Sigma Percentile
JEE Main 2023 (25 January Shift 2)
LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: If and respectively are the numbers of positive and negative value of in the interval that satisfy the equation , then is equal to _____.

Enter Numerical Value:

Visualized Solution

Initial Equation Setup

  • Given equation:
  • Multiply both sides by to prepare for product-to-sum identities:

Applying Product-to-Sum Identity

  • Use identity:
  • LHS:
  • RHS:

Simplifying the Equation

  • Equating LHS and RHS:
  • Subtract from both sides:

General Solution for Cosine

  • General solution for is
  • Applying this to our equation:

Case 1: Positive Sign

  • Taking the positive sign:

Case 2: Negative Sign

  • Taking the negative sign:

Visualizing the Interval

  • Interval:
  • General Solution:
  • Substitute such that
  • This implies

Finding Positive Values ()

  • Positive values of ():
  • For
  • So,

Finding Negative Values ()

  • Negative values of ():
  • For
  • So,

Final Calculation of

  • Final calculation:
  • Final Answer: 25

The Sigma Insight: General Solution of Trigonometric Equations

Solution Diagram

Analyzing the Setup

Welcome, fellow traveler of the JEE path. Today, we are not just solving an equation; we are uncovering a hidden symmetry in the world of trigonometry.
The problem before us is:
At first glance, it looks like a chaotic mess of angles. But remember, in the realm of JEE Advanced, chaos is often just order in disguise.

The Transformation

We start with the equation . When you see products of cosines, your mind should immediately race to the product-to-sum identities.
To use them, we need a factor of . We multiply both sides by :
Now, we apply the identity .
On the left-hand side, let and . Adding them gives , and subtracting them gives . Thus, the left side becomes:
On the right-hand side, let and . Adding them gives , and subtracting them gives . Thus, the right side becomes:

The Elegant Cancellation

Look at what we have now:
The term is waiting on both sides like a silent observer. We subtract it away, and suddenly, the equation collapses into something beautiful:
This is the moment where the complexity vanishes, leaving us with a clean, solvable structure.

The General Solution

We know that the general solution for is , where . Applying this to our equation, we get:
This splits into two cases.
Case 1 (the positive sign):
Case 2 (the negative sign):
Since is just a subset of , our general solution is simply:

Final Calculation

We are restricted to the interval . Setting , we find that must be an integer in the range .
The possible values for are .
The positive values of occur for , so . The negative values of occur for , so .
The product is:
We have navigated the storm and arrived at the shore. The beauty of this problem lies not in the final number, but in the elegance of the path we took to find it.

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