Problem 27 - Algebra 2 and Trigonometry

Question

In \(27-32,\) for each of the given inverse trigonometric function values, find the exact function value. \(\sin (\arccos 1)\)

Step-by-Step Solution

Answer

The exact function value is \( \sin(\arccos(1)) = 0 \).

Step 1: Understand the Problem

The problem asks us to find the exact value of the trigonometric function \( \sin(\arccos(1)) \). This involves understanding what \( \arccos(1) \) represents.

Step 2: Evaluate \( \arccos(1) \)

The function \( \arccos(x) \) gives the angle whose cosine is \( x \). We need to determine the angle \( \theta \) such that \( \cos(\theta) = 1 \). Since \( \cos(0) = 1 \), we have \( \theta = 0 \). Thus, \( \arccos(1) = 0 \).

Step 3: Compute \( \sin(0) \)

Now that we know \( \arccos(1) = 0 \), we need to find \( \sin(0) \). Recall the sine of an angle is the y-coordinate of the unit circle at that angle. Thus, \( \sin(0) = 0 \).

Key Concepts

Understanding Arccos

Arccos, short for arc cosine, is one of the inverse trigonometric functions. It specifically deals with finding the angle whose cosine value is that given number. If we have \( \arccos(x) \), it means "what angle has a cosine of \( x \)?".
Let's take it step by step:

• The range of \( \arccos \) is from 0 to \( \pi \) radians, which is about 0 to 180 degrees. This is because cosine is positive in the first quadrant and negative in the second.
• When you see \( \arccos(1) \), you are looking for an angle where the cosine value is 1. The only place on the unit circle where cosine equals 1 is at 0 radians (or 0 degrees).
• This makes \( \arccos(1) = 0 \).

These characteristics make \( \arccos \) really useful in trigonometry problems, especially when trying to solve equations involving cosine functions.

Exploring Trigonometric Identities

Trigonometric identities are like tools that make solving math problems much smoother. They are equations that involve trigonometric functions and are true for any value of the included variables.
Some key identities include:

• \( \sin^2(\theta) + \cos^2(\theta) = 1 \)
• \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \)
• \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) \)

For our problem, observe that since \( \arccos(1) = 0 \), it's straightforward to check identity:
When \( \theta = 0 \), \( \cos(0) = 1 \) holds true. Also, knowing basic identities can often confirm your solution is correct. They are foundational because they help establish relationships between different trigonometric functions like sine, cosine, and tangent.

The Power of the Unit Circle

The unit circle is a crucial concept in trigonometry. It's a circle with a radius of 1, centered at the origin of a coordinate plane. This circle helps us visualize and understand angles' sine and cosine values visually.
Some key points to understand include:

• The unit circle allows us to define sine and cosine for all real numbers \( \theta \).
• The x-coordinate of a point on the unit circle at angle \( \theta \) represents \( \cos(\theta) \), while the y-coordinate represents \( \sin(\theta) \).
• For example, at \( \theta = 0 \), the point on the unit circle is (1, 0), which means \( \cos(0) = 1 \) and \( \sin(0) = 0 \).

The unit circle is an excellent tool to remember key angle values and to see symmetries between different trigonometric functions. It simplifies calculating function values, just like in our \( \sin(\arccos(1)) = 0 \) problem.

Insights into the Sine Function

The sine function is a primary trigonometric function that relates an angle of a right triangle to the ratio of the length of the opposite side over the hypotenuse. In the context of the unit circle, sine represents the y-coordinate
Here are some essential features:

• Sine is periodic with a period of \( 2\pi \), meaning it repeats its values every \( 2\pi \) radians.
• Its range is from -1 to 1.
• The function is an odd function, which means \( \sin(-\theta) = -\sin(\theta) \).

When evaluating \( \sin(\arccos(1)) \), remember you're essentially looking at the y-value of the specified angle on the unit circle, which is 0. That's because \( \arccos(1) \) gives us 0 radians and consequently using the fact that \( \sin(0) = 0 \), helps you understand why \( \sin(\arccos(1)) = 0 \). This highlights how interconnected trigonometric functions and their inverses can be visualized and solved.