Question

$\frac{\cos 75^{\circ} - \cos 15^{\circ}}{\cos 75^{\circ} + \cos 15^{\circ}} =$

• A. $\frac{-1}{\sqrt{3}}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{-1}{\sqrt{2}}$
• E. $\sqrt{3}$

Hint:

$\frac{\cos A - \cos B}{\cos A + \cos B} = -\tan(\frac{A+B}{2}) \tan(\frac{A-B}{2})$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to simplify a trigonometric expression which is a ratio of the difference and sum of two cosine values. This is a direct application of the sum-to-product formulas.
Step 2: Key Formula or Approach:
We will use the following sum-to-product identities:
1. \( \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \)
2. \( \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right) \)
Step 3: Detailed Explanation:
Let \( A = 75^\circ \) and \( B = 15^\circ \).
First, calculate the average and half-difference of the angles:
\[ \frac{A+B}{2} = \frac{75^\circ + 15^\circ}{2} = \frac{90^\circ}{2} = 45^\circ \] \[ \frac{A-B}{2} = \frac{75^\circ - 15^\circ}{2} = \frac{60^\circ}{2} = 30^\circ \] Now apply the formulas to the numerator and denominator of the given expression.
Numerator:
\[ \cos 75^\circ - \cos 15^\circ = -2 \sin(45^\circ) \sin(30^\circ) \] Denominator:
\[ \cos 75^\circ + \cos 15^\circ = 2 \cos(45^\circ) \cos(30^\circ) \] Now, form the fraction:
\[ \frac{-2 \sin(45^\circ) \sin(30^\circ)}{2 \cos(45^\circ) \cos(30^\circ)} \] Cancel the 2s:
\[ - \frac{\sin(45^\circ)}{\cos(45^\circ)} \cdot \frac{\sin(30^\circ)}{\cos(30^\circ)} \] This simplifies to:
\[ - \tan(45^\circ) \tan(30^\circ) \] Substitute the known values for the tangent function:
\( \tan(45^\circ) = 1 \)
\( \tan(30^\circ) = \frac{1}{\sqrt{3}} \)
The final result is:
\[ - (1) \left(\frac{1}{\sqrt{3}}\right) = -\frac{1}{\sqrt{3}} \] Step 4: Final Answer:
The value of the expression is \( -\frac{1}{\sqrt{3}} \).