Problem 6 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\cos 25^{\circ}+\sin 25^{\circ}=k\), then \(\cos 50^{\circ}\) is equal to (A) \(k \sqrt{2-k^{2}}\) (B) \(-\sqrt{2-k^{2}}\) (C) \(\sqrt{2-k^{2}}\) (D) \(-k \sqrt{2-k^{2}}\)

Step-by-Step Solution

Answer

(C) \( \sqrt{2-k^{2}} \).

Step 1: Express cos(50°) Using Double Angle Formula

We know the double angle identity for cosine: \( \cos(2\theta) = 1 - 2\sin^2(\theta) \). Since we have \( \cos(50^{\circ}) = \cos(2 \cdot 25^{\circ}) \), let's use this formula: \( \cos(50^{\circ}) = 1 - 2 \sin^2(25^{\circ}) \).

Step 2: Express sin²(25°) in Terms of k

From the given \( \cos(25^{\circ}) + \sin(25^{\circ}) = k \), we need to find expressions for \( \cos(25^{\circ}) \) and \( \sin(25^{\circ}) \). Let's consider \( (\cos(25^{\circ}) + \sin(25^{\circ}))^2 = k^2 \), leading to \( \cos^2(25^{\circ}) + 2\cos(25^{\circ})\sin(25^{\circ}) + \sin^2(25^{\circ}) = k^2 \).

Step 3: Simplify the Expression

Since \( \cos^2(25^{\circ}) + \sin^2(25^{\circ}) = 1 \), we have \( 1 + 2\cos(25^{\circ})\sin(25^{\circ}) = k^2 \). Therefore, \( \cos(25^{\circ})\sin(25^{\circ}) = \frac{k^2 - 1}{2} \).

Step 4: Calculate sin²(25°)

Substituting back into the identity \( \cos^2(50^{\circ}) = 1 - 2\sin^2(25^{\circ}) \), first express \( \sin^2(25^{\circ}) \) using the complement: \( \sin^2(25^{\circ}) = 1 - \cos^2(25^{\circ}) \). \( \cos(25^{\circ}) = \frac{k}{\sqrt{2}} \), then \( \sin^2(25^{\circ}) = 1 - \left(\frac{k}{\sqrt{2}}\right)^2 = \frac{2-k^2}{2} \).

Step 5: Calculate cos(50°)

Using \( \cos^2(50^{\circ}) = 1 - \frac{2-k^2}{2} \), we have \( \cos^2(50^{\circ}) = \frac{k^2}{2} \). Consequently, \( \cos(50^{\circ}) = \sqrt{2-k^2} \).

Step 6: Match the Result to the Options

The calculated \( \cos(50^{\circ}) \) matches option (C), which is \( \sqrt{2-k^{2}} \). This confirms our calculation is correct.

Key Concepts

Double Angle Formula

The Double Angle Formula is a powerful tool in trigonometry. It allows you to express trigonometric functions at double angles in terms of single angles. One common version for the cosine is given by \( \cos(2\theta) = 1 - 2\sin^2(\theta) \). This is particularly useful because it helps simplify equations and solve problems involving angles that are double or half of each other.

• For example, in the problem provided, the double angle formula for cosine is used to express \( \cos(50^{\circ}) \). Noticing that \( 50^{\circ} \) is \( 2 \times 25^{\circ} \) suggests utilizing the formula \( \cos(2 \cdot 25^{\circ}) = 1 - 2\sin^2(25^{\circ})\).

Using this formula involves knowing \( \sin(\theta) \) or \( \cos(\theta) \) or being able to calculate them from given information or assumptions. Double Angle Formulas can also be used for sine and tangent functions, which similarly allow transformation of angle measures.

Sum and Difference Formulas

The Sum and Difference Formulas for trigonometry add great flexibility in handling trigonometric problems, especially for non-standard angles or when combining multiple angles. These formulas involve the sine and cosine of sums or differences of angles. For instance:

• For sine: \( \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B \)
• For cosine: \( \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B \)

In our specific problem, while these formulas aren't explicitly used, understanding them could help derive or verify the answers. When given key values, these formulas can be used to break down complex expressions into simpler parts, often reducing the need for direct, cumbersome calculations. These formulas are vital in scenarios involving angle summations, providing ready responses for expressions like \( \cos(50^{\circ}) \) in terms of easier or known angles.

Trigonometric Equations

Solving trigonometric equations means finding angle measures (or equivalently, function values) that satisfy given trigonometric relationships. It typically involves the use of identities and algebraic manipulation. Consider a scenario like the exercise, where you have an equation involving trigonometric functions such as:

• \( \cos(\theta) + \sin(\theta) = k \)

To solve it, you often need to employ identities like \( \sin^2(\theta) + \cos^2(\theta) = 1 \). Here, factoring, squaring both sides, or introducing auxiliary angles can help. For example, squaring \( \cos(25^{\circ}) + \sin(25^{\circ}) = k \) is a trick to reveal expressions in terms of \( k \), simplifying a multi-step solution. Finally, matching derived expressions with given options, as in our problem's solution process, is a common end goal in problem sets. Aligning these solutions and verifying them with known options can give confidence in your approach and results.