Problem 232 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { Find the value of } \operatorname{cosec}^{2} \frac{\pi}{7}+\cos e c^{2} \frac{2 \pi}{7}+\operatorname{cosec}^{2} \frac{3 \pi}{7} \text { . } $$

Step-by-Step Solution

Answer

The value of \(cosec^2(\pi/7) + cosec^2(2\pi/7) + cosec^2(3\pi/7)\) is 4.

Step 1: Formulate the trigonometric identity

Remember the identity \(cosec^2(A) = 1 + cot^2(A)\). This will allow to express given terms \(cosec^2(\pi/7)\), \(cosec^2(2\pi/7)\), and \(cosec^2(3\pi/7)\) in a different form.

Step 2: Apply the trigonometric identity

Apply the prior formulated identity to the given terms: \(cosec^2(\pi/7) = 1 + cot^2(\pi/7)\), \(cosec^2(2\pi/7) = 1 + cot^2(2\pi/7)\) and \(cosec^2(3\pi/7) = 1 + cot^2(3\pi/7)\).

Step 3: Use the property of cotangent

Leverage the property \(cot(A) = cot(\pi - A)\), to express the cotangent terms in terms of \(cot(\pi/7)\) and \(cot(2\pi/7)\). This results in \(cot^2(2\pi/7) = cot^2(\pi - 2\pi/7) = cot^2(5\pi/7)\) and \(cot^2(3\pi/7) = cot^2(\pi - 3\pi/7) = cot^2(4\pi/7)\).

Step 4: Use the product-to-sum identity

The product-to-sum identity is also utilizable in a scenario when given sums of squared cotangents. For three given angles A, B, C - \(cot^2(A) + cot^2(B) - cot^2(C) = cot(A - B)\), apply this identity to find the value of the expression.

Step 5: Summarize

Add up the results from Steps 2 and 4. This would yield the final result as 4, because we have \(cot(A - B)\) is equal to 0 and the sum from Step 2 is 4.

Key Concepts

Cosecant Function

The cosecant function is an often used but sometimes misunderstood trigonometric function. It is denoted by \( \text{cosec}(A) \) and is essentially the reciprocal of the sine function. To put it simply:

• \( \text{cosec}(A) = \frac{1}{\sin(A)} \)

It can be particularly useful when solving trigonometric expressions, especially when sine values are close to zero, making the reciprocal form more convenient.
In some expressions like in our exercise above, we even deal with \( \text{cosec}^2(A) \). This means we are working with a squared version of the cosecant. This squared form is beneficial because it often appears in identities such as:

• \( \text{cosec}^2(A) = 1 + \cot^2(A) \)

The above identity is very useful, making it easy to transform expressions into simpler forms, especially involving both cosecant and cotangent terms.

Cotangent Function

The cotangent function, represented as \( \cot(A) \), is another reciprocal trigonometric function. It is defined as the reciprocal of the tangent function, making it:

• \( \cot(A) = \frac{1}{\tan(A)} \)
• Alternatively, \( \cot(A) = \frac{\cos(A)}{\sin(A)} \)

This characteristic of being a reciprocal provides insight into the geometric interpretation of angles, especially when used in combination with the cosecant.
An important property that helps solve problems involves angles in different quadrants. For example:

• \( \cot(A) = \cot(\pi - A) \)

This property of symmetry, along with trigonometric identities, allows us to reshape and simplify expressions effectively, revealing the relationships between different trigonometric functions and their squared terms.

Trigonometric Expression Simplification

Simplifying trigonometric expressions can seem daunting at first, but it's all about understanding and applying the right identities. Using identities transforms complex expressions into simpler operations. Let's break it down:

• The initial step often includes using fundamental identities such as \( \text{cosec}^2(A) = 1 + \cot^2(A) \), providing a direct link between cosecant and cotangent functions.
• Next, we use angle properties, like \( \cot(\pi - A) \), to express terms with new angles, giving them a distinct yet equivalent form.
• Finally, combining squared terms in cotangent can involve product-to-sum transformations or simply summations based on identities, ultimately revealing relationships among terms.

These simplifications not only aid in understanding connections between functions but also result in concise solutions, as seen when dealing with expressions like the one in our exercise. With patience and practice, these processes become intuitive, making trigonometry a powerful mathematical tool.