Problem 20 - Algebra 2 and Trigonometry

Question

a. Find the exact value of \(\cos 315^{\circ}\) by using \(\cos \left(270^{\circ}+45^{\circ}\right) .\) b. Find the exact value of \(\sin 315^{\circ}\) by using \(\cos ^{2} \theta+\sin ^{2} \theta=1\) and the value of \(\cos 315^{\circ}\) found in a. c. Find the exact value of \(\cos 345^{\circ}\) by using \(\cos \left(315^{\circ}+30^{\circ}\right) .\) d. Explain why \(\cos 405^{\circ}=\cos 45^{\circ}\)

Step-by-Step Solution

Answer

a. \( \cos 315^{\circ} = \frac{\sqrt{2}}{2} \); b. \( \sin 315^{\circ} = -\frac{\sqrt{2}}{2} \); c. \( \cos 345^{\circ} = \frac{\sqrt{6}+\sqrt{2}}{4} \); d. \( \cos 405^{\circ} = \cos 45^{\circ} \).

Step 1: Use the Cosine Addition Formula for 315°

We can find \( \cos 315^{\circ} \) by noting that \( 315^{\circ} = 270^{\circ} + 45^{\circ} \). Use the cosine addition formula: \( \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \).For \( \cos(270^{\circ} + 45^{\circ}) \):- \( \alpha = 270^{\circ} \) and \( \beta = 45^{\circ} \)- \( \cos 270^{\circ} = 0 \) and \( \sin 270^{\circ} = -1 \)- \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \) and \( \sin 45^{\circ} = \frac{\sqrt{2}}{2} \)Thus, \[ \cos 315^{\circ} = \cos 270^{\circ} \cos 45^{\circ} - \sin 270^{\circ} \sin 45^{\circ} = 0 \times \frac{\sqrt{2}}{2} - (-1) \times \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \].

Step 2: Use Pythagorean Identity to Find \( \sin 315^{\circ} \)

Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \) and knowing \( \cos 315^{\circ} = \frac{\sqrt{2}}{2} \), substitute \( \cos 315^{\circ} \) into the equation:\[ \left(\frac{\sqrt{2}}{2}\right)^2 + \sin^2 315^{\circ} = 1 \]This simplifies to:\[ \frac{2}{4} + \sin^2 315^{\circ} = 1 \]\[ \frac{1}{2} + \sin^2 315^{\circ} = 1 \]Subtract \( \frac{1}{2} \) from both sides:\[ \sin^2 315^{\circ} = \frac{1}{2} \]\[ \sin 315^{\circ} = -\frac{\sqrt{2}}{2} \]Since \( 315^{\circ} \) is in the fourth quadrant, where sine is negative.

Step 3: Use Cosine Addition Formula for 345°

To find \( \cos 345^{\circ} \), express it as \( \cos(315^{\circ} + 30^{\circ}) \). Use the cosine addition formula:\[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \]Substitute known values:- \( \alpha = 315^{\circ} \) and \( \beta = 30^{\circ} \)- \( \cos 315^{\circ} = \frac{\sqrt{2}}{2} \), \( \sin 315^{\circ} = -\frac{\sqrt{2}}{2} \)- \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \), \( \sin 30^{\circ} = \frac{1}{2} \)Calculate \( \cos 345^{\circ} \):\[ \cos 345^{\circ} = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} - \left(-\frac{\sqrt{2}}{2}\right) \times \frac{1}{2} \]\[ = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4} \]

Step 4: Explain Why \( \cos 405^{\circ} = \cos 45^{\circ} \)

Angles that differ by a full circle (e.g., 360°) have identical trigonometric values. Since \( 405^{\circ} = 360^{\circ} + 45^{\circ} \), \( \cos 405^{\circ} \) is the same as \( \cos 45^{\circ} \).Therefore, regardless of how many full circles you rotate, cosine remains unaffected:\( \cos 405^{\circ} = \cos 45^{\circ} = \frac{\sqrt{2}}{2} \).

Key Concepts

Cosine Addition Formula

The cosine addition formula is a fundamental tool in trigonometry that helps us find the cosine of the sum of two angles. The formula is expressed as:

• \( \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \)

Imagine needing to find \( \cos 315^{\circ} \), which can be expressed as \( \cos(270^{\circ} + 45^{\circ}) \). Here, the angle 270° is a special angle with known values: \( \cos 270^{\circ} = 0 \) and \( \sin 270^{\circ} = -1 \). For 45°, a common angle, we have \( \cos 45^{\circ} = \sin 45^{\circ} = \frac{\sqrt{2}}{2} \). Using these, the cosine addition formula becomes useful:

• \( \cos 315^{\circ} = \cos 270^{\circ} \cos 45^{\circ} - \sin 270^{\circ} \sin 45^{\circ} \)
• = \( 0 \times \frac{\sqrt{2}}{2} - (-1) \times \frac{\sqrt{2}}{2} \)
• = \( \frac{\sqrt{2}}{2} \)

Thus, through the cosine addition formula, we've found that \( \cos 315^{\circ} = \frac{\sqrt{2}}{2} \). This formula is not only handy for evaluating precise trigonometric values but also for solving complex trigonometric equations.

Pythagorean Identity

The Pythagorean identity is a key concept in trigonometry. It is expressed as:

• \( \cos^2 \theta + \sin^2 \theta = 1 \)

This identity illustrates a fundamental relationship between the cosine and sine of an angle. For example, having found \( \cos 315^{\circ} = \frac{\sqrt{2}}{2} \), we can use this identity to find \( \sin 315^{\circ} \):

• \( \left(\frac{\sqrt{2}}{2}\right)^2 + \sin^2 315^{\circ} = 1 \)
• \( \frac{2}{4} + \sin^2 315^{\circ} = 1 \)
• \( \frac{1}{2} + \sin^2 315^{\circ} = 1 \)

Solving for \( \sin 315^{\circ} \) gives:

• \( \sin^2 315^{\circ} = \frac{1}{2} \)
• \( \sin 315^{\circ} = -\frac{\sqrt{2}}{2} \) \(\text{(since 315° is in the fourth quadrant where sine is negative)}\)

The Pythagorean identity is invaluable in trigonometry, consistently verifying relationships between sine and cosine, and aiding in the determination of unknown trigonometric values.

Angle Reduction

Angle reduction simplifies dealing with angles larger than 360° or negative angles by associating them with "simpler" angles. A remarkable property of trigonometric functions is their periodic nature. For cosine, a primary period is 360°:

• Given \( \theta = 405^{\circ} \), the reduction is \( \theta - 360^{\circ} = 45^{\circ} \)
• Thus, \( \cos 405^{\circ} = \cos 45^{\circ} \)

This simplification works because moving around the unit circle in full circles (like adding or subtracting 360°) doesn't affect the trigonometric values. Consequently, \( \cos 405^{\circ} = \frac{\sqrt{2}}{2} \), the same as \( \cos 45^{\circ} \). When you remember that trigonometric functions are periodic, angle reduction becomes an incredibly useful tool. It allows us to easily evaluate trigonometric functions at seemingly complex angles by returning to known values.

Special Angles

In trigonometry, special angles such as 30°, 45°, and 60° have well-defined trigonometric values. They are often used as reference angles due to their simplicity and frequent occurrence in problems:

• \( \cos 45^{\circ} = \sin 45^{\circ} = \frac{\sqrt{2}}{2} \)
• \( \cos 30^{\circ} = \frac{\sqrt{3}}{2}, \; \sin 30^{\circ} = \frac{1}{2} \)
• \( \cos 60^{\circ} = \frac{1}{2}, \; \sin 60^{\circ} = \frac{\sqrt{3}}{2} \)

These values are derived from equilateral and isosceles triangles and are essential in solving trigonometric equations and simplifying expressions. When tackling trigonometric problems involving addition or subtraction of known angles, we often depend on these special angles to calculate exact values, like finding \( \cos 345^{\circ} \) through \( \cos(315^{\circ} + 30^{\circ}) \):

• \( \cos 345^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4} \)

Understanding these special angles will boost your ability to solve a wide range of trigonometric problems with precision and ease.