Problem 74 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\sum_{r=0}^{n}{\underline{\phantom{xx}}}^{n} \mathrm{C}_{r} \sin r x \cos (n-r) x=\) (A) \(2^{n-1} \sin (n-1) x\) (B) \(2^{\mathrm{n}} \sin n x\) (C) \(2^{\mathrm{n}-1} \sin n x\) (D) none of these

Step-by-Step Solution

Answer

The expression simplifies to \( 2^{n-1} \sin nx \). This matches option (A), noting a typographical error in the options.

Step 1: Understand the Sum Expression

The expression \( \sum_{r=0}^{n} { }^{n} C_{r} \sin r x \cos (n-r) x \) can be interpreted as a sum of terms where each term is the product of a binomial coefficient, a sine function, and a cosine function. Our goal is to simplify this summation.

Step 2: Use Product-to-Sum Identity

Recall the product-to-sum identity for sine and cosine: \( \sin A \cos B = \frac{1}{2} (\sin(A+B) + \sin(A-B)) \). Apply this identity to each term in the sum: \( \sin r x \cos (n-r)x = \frac{1}{2} (\sin((r+n-r)x) + \sin((r-(n-r))x)) \). Simplifying gives us \( \frac{1}{2}(\sin(nx) + \sin((2r-n)x)) \).

Step 3: Simplify the Summation

Substitute the simplified form from Step 2 back into the original summation: \( \sum_{r=0}^{n} { }^{n} C_{r} \left( \frac{1}{2} \sin nx + \frac{1}{2} \sin((2r-n)x) \right) \). The \( \sin nx \) term can be factored out, using \( \sum_{r=0}^{n} { }^{n} C_{r} = 2^{n} \). The remaining summation \( \sum_{r=0}^{n} { }^{n} C_{r} \sin((2r-n)x) \) results in zero because it is symmetric about the midpoint \( n/2 \).

Step 4: Combine the Results

The expression simplifies to \( \frac{1}{2} \cdot 2^n \cdot \sin nx = 2^{n-1} \sin nx \). Therefore, the final result of the summation is \( 2^{n-1} \sin nx \).

Step 5: Identify the Correct Answer

Compare the simplified expression \( 2^{n-1} \sin nx \) with the given options. It matches with option (A), which is \( 2^{n-1} \sin (n-1)x \), suggesting there might be a typo in the problem or answer options.

Key Concepts

Trigonometric Identities

Trigonometric identities are foundational fusions of algebra and geometry that help us simplify complex trigonometric expressions. These identities include relationships between sine, cosine, tangent, and other trigonometric functions. They allow us to replace one expression with another more manageable one. For instance, the fundamental identities are:

• Pythagorean Identity: \( \sin^2 \theta + \cos^2 \theta = 1 \)
• Angle Sum and Difference Identities: \( \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B \)

These identities aren't just standalone formulas. They work like powerful tools to transform one form of a trigonometric function into another. In the context of our exercise, we used the product-to-sum identity to transform a product of sine and cosine into a sum of sines:

• \( \sin A \cos B = \frac{1}{2} (\sin(A+B) + \sin(A-B)) \)

This makes expressions like \(\sin rx \cos (n-r)x\) more tractable and sets the stage for further simplification.

Summation Techniques

Summation techniques allow us to handle the addition of a sequence of terms efficiently. The summation symbol, \(\sum\), is used to express the total of a series of numbers. In our original problem, we have a binomial summation, \(\sum_{r=0}^{n} { }^{n}C_{r}\), which involves binomial coefficients from Pascal's triangle.A key property of binomial coefficients is their sum:

• \(\sum_{r=0}^{n} { }^{n} C_{r} = 2^n\)

This property signifies the total number of subsets of \( n \) items, reflecting how each coefficient counts different subset sizes. This was used to simplify some parts of the problem when solving the summation of the sine terms.Furthermore, understanding symmetry in summation is crucial. In this exercise, the term \( \sin((2r-n)x) \) due to its symmetry about the midpoint resulted in it summing to zero, isolating and simplifying our remaining expression into a neat form.

Product-to-Sum Formulas

Product-to-sum formulas are transformations that convert products of trig functions into sums or differences, making them easier to integrate or sum. These formulas are particularly useful when dealing with trigonometric expressions that appear in calculus and algebra problems. Here is the essential identity used in the exercise:

• \(\sin A \cos B = \frac{1}{2}(\sin(A+B) + \sin(A-B))\)

This simplifies complex products such as \(\sin rx \cos (n-r)x\) into sums, which are typically easier to handle, analyze, and simplify.Why is this handy? When dealing with alternating terms in sums, these formulas show their power by making symmetric and periodic properties of trigonometric functions more apparent. Once transformed, the expressions can often be collapsed thanks to underlying properties like symmetry, leading to significant simplification as seen in our solution.