functions—sine, cosine, tangent, and their reciprocals (cosecant, secant, and cotangent)—are defined using the ratios of sides in a right-angled triangle or points on a unit circle.

Specifically, the cosine function, which measures the horizontal distance from the origin to a point on a unit circle, is at the heart of solving our exercise. Knowing that the cosine of an angle takes values between -1 and 1, and that it equals 1 when the angle is an integer multiple of \[\begin{equation}2\theta = 2n\theta\end{equation}\]is essential. This piece of information assists in finding all possible solutions for the variable \[\begin{equation}\theta\end{equation}\]and is a testament to the importance of understanding these functions when solving trigonometric equations.

Sum of Solutions in Trigonometry

In trigonometry, finding the sum of solutions can be straightforward for finite sets of angles. However, complications arise when dealing with infinite sets, as in the case of our trigonometric equation. The sum mentioned in this context refers to the algebraic sum of all the solutions of the equation over one period or multiple periods for periodic functions.

The challenge in the provided exercise lies in the infinite nature of the solutions. Since \[\begin{equation}\theta = n\theta\end{equation}\]represents an infinite set of multiples of \[\begin{equation}\theta\end{equation}\](summing over all integers \[\begin{equation}n\end{equation}\]), we encounter an arithmetic series without a finite sum. Therefore, even if we can list individual solutions, their sum does not converge to a specific value, making it impossible to provide a finite answer for the sum of all solutions. This concept is pivotal in understanding when and how we can determine the sum of solutions for trigonometric equations.