Problem 13 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve: \(r \sin \theta=3\) and \(r=4(1+\sin \theta), 0 \leq \theta \leq 2 \pi\)

Step-by-Step Solution

Answer

The solutions are \((r, \theta) = (2, 0.848)\) and \((r, \theta) = (4, \pi), (4, 2\pi)\).

Step 1: Express r from first equation

From the first equation \(r \sin \theta = 3\), express \(r\) as \(r = \frac{3}{\sin \theta}\).

Step 2: Substitute r into the second equation

Substitute \(r = \frac{3}{\sin \theta}\) into the second equation, thus obtaining \( \frac{3}{\sin \theta} = 4(1 + \sin \theta)\). Multiply both sides by \(\sin \theta\) to arrive at \(3 = 4 \sin \theta(1 + \sin \theta)\). Expanding gives \(3 = 4 \sin \theta + 4 \sin^2 \theta\).

Step 3: Re-arrange the equation

Rearrange the equation to have the form of a quadratic equation \(4 \sin^2 \theta + 4 \sin \theta - 3 = 0\). Divide by 4 to simplify the equation to the form \(\sin^2 \theta + \sin \theta - \frac{3}{4} = 0\).

Step 4: Find the roots

We can now solve for \(\sin \theta\) using the quadratic formula. This gives us \(\sin \theta = -1\) or \(\sin \theta = \frac{3}{4}\).

Step 5: Substitute back into the equation to find the solutions

The angles \(\theta\) that satisfy these equations are given by the inverse sine function. For \(\sin \theta = \frac{3}{4}\), we find \(\theta = \sin^{-1} (\frac{3}{4})\) which yields \(\theta = 0.848\), and for \(\sin \theta = -1\), we find \(\theta = \sin^{-1}(-1)\), yielding \(\theta = \pi, 2\pi\). In addition, by using these angles in the second equation, we get \(r = 2\), for \(\theta = 0.848\), and \(r = 4\), for \(\theta = \pi, 2\pi\) respectively.

Key Concepts

Polar Coordinates

Polar coordinates are a way of representing points in a plane using a distance and an angle. Unlike the Cartesian system which uses x and y coordinates, polar coordinates use two values: radius (r) and angle (\( \theta \)). Understanding how these two values interact is essential in problems dealing with circles, curves, and spirals.

• The radius \( r \) represents how far a point is from the origin (the center of the coordinate system).
• The angle \( \theta \) is measured in radians from the positive x-axis. It tells us the direction of the point from the origin.

Expressions in polar coordinates often relate \( r \) and \( \theta \) to trigonometric functions, such as in the exercise example of \( r \sin \theta = 3 \). Here, \( \sin \theta \) dictates how much of the radius is projected vertically across a given angle. When solving polar equations, converting between polar and Cartesian can often simplify the manipulation of equations.

One way polar coordinates are particularly useful is in translating complex geometrical figures into manageable equations. For instance, equations of circles or spirals can become linear when devised in polar rather than Cartesian form.

Quadratic Equations

Quadratic equations are polynomial equations of degree two. They follow the general formula \( ax^2 + bx + c = 0 \). In the exercise's solution, a quadratic form emerges as \( 4 \sin^2 \theta + 4 \sin \theta - 3 = 0 \), indicating the need to solve for \( \sin \theta \) using quadratic methods.

• Quadratic equations can be solved by factoring, completing the square, or using the quadratic formula.
• The quadratic formula is \( \sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). It provides a systematic means to find both real and complex roots.
• The discriminant \( (b^2 - 4ac) \) indicates the nature of the solutions: two distinct real roots, one real double root, or two complex roots.

In this exercise, simplifying the equation \( \sin^2 \theta + \sin \theta - \frac{3}{4} = 0 \) leads to using the quadratic formula in terms of \( \sin \theta \). Understanding how trigonometric solutions fit into quadratic frameworks is crucial for advancing with such equations.

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to determine angles given a trigonometric ratio. They are the 'reverse' operations of the standard trigonometric functions, providing angles where function values like \( \sin \theta, \cos \theta, \) and \( \tan \theta \) are known.

• The inverse sine function \( \sin^{-1}(x) \) yields an angle \( \theta \) such that \( \sin \theta = x \).
• This is indicated in the exercise with solutions like \( \theta = \sin^{-1}(\frac{3}{4}) \) and \( \theta = \sin^{-1}(-1) \).
• Similarly, inverse cosine \( \cos^{-1}(x) \) and inverse tangent \( \tan^{-1}(x) \) do the same for the respective functions.

The usefulness of inverse trigonometric functions comes from their ability to translate ratios back into angles, which is crucial in solving geometrical and trigonometric equations. They help us find specific angles that satisfy conditions across typical angle ranges, often between \( 0 \) and \( 2\pi \), as seen in many exercises including our example above.