Problem 190 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ x=k \sin t-\sin k t, y=k \cos t+\cos k t $$

Step-by-Step Solution

Answer

The expressions can be simplified by using the product-to-sum identities to \(x=2 \cos \big(\frac{kt+t}{2}\big) \sin \big(\frac{kt-t}{2}\big)\) and \(y=2 \cos \big(\frac{kt+t}{2}\big) \cos \big(\frac{kt-t}{2}\big)\).

Step 1: Identify the Relevant Trigonometric Identity

We start by recognizing that the form of the given expressions resembles the right side of the product-to-sum identities, which are \(\sin A - \sin B = 2 \cos \big(\frac{A+B}{2}\big) \sin \big(\frac{A-B}{2}\big)\) and \(\cos A + \cos B = 2 \cos \big(\frac{A+B}{2}\big) \cos \big(\frac{A-B}{2}\big)\). We will utilize these identities to rewrite the given expressions.

Step 2: Apply the Trigonometric Identity to the x-expression

The expressed formula for x is \(x=k \sin t-\sin k t\). If we think of \(A = kt\) and \(B = t\), our formula matches the form \(\sin A - \sin B\. Applying the identity we get for x that \(x=2 \cos \big(\frac{kt+t}{2}\big) \sin \big(\frac{kt-t}{2}\big)\).

Step 3: Apply the Trigonometric Identity to the y-expression

The expressed formula for y is \(y=k \cos t+\cos k t\). Using the same choice for \(A\) and \(B\), the formula matches the form \(\cos A + \cos B\). Applying the identity we get for y that \(y=2 \cos \big(\frac{kt+t}{2}\big) \cos \big(\frac{kt-t}{2}\big)\).

Key Concepts

Understanding Product-to-Sum Identities

Trigonometric identities are like mathematical shortcuts that help simplify complex expressions involving trigonometric functions. One set of these identities is called product-to-sum identities. These identities allow us to convert products of sine and cosine into sums or differences, which are often easier to work with.
For example:

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

These identities are particularly useful when dealing with expressions involving multiple angles. In the exercise, they help rewrite the given parametric equations into a simpler form. Understanding how to apply these identities is key to analyzing and solving trigonometric expressions effectively.

Introducing Differential Calculus

Differential calculus is a branch of mathematics focused on how functions change and how rates are determined. It involves the concept of derivatives, which measure the rate at which a quantity changes.
If you have a function, the derivative can tell you the slope of the function at any point. This is crucial in finding out how quickly things are changing, such as velocity and acceleration in physics.
For a parametric equation like the one in this exercise, differential calculus lets us explore changes in \(x\) and \(y\) as \(t\) changes. This can help us understand the shape and nature of the curve the parametric equations describe. Being able to apply differentiation to parametric equations is a powerful tool in mathematics.

Exploring Parametric Equations

Parametric equations are unique because they define both \(x\) and \(y\) as functions of a third variable, typically \(t\), called the parameter. This is different from the regular \(y = f(x)\) format. These equations allow us to describe a variety of curves, which might be difficult or impossible to express as a single function of \(x\).
With parametric equations, each value of \(t\) gives a point \((x, y)\) on the curve. By varying \(t\), we can see the path traced out by these points.
This is especially useful in fields like physics and engineering, where objects follow paths based on time or other parameters. In this exercise, using parametric equations helps visualize the relationships between trigonometric functions and the geometric shapes they produce.