Problem 68 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let A and B denote the statements A: \(\cos \alpha+\cos \beta+\cos \lambda=0\) B: \(\sin \alpha+\sin \beta+\sin \lambda=0\) If \(\cos (\beta-\lambda)+\cos (\beta-\alpha)+\cos (\alpha-\beta)=-\frac{3}{2}\), then (A) \(\mathrm{A}\) is true and \(\mathrm{B}\) is false (B) A is false and \(\mathrm{B}\) is true (C) both \(\mathrm{A}\) and \(\mathrm{B}\) are true (D) both \(\mathrm{A}\) and \(\mathrm{B}\) are false

Step-by-Step Solution

Answer

C: Both \(\mathrm{A}\) and \(\mathrm{B}\) are true.

Step 1: Understand Given Equation

We are given that \( \cos (\beta-\lambda) + \cos (\beta-\alpha) + \cos (\alpha-\beta) = -\frac{3}{2} \). We need to understand what this implies about the angles \(\alpha, \beta, \lambda\).

Step 2: Use Cosine of Difference Formula

The cosine of the difference formula is \( \cos (X - Y) = \cos X \cos Y + \sin X \sin Y \). Apply this formula to each term in the equation: \(\cos (\beta-\lambda), \cos (\beta-\alpha), \text{and } \cos (\alpha-\beta)\).

Step 3: Calculate Each Term

For \( \cos (\beta-\lambda) = \cos \beta \cos \lambda + \sin \beta \sin \lambda \), \( \cos (\beta-\alpha) = \cos \beta \cos \alpha + \sin \beta \sin \alpha \), and \( \cos (\alpha-\beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \). Summing these gives \[ 2(\cos \beta \cos \alpha + \sin \beta \sin \alpha) + \cos(\alpha + \lambda - \beta) = -\frac{3}{2} \].

Step 4: Analyze Special Angles

The equation \( \cos (\beta-\lambda) = \cos(\beta-\alpha) = \cos(\alpha-\beta) = -\frac{1}{2} \) suggests that we are dealing with specific angle differences. This might imply that \( \alpha, \beta, \lambda \) are separated by \(120^\circ\) if assumed equal distribution on a circle, given that \(\cos 120^\circ = -\frac{1}{2} \).

Step 5: Verify Conditions A and B

If this separation holds, \( \cos \alpha + \cos \beta + \cos \lambda = 0 \) and \( \sin \alpha + \sin \beta + \sin \lambda = 0 \) must hold true because the sum of vectors at \(120^\circ\) separation results in zero. Verify this by calculation or geometric consideration.

Step 6: Choose Correct Option

Since both conditions derived are true, we conclude that both \(\mathrm{A}\) and \(\mathrm{B}\) are true.

Key Concepts

Cosine of Difference

When tackling trigonometric identities, one crucial formula is the cosine of the difference, expressed as \( \cos(X - Y) = \cos X \cos Y + \sin X \sin Y \). This formula helps us understand how the cosine of an angle difference is built from the cosines and sines of the two original angles.
It breaks down the interaction between two angles and creates a relationship based on their trigonometric functions. Applying this formula allows us to decompose complex trigonometric expressions into manageable parts.
If given angles like \( \beta \), \( \lambda \), and \( \alpha \), substituting them into this formula helps us derive other useful expressions, such as in sums involving multiple cosine differences. This reveals deeper patterns or symmetries in trigonometric problems.

Angle Sum

Trigonometric problems frequently involve understanding the positions of angles relative to each other, especially in terms of their sums. In exercises like ours, it's often the scenario that angles can be symmetrically placed on a circle.
When angles \( \alpha \), \( \beta \), and \( \lambda \) are set at specific increments (like multiples of \( 120^\circ \)), both their sine and cosine components can contribute to a nuanced pattern.

• If angles are positioned such that their sum or difference leads to certain symmetric separations (e.g., \(120^\circ\) apart), then identities such as \( \cos \alpha + \cos \beta + \cos \lambda = 0 \) become evident through geometric reasoning.
• Similarly, \( \sin \alpha + \sin \beta + \sin \lambda = 0 \) can arise due to the circular distribution of these angles.

Recognizing such patterns often confirms the relationship between trigonometric sums and geometry.

Vector Addition

In trigonometry, visualizing the addition of vectors can simplify complex scenarios with angles. When vectors are set \( 120^\circ \) apart, such as with angles displayed around a circle, they often perfectly balance each other.
When you talk about vectors in trigonometry:

• Each angle like \( \alpha \), \( \beta \), and \( \lambda \) can be visualized as a vector on the unit circle.
• The coordinates of these vectors are given by \( (\cos \theta, \sin \theta) \) for angle \( \theta \).

In situations where the vector angles are distributed evenly, like \( 120^\circ \) apart, they can add up to zero. This is because the components of these vectors (both sine and cosine parts) exactly cancel out, reflecting the symmetry of the circular arrangement. It's a powerful graphical approach to solving trigonometric identities and understanding their implications.