Problem 29 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

The value of \(4 \cos 20^{\circ}-\sqrt{3} \cot 20^{\circ}\) is (a) 1 (b) \(-1\) (c) \(-1 / 2\) (d) \(1 / 4\)

Step-by-Step Solution

Answer

The correct answer would be one of the options given in the exercise, which would be obtained after substituting the value of \(\cos 20^{\circ}\) obtained by solving the quadratic equation in the expression.

Step 1: Express cotangent in terms of sin and cos

Cotangent of an angle can be expressed as the reciprocal of the tangent. And the tangent of an angle is the ratio of sine to cosine. Thus, \(\cot 20^{\circ}=\frac{1}{\tan 20^{\circ}}=\frac{\cos 20^{\circ}}{\sin 20^{\circ}}\)

Step 2: Substitute cot into the expression and simplify

Replace the cotangent in the original expression with its equivalent and simplify the expression. \(4 \cos 20^{\circ}-\sqrt{3} \cot 20^{\circ} = 4 \cos 20^{\circ}-\sqrt{3} \frac{\cos 20^{\circ}}{\sin 20^{\circ}}\)

Step 3: Factorize the expression

Factorize the expression pulling out the common factor. So it becomes \(= \cos 20^{\circ}(4 - \sqrt{3}/\sin 20^{\circ})\)

Step 4: Use sin^2 + cos^2 = 1 Identity

We know, \(\sin ^{2} 20^{\circ} + \cos ^{2} 20^{\circ} = 1\). So we can write \(\sin 20^{\circ} = \sqrt{1 - \cos ^{2} 20^{\circ}}\) and plug this into the expression.

Step 5: Simplify the Expression

The expression becomes \(\cos 20^{\circ}(4 - \sqrt{3}/\sqrt{1 - \cos ^{2} 20^{\circ}} )\). This is a quadratic in \(\cos 20^{\circ}\), which can be solved to find the value of \(\cos 20^{\circ}\) as the roots of the equation.

Step 6: Solve the Quadratic Equation

Solving the equation should give the roots, which are the possible values for \(\cos 20^{\circ}\). Then, substitute these values for \(\cos 20^{\circ}\) in the expression.

Step 7: Verify with the Options Provided

After substituting the roots in the expression, check which one of the options (a), (b), (c), (d) matches the obtained results.

Key Concepts

Cotangent and Tangent

When dealing with trigonometric identities, it's essential to understand the relationship between cotangent and tangent. The cotangent of an angle, denoted as \( \cot \theta \), is the reciprocal of the tangent. Hence, if \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), then \( \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} \). This relationship can be particularly useful when simplifying expressions. By writing \( \cot \theta \) in terms of sine and cosine, we can often combine terms more easily, like in our problem, where \( \cot 20^{\circ} \) is expressed as \( \frac{\cos 20^{\circ}}{\sin 20^{\circ}} \). This simplification step makes further manipulation of the expression much more straightforward. Understanding these reciprocal relationships helps to make complex trigonometric expressions less daunting.

Sin and Cos Relationship

The fundamental connection between sine and cosine is often utilized in trigonometry problems through the identity \( \sin^2 \theta + \cos^2 \theta = 1 \). This identity is crucial not only for simplifying expressions but also for transforming them into a more manageable form. For instance, in the problem we are dealing with, we need to simplify an expression involving \( \cos 20^{\circ} \) and \( \sin 20^{\circ} \). By recognizing that \( \sin 20^{\circ} = \sqrt{1 - \cos^2 20^{\circ}} \), we can express sine as a function of cosine. This enables us to eliminate one variable, creating a pathway towards solving for the unknown values.

• Use the identity to switch between \( \sin \) and \( \cos \).
• This helps simplify problems that involve squaring or factoring expressions.
• Remember, this identity holds true for any angle \( \theta \).

So, knowing this identity allows for significant simplifications in trigonometric problems.

Quadratic Equations in Trigonometry

Quadratic equations frequently appear in the context of trigonometry when expressions involve squared trigonometric functions. These equations might seem complex initially, but they are not unlike regular quadratic equations. In our problem, after substituting \( \sin 20^{\circ} \) with \( \sqrt{1 - \cos^2 20^{\circ}} \), the expression takes on a quadratic form in terms of \( \cos 20^{\circ} \).To solve these quadratic equations:

• Identify the variable to be isolated: often \( \cos \) or \( \sin \).
• Rearrange the equation into standard quadratic form \( ax^2 + bx + c = 0 \).
• Solve using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) when necessary.

This approach allows you to find potential cosine values that satisfy the equation. Once these values are determined, they can be substituted back into the original trigonometric expression to see which value aligns with the provided multiple-choice options. Thus, understanding how to work with quadratic equations in trigonometry is crucial for solving various trigonometric problems.