Problem 57 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\alpha\) is a root of \(25 \cos ^{2} \theta+5 \cos \theta-12=0 \frac{\pi}{2}<\alpha<\pi\) then \(\sin 2 \alpha\) is equal to : (A) \(\frac{24}{25}\) (B) \(-\frac{24}{25}\) (C) \(\frac{13}{18}\) (D) \(-\frac{13}{18}\)

Step-by-Step Solution

Answer

\(-\frac{24}{25}\)

Step 1: Identify the Equation

The given equation is a quadratic in terms of \(\cos \theta\): \(25 \cos^2 \theta + 5 \cos \theta - 12 = 0\). We shall solve for \(\cos \theta\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 25\), \(b = 5\), and \(c = -12\).

Step 2: Apply the Quadratic Formula

Substitute \(a\), \(b\), and \(c\) into the quadratic formula: \(\cos \theta = \frac{-5 \pm \sqrt{5^2 - 4 \times 25 \times (-12)}}{2 \times 25}\).This simplifies to \(\cos \theta = \frac{-5 \pm \sqrt{25 + 1200}}{50}\), giving \(\cos \theta = \frac{-5 \pm \sqrt{1225}}{50}\).

Step 3: Simplify the Equation

Calculate \(\sqrt{1225} = 35\), leading to two possible solutions for \(\cos \theta\):1. \(\cos \theta = \frac{-5 + 35}{50} = \frac{30}{50} = \frac{3}{5}\)2. \(\cos \theta = \frac{-5 - 35}{50} = \frac{-40}{50} = -\frac{4}{5}\).

Step 4: Choose the Correct Solution Based on \(\alpha\)'s Range

Since the root \(\alpha\) is constrained by \(\frac{\pi}{2} < \alpha < \pi\), in this quadrant, \(\cos \alpha\) is negative. Therefore, \(\cos \alpha = -\frac{4}{5}\).

Step 5: Calculate \(\sin \alpha\)

Use the Pythagorean identity \(\sin^2 \alpha + \cos^2 \alpha = 1\). Substitute \(\cos \alpha = -\frac{4}{5}\) into the identity:\(\sin^2 \alpha + \left(-\frac{4}{5}\right)^2 = 1\).This gives \(\sin^2 \alpha = 1 - \frac{16}{25}\) which simplifies to \(\sin^2 \alpha = \frac{9}{25}\), resulting in \(\sin \alpha = \frac{3}{5}\) (note: since \(\alpha\) is in the second quadrant, \(\sin \alpha > 0\)).

Step 6: Calculate \(\sin 2\alpha\)

Use the double angle formula \(\sin 2\alpha = 2 \sin \alpha \cos \alpha\). Substituting the values \(\sin \alpha = \frac{3}{5}\) and \(\cos \alpha = -\frac{4}{5}\), we have:\(\sin 2\alpha = 2 \times \frac{3}{5} \times \left(-\frac{4}{5}\right) = -\frac{24}{25}\).

Step 7: Conclusion: Final Answer

The value of \(\sin 2\alpha\) is \(-\frac{24}{25}\).

Key Concepts

Quadratic Equations

When tackling trigonometric problems that involve quadratic equations, it's crucial to recognize the structure of a quadratic. Quadratic equations usually take the general form:

• \[ ax^2 + bx + c = 0 \]

In trigonometric terms, like our example problem, we often replace \(x\) with a trigonometric function, such as \(\cos \theta\). Using the quadratic formula

• \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

we can solve for the unknown. Here, identifying the coefficients \(a\), \(b\), and \(c\) is the first step. Once determined, substitute them into the quadratic formula. This may lead to two potential solutions for \(\cos \theta\). However, based on the context, we'll only choose the solution fitting the specified conditions in the problem. Make sure to double-check calculations during this process to avoid common pitfalls.

Quadrants in Trigonometry

In trigonometry, understanding the unit circle and the quadrants is tremendously important. Each quadrant in the unit circle corresponds to a specific range of angles and affects the sign of sine, cosine, and tangent functions. For example:

• First Quadrant: \(0 < \theta < \frac{\pi}{2}\), both sine and cosine are positive.
• Second Quadrant: \(\frac{\pi}{2} < \theta < \pi\), sine is positive, but cosine is negative.
• Third Quadrant: \(\pi < \theta < \frac{3\pi}{2}\), sine and cosine are negative.
• Fourth Quadrant: \(\frac{3\pi}{2} < \theta < 2\pi\), sine is negative, but cosine is positive.

This information helps to determine the correct sign of trigonometric functions. Hence, for our problem, where \(\alpha\) is in the second quadrant (\(\frac{\pi}{2} < \alpha < \pi\)), we know \(\cos \alpha\) must be negative, and \(\sin \alpha\) should be positive.

Trigonometric Identities

A foundation of trigonometry lies in understanding and applying various trigonometric identities. These identities simplify complex expressions and help solve equations more efficiently. Here's a simple breakdown of key identities often used:

• Pythagorean Identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \]This is invaluable when you know one trigonometric function and need the other.
• Double Angle Identity for Sine: \[ \sin 2\theta = 2 \sin \theta \cos \theta \]This is particularly useful when calculation for angles like \(2\alpha\) in our problem.

By understanding these identities, we could determine \(\sin \alpha\) using the Pythagorean identity, and \(\sin 2\alpha\) using the double angle formula. Remember, identities not only aid in solving equations but also in verifying solutions.