Problem 1 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve the following equations and tick the correct one. \(\sin ^{2} \theta-\cos \theta=\frac{1}{2}, 0 \leq \theta \leq 2 \pi\) (a) \(\frac{2 \pi}{3}, \frac{\pi}{3}\) (b) \(\frac{\pi}{3}, \frac{5 \pi}{3}\) (c) \(-\frac{\pi}{3}, \frac{2 \pi}{3}\) (d) \(\frac{2 \pi}{3}, \frac{5 \pi}{3}\)

Step-by-Step Solution

Answer

The correct option is (b) \(\frac{\pi}{3}, \frac{5 \pi}{3}\).

Step 1: Apply Pythagorean Identity

First, let's simplify the given equation using the Pythagorean Identity, \(\sin^{2}\theta + \cos^{2}\theta = 1\); this implies \(\sin^{2}\theta = 1 - \cos^{2}\theta\). Substituting this in the equation gives us: \(1 - \cos^{2}\theta - cos\theta = \frac{1}{2}\).

Step 2: Form Quadratic Equation

Rearrange the equation to form a quadratic equation in terms of cos\theta. Thus the equation becomes \(\cos^{2}\theta + \cos\theta - \frac{1}{2} = 0\).

Step 3: Solve Quadratic Equation

Now solve the quadratic equation using the quadratic formula \(\cos\theta=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\). Where \(a = 1\), \(b = 1\), and \(c = -\frac{1}{2}\). Solving will yield two values for \(\cos\theta\), say \(\cos\theta_{1}\) and \(\cos\theta_{2}\).

Step 4: Calculate Corresponding Theta

For each of the two cos\theta values, calculate the corresponding \(\theta\) in radians (use trigonometric tables or a calculator). Remember, only \(\theta\) values in the interval \(0 \leq \theta \leq 2 \pi\) should be considered.

Step 5: Comparing Answer with Options

Finally, the calculated \(\theta\) values should be compared with the ones provided in the options to determine the correct one.

Key Concepts

Pythagorean Identity

When solving trigonometric equations, understanding the Pythagorean Identity is crucial. This identity, which states that \(\sin^2\theta + \cos^2\theta = 1\), provides a relationship between the sine and cosine of the same angle. This allows us to express one trigonometric function in terms of the other, simplifying equations that involve both sine and cosine functions.

For instance, if an equation includes \(\sin^2\theta\) and we want to eliminate it, we can use the Pythagorean Identity to substitute \(\sin^2\theta\) with \(1 - \cos^2\theta\). This is a powerful technique, as it often transforms a trigonometric equation into a more manageable form, such as a quadratic equation, which can then be solved using familiar algebraic methods.

Quadratic Equations in Trigonometry

Quadratic equations in trigonometry appear when we have an expression of the form \(a\cos^2\theta + b\cos\theta + c = 0\) or a similar sine-based equation. The approach to solving these equations mirrors that of solving any quadratic equation, but with trigonometric functions taking the place of the variable.

To solve for \(\theta\), we would usually rearrange the equation so that one side equals zero, and then apply the quadratic formula \(\cos\theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where the coefficients \(a\), \(b\), and \(c\) come from the original equation. The solutions to this formula give us possible values for \(\cos\theta\) or \(\sin\theta\), depending on the equation. It's important to check that the resulting angles are within the specified range, and to remember that trigonometric functions are periodic, meaning they repeat values over intervals, so there can be multiple correct angles that satisfy the quadratic equation.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are fundamental in solving trigonometric equations. They relate the angles of triangles to the lengths of their sides in a right-angled triangle. But more broadly, these functions describe relationships involving angles and define wave-like patterns, which are applicable in various areas of mathematics and physics.

Sine and cosine functions also have certain properties and values that are key to finding solutions to trigonometric equations. They are periodic, have specific ranges \(\sin\theta:\,[-1, 1]\) and \(\cos\theta:\,[-1, 1]\), and can be represented on the unit circle, where the angle \(\theta\) corresponds to a point on the circle's circumference. By understanding these properties, and how to evaluate or inversely determine these functions at certain values, students can solve for angles \(\theta\) within a given domain, typically between \(0\) and \(2\pi\) radians in most textbooks, which is equivalent to a full rotation around the circle.