Chapter 25: Problem 6
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) \(-k \sqrt{2-k^{2}}\)
Short Answer
Expert verified
(C) \( \sqrt{2-k^{2}} \).
Step by step solution
01
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: \( \cos(50^{\circ}) = 1 - 2 \sin^2(25^{\circ}) \).
02
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 \( \cos^2(25^{\circ}) + 2\cos(25^{\circ})\sin(25^{\circ}) + \sin^2(25^{\circ}) = k^2 \).
03
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, \( \cos(25^{\circ})\sin(25^{\circ}) = \frac{k^2 - 1}{2} \).
04
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 \( \sin^2(25^{\circ}) = 1 - \left(\frac{k}{\sqrt{2}}\right)^2 = \frac{2-k^2}{2} \).
05
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, \( \cos(50^{\circ}) = \sqrt{2-k^2} \).
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( \cos(2 \cdot 25^{\circ}) = 1 - 2\sin^2(25^{\circ})\).
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: \( \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B \)
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:
- \( \cos(\theta) + \sin(\theta) = k \)