Problem 57 - A Complete Resource Book in Mathematics for JEE Main 2017

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

The correct answer is (B) \(-\frac{24}{25}\).

Step 1: Understand the Problem

The given equation is a quadratic in terms of \(\cos \theta\). We need to find \(\alpha\) such that it lies in the interval \(\frac{\pi}{2}<\alpha<\pi\) and then determine \(\sin 2\alpha\).

Step 2: Express the Equation as a Quadratic

The given equation is \(25 \cos^2 \theta + 5 \cos \theta - 12 = 0\). We express it as a standard quadratic equation by letting \(x = \cos \theta\), giving us \(25x^2 + 5x - 12 = 0\).

Step 3: Solve the Quadratic Equation

We use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve for \(x\). Here, \(a = 25\), \(b = 5\), and \(c = -12\). Substituting these values, we have:\[x = \frac{-5 \pm \sqrt{5^2 - 4 \times 25 \times (-12)}}{2 \times 25} \] \[x = \frac{-5 \pm \sqrt{25 + 1200}}{50}\] \[x = \frac{-5 \pm \sqrt{1225}}{50}\]\[x = \frac{-5 \pm 35}{50}\]The solutions are \(x = \frac{30}{50} = \frac{3}{5}\) and \(x = \frac{-40}{50} = -\frac{4}{5}\).

Step 4: Determine Which Root is Valid for \(\alpha\)

Since \(\cos \alpha = -\frac{4}{5}\) must lie in the domain \(\frac{\pi}{2}<\alpha<\pi\) (where \(\cos\) is negative), the valid root of the quadratic that satisfies this condition is \(-\frac{4}{5}\).

Step 5: Calculate \(\sin 2\alpha\)

Knowing \(\cos \alpha = -\frac{4}{5}\), we can find \(\sin \alpha\) using \(\sin^2 \alpha + \cos^2 \alpha = 1\) to find \(\sin \alpha\): \[\sin^2 \alpha = 1 - \left(-\frac{4}{5}\right)^2\] \[\sin^2 \alpha = 1 - \frac{16}{25} = \frac{9}{25}\] Using \(\frac{\pi}{2}<\alpha<\pi\), \(\sin \alpha\) is positive, so \(\sin \alpha = \frac{3}{5}\). Now, calculate \(\sin 2\alpha\):\(\sin 2\alpha = 2 \sin \alpha \cos \alpha = 2 \times \frac{3}{5} \times -\frac{4}{5} = -\frac{24}{25}\).

Step 6: Choose the Correct Answer

From the options given, \(-\frac{24}{25}\) corresponds to option (B).

Key Concepts

Quadratic Equation

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown variable. These equations are fundamental in algebra and are solved using various methods including factoring, completing the square, and the quadratic formula.

The quadratic formula, \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), is a reliable way to find the roots of any quadratic equation. The expression inside the square root, \(b^2 - 4ac\), is known as the discriminant. It indicates the nature of the roots:

• If the discriminant is positive, the equation has two distinct real roots.
• If it is zero, there is exactly one real root, also called a repeated or double root.
• If the discriminant is negative, the roots are complex or imaginary.

In this exercise, solving the quadratic \(25x^2 + 5x - 12 = 0\) using the quadratic formula gives us the possible roots for \(x = \cos \theta\). The solution requires careful substitution and arithmetic operations to ensure accuracy in finding these roots.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables within their domains. They are essential tools in simplifying expressions and solving trigonometric equations. One of the most important identities is the Pythagorean identity: \(\sin^2 \theta + \cos^2 \theta = 1\).

This identity helps in determining one trigonometric function if the other is known. For example, if \(\cos \alpha\) is given, \(\sin^2 \alpha\) can be found by rearranging the Pythagorean identity to \(\sin^2 \alpha = 1 - \cos^2 \alpha\).

In our problem, after finding \(\cos \alpha = -\frac{4}{5}\), we use this identity to find \(\sin \alpha\). Since \(\alpha\) is in the interval \(\frac{\pi}{2} < \alpha < \pi\), where sine is positive, we determine that \(\sin \alpha = \frac{3}{5}\). Utilizing these identities allows us to solve trigonometric equations efficiently and accurately.

Trigonometric Functions

Trigonometric functions such as sine and cosine are fundamental in mathematics, describing relationships in right-angle triangles and periodic phenomena. These functions have specific values depending on the angle, and each function is associated with an angle's ratio in a right triangle.

The cosine function, \(\cos \theta\), represents the adjacent side over the hypotenuse in a right triangle, while the sine function, \(\sin \theta\), represents the opposite side over the hypotenuse. Trigonometric functions are periodic, cycling values at regular intervals, typically on the unit circle.

• Cosine is negative in the second quadrant, which is relevant in this problem as \(\alpha\) lies within \(\frac{\pi}{2} < \alpha < \pi\).
• Sine is positive in both the first and second quadrants.

When solving for \(\sin 2\alpha\), we use the identity \(\sin 2\alpha = 2 \sin \alpha \cos \alpha\). With \(\cos \alpha = -\frac{4}{5}\) and \(\sin \alpha = \frac{3}{5}\), this calculation gives \(\sin 2\alpha = -\frac{24}{25}\), which matches the answer option \(B\). Understanding these functions allows us to work with trigonometric equations and identities effectively.