Problem 39 - Precalculus with Calculus Previews

Question

Use a half-angle formula to find the exact value of the given trigonometric function. Do not use a calculator. $$ \cos 67.5^{\circ} $$

Step-by-Step Solution

Answer

The exact value is \( \frac{\sqrt{2 - \sqrt{2}}}{2} \).

Step 1: Understand the Problem

We need to find the exact value of \( \cos 67.5^{\circ} \) using a half-angle formula.

Step 2: Identify a Relevant Half-Angle Formula

We use the formula for cosine of half an angle: \[ \cos \theta = \sqrt{\frac{1 + \cos 2\theta}{2}} \] to find \( \cos 67.5^{\circ} \). Since \( 67.5^{\circ} = \frac{135^{\circ}}{2} \), we identify that \( 2\theta = 135^{\circ} \).

Step 3: Find \( \cos 135^{\circ} \)

We know that \( 135^{\circ} = 180^{\circ} - 45^{\circ} \), and in the second quadrant, cosine is negative, so \( \cos 135^{\circ} = -\cos 45^{\circ} = -\frac{\sqrt{2}}{2} \).

Step 4: Apply the Half-Angle Formula

Substitute \( \cos 135^{\circ} = -\frac{\sqrt{2}}{2} \) into the half-angle formula: \[ \cos 67.5^{\circ} = \sqrt{\frac{1 + \left(-\frac{\sqrt{2}}{2}\right)}{2}} = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}} \].

Step 5: Simplify the Expression

Simplify the fraction from the half-angle formula: \[ \cos 67.5^{\circ} = \sqrt{\frac{2}{2} - \frac{\sqrt{2}}{2}}{2} = \sqrt{\frac{2 - \sqrt{2}}{4}} = \frac{\sqrt{2 - \sqrt{2}}}{2} \].

Step 6: Conclude the Exact Value

The exact value of \( \cos 67.5^{\circ} \) is \( \frac{\sqrt{2 - \sqrt{2}}}{2} \).

Key Concepts

Trigonometric Functions

Trigonometric functions are an essential part of mathematics, especially in fields such as geometry, physics, and engineering. They relate angles of a triangle to the ratios of its sides. The main trigonometric functions include sine (\( \sin \theta \)), cosine (\( \cos \theta \)), and tangent (\( \tan \theta \)). Each function has its unique characteristics and relationships with one another.

Understanding these relationships helps when solving problems involving angles and lengths.Some notable aspects of trigonometric functions are:

• They are periodic, meaning they repeat their values at regular intervals.
• They can take values between -1 and 1.
• They are fundamental in modeling oscillatory processes like sound waves or tides.

In the context of our exercise, the cosine function is vital in helping us solve for specific angles using the half-angle formula. Mastering the trigonometric functions is key to solving such mathematical problems.

Cosine

Cosine is a primary trigonometric function, expressed as \( \cos \theta \). It is defined as the ratio of the adjacent side to the hypotenuse in a right triangle. Cosine values are crucial in determining the magnitude of an angle's effect in equations and modeling various natural phenomena.With cosine:

• The values range from -1 to 1.
• It is even, meaning \( \cos(-\theta) = \cos(\theta) \)
• It helps determine the horizontal component of a vector in physics.

The half-angle formulas, like the one used in our exercise, simplify the calculation of cosine for specific angles without using a calculator. Knowing how to work with cosine, including its properties and formulas, is essential for various mathematical applications, such as finding exact angle values.

Exact Value of Angles

Finding the exact value of angles like \( 67.5^{\circ} \) involves using specific trigonometric identities and formulas. This task can be extremely fruitful when the goal is to avoid decimal approximations that calculators might offer. Using half-angle formulas lets you derive precise values by knowing only some key angles in the unit circle.For example, the half-angle formulas offer a pathway to express trigonometric functions for new angles that are not part of the common ones, like \( 30^{\circ}, 45^{\circ}, 60^{\circ} \).

The process follows these steps:

• Find a related known angle that's double or half the required angle.
• Use trigonometric identities to derive the value.
• Simplify the expression to get the exact value.

This procedure results in an elegant expression of an otherwise complex value, maintaining mathematical purity and precision. Being comfortable with such transformations paves the way for solving broader classes of trigonometric problems effortlessly.