Problem 211 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { Prove that } \sin ^{2} \alpha+\sin ^{2} 2 \alpha+\sin ^{2} 3 \alpha+\ldots \ldots \text { to } n \text { terms }=\frac{1}{4}[(2 n+1) \sin \alpha-\sin (2 n+1) \alpha] \operatorname{cosec} \alpha $$

Step-by-Step Solution

Answer

This question involves using trigonometric identities to simplify the expressions, and consider each term in the series to finally build up to prove the given equation. After simplifying, we add each term so that the cosine terms get cancelled out and convert the remaining sum to the right hand side form.

Step 1: Identify the pattern of the terms

We notice the $\sin^2$ terms for $1\alpha, 2\alpha, 3\alpha, ..., n\alpha$. This will help us understand what to add for each new term.

Step 2: Simplify the term

We'll start by simplifying $\sin^2\alpha$. We use the identity $\sin^2x = \frac{1-\cos2x}{2}$ to simplify $\sin^2\alpha$ where $x = \alpha$.

Step 3: Consider another term

Now, we consider the next term $\sin^22\alpha$. Again, we use the identity $\sin^2x = \frac{1-\cos2x}{2}$ to simplify $\sin^22\alpha$ where $x = 2\alpha$.

Step 4: Keep adding terms

Continuing in the same way, add all the remaining terms upto $n$ terms. They should be added in a way that the cosine terms will get cancelled out.

Step 5: Try to convert the obtained sum to the RHS form

In the remaining sum, we have to write $\cos(2n\alpha)$ in terms of $\sin\alpha$ using the Pythagorean identity. We rearrange the terms slightly, and use the identity $\cosec{x}= \frac{1}{\sin x}$.

Key Concepts

Sum of Sines

The sum of sines is an interesting topic in trigonometry that involves adding sine functions of different angles. In the given problem, we are dealing with sine terms like \( \sin^2 \alpha, \sin^2 2\alpha, \sin^2 3\alpha, \ldots \). Managing the sum of these terms requires a strategic approach to leverage identities.
To tackle this, we use a trigonometric identity to transform and manage the terms. The identity is \( \sin^2 x = \frac{1 - \cos 2x}{2} \), which converts a squared sine term into an expression that includes cosine. This conversion is pivotal as it allows easier summation and manipulates the terms in such a way that cosine terms can compost themselves out.

• This step involves expressing each squared sine as a half-adjusted cosine.
• As each term is added, cosine contributions start to interact, often cancelling one another.

The result is a sum that is much simpler to handle and one that can potentially be transformed into the required form on the right-hand side of the equation.

Double Angle Formula

The double angle formula is another essential tool used to simplify expressions involving trigonometric functions. Specifically, it connects the sine and cosine of double angles through these identities:

• \( \sin 2x = 2\sin x \cos x \)
• \( \cos 2x = \cos^2 x - \sin^2 x \)

For our problem, the double angle identity \( \sin^2 x = \frac{1 - \cos 2x}{2} \) was repeatedly utilized. This formula is particularly useful because it converts a squared term \( \sin^2 x \) into a function of \( \cos 2x \).
This transformation helps when summing multiple terms like \( \sin^2 \alpha + \sin^2 2\alpha + \sin^2 3\alpha + \ldots \), because it can neutralize the cosine terms, allowing for a simplification of the cumulative expression. These actions taken using the double angle formula are essential to putting the overall sum into a manageable form.

Pythagorean Identity

The Pythagorean Identity is a cornerstone of trigonometry, needed to relate sine and cosine in equations. The main identity states:

• \( \sin^2 x + \cos^2 x = 1 \)

In the exercise, this identity was implied when rearranging terms to express certain cosine terms in terms of their equivalent sine expressions. For instance, knowing that \( \cos^2 x = 1 - \sin^2 x \) can be utilized to manipulate the term \( \cos(2n\alpha) \)
. This arithmetic is useful when the task is to rewrite the original sum expression into another form involving \( \csc \alpha = \frac{1}{\sin \alpha} \) or similar terms.
Ultimately, these operations enable the algebraic shift from complex trigonometric sums to a solution resembling the problem's right-hand side structure.