Problem 15 - A Graphical Approach to Precalculus with Limits

Question

Use identities to find the exact value of each expression. Do not use a calculator. $$\cos \left(-15^{\circ}\right)$$

Step-by-Step Solution

Answer

\(\cos(-15^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}\).

Step 1: Understanding the Negativity Property

The cosine function is even, which means that for any angle \(\theta\), \(\cos(-\theta) = \cos(\theta)\). Therefore, \(\cos(-15^\circ) = \cos(15^\circ)\). So, we can focus on finding \(\cos(15^\circ)\).

Step 2: Use the Angle Difference Identity

To find \(\cos(15^\circ)\), we can use the cosine angle difference identity: \(\cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b)\). Let's choose \(a = 45^\circ\) and \(b = 30^\circ\). Thus, \(\cos(15^\circ) = \cos(45^\circ - 30^\circ)\).

Step 3: Apply the Values for the Identity

Using known values, \(\cos(45^\circ) = \frac{\sqrt{2}}{2}\), \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\), \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\), and \(\sin(30^\circ) = \frac{1}{2}\). Plug these into the identity: \(\cos(15^\circ) = \cos(45^\circ)\cos(30^\circ) + \sin(45^\circ)\sin(30^\circ)\).

Step 4: Calculate the Expression

Substitute the values: \(\cos(15^\circ) = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)\). This simplifies to \(\frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}\).

Step 5: Simplify the Result

Combine the fractions: \(\cos(15^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}\). Therefore, \(\cos(-15^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4}\).

Key Concepts

Cosine Function

The cosine function is one of the primary trigonometric functions and is particularly useful in understanding how angles work in different contexts. An interesting property of the cosine function is that it is an "even function." This means that the cosine of a negative angle is the same as the cosine of the positive angle. Mathematically, this is expressed as

• \(\cos(-\theta) = \cos(\theta)\).

So, for example, \(\cos(-15^\circ)\) simplifies to \(\cos(15^\circ)\). This property is helpful because it allows us to focus on the absolute angle value without concern for direction. The feature of being an even function makes cosine particularly useful for problems involving symmetry and reflections. The cosine function describes the horizontal component of a unit circle, and this symmetry across the y-axis makes it quite powerful in trigonometry.

Angle Difference Identity

The angle difference identity for cosine is a crucial tool in trigonometry that enables us to find the exact values of specific angles. It is given by the formula:

• \(\cos(a - b) = \cos(a)\cos(b) + \sin(a)\sin(b)\).

This identity is especially useful when you need to find the cosine of angles that are not standard angles, like \(45^\circ\) or \(30^\circ\), by breaking them down into these more manageable components.
For example, finding \(\cos(15^\circ)\) involves expressing it as \(\cos(45^\circ - 30^\circ)\). By using the angle difference identity, we leverage known values of cosine and sine for \(45^\circ\) and \(30^\circ\) to construct the value for \(15^\circ\). This decomposition is a clever way to solve trigonometric problems where direct measurement or calculation might be cumbersome, and it helps in constructing exact values without approximation.

Exact Trigonometric Values

Exact trigonometric values apply to very specific angles that have been established through geometric constructions. These include angles like \(0^\circ\), \(30^\circ\), \(45^\circ\), \(60^\circ\), and \(90^\circ\), among others. Knowing the exact trigonometric values for these angles lets us solve more complicated problems by breaking them down into these simpler known values.

• \(\cos(45^\circ) = \frac{\sqrt{2}}{2}\)
• \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\)
• \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\)
• \(\sin(30^\circ) = \frac{1}{2}\)

These values are derived from the properties of an equilateral triangle and the unit circle. Memorizing these core values allows for quick computation and understanding of complex problems by decomposing them into manageable parts. They also provide a foundation for using identities like the angle difference identity to determine the values for less common angles by piecing together the known pieces.