Problem 4
Question
How would the six trigonometric functions of \(90^{\circ}\) and \(-270^{\circ}\) compare? Why?
Step-by-Step Solution
Verified Answer
The functions for both angles are the same because \(-270^{\circ} = 90^{\circ}\) when adding 360°.
1Step 1: Understand the Angle Relation
First, note the equivalence of the angles: \(90^{\circ}\) and \(-270^{\circ}\). To find if they coincide, add \(360^{\circ}\) to \(-270^{\circ}\): \(-270^{\circ} + 360^{\circ} = 90^{\circ}\). Thus, \(90^{\circ}\) and \(-270^{\circ}\) refer to the same position on the unit circle.
2Step 2: Calculate Trigonometric Functions for 90°
For the angle \(90^{\circ}\) or \(\frac{\pi}{2}\) radians on the unit circle:\(\sin 90^{\circ} = 1\)\(\cos 90^{\circ} = 0\)\(\tan 90^{\circ} = \text{undefined (since you divide by zero)}\)\(\csc 90^{\circ} = 1\)\(\sec 90^{\circ} = \text{undefined (since you divide by zero)}\)\(\cot 90^{\circ} = 0\).
3Step 3: Relate Functions for -270°
Because \(-270^{\circ}\) is equivalent to \(90^{\circ}\), the trigonometric functions are the same for both angles:\(\sin (-270^{\circ}) = 1\)\(\cos (-270^{\circ}) = 0\)\(\tan (-270^{\circ}) = \text{undefined}\)\(\csc (-270^{\circ}) = 1\)\(\sec (-270^{\circ}) = \text{undefined}\)\(\cot (-270^{\circ}) = 0\).
Key Concepts
Understanding the Unit CircleAngle EquivalenceExploring Undefined ValuesSine and Cosine Functions
Understanding the Unit Circle
The unit circle is a fundamental tool in trigonometry. It's simply a circle centered at the origin of a coordinate plane, with a radius of one unit. This allows us to conveniently calculate trigonometric functions for any angle by relying on the coordinates of points on the circle.
Key aspects of the unit circle include:
Key aspects of the unit circle include:
- The x-coordinate represents the cosine of the angle.
- The y-coordinate represents the sine of the angle.
- Angles are usually measured from the positive x-axis.
Angle Equivalence
Angle equivalence is a concept where angles that differ by multiples of 360° lead to the same trigonometric function values. This is because a full rotation around the circle doesn't change the position of points, so the trigonometric quantities remain constant.
For instance, 90° and -270° are equivalent because:
For instance, 90° and -270° are equivalent because:
- Add 360° to -270°: \(-270° + 360° = 90°\)
- They both describe the same position on the circle at \(\frac{\pi}{2}\) radians.
Exploring Undefined Values
Trigonometric functions sometimes become undefined, notably when involving division by zero. In the context of our examples, the \(\tan\) and \(\sec\) functions become undefined at specific angles.
Consider \(90^{\circ}\):
Consider \(90^{\circ}\):
- \(\tan 90^{\circ},\) equivalent to dividing the sine by the cosine, is undefined because \(\cos 90^{\circ} = 0\).
- \(\sec 90^{\circ},\) evaluated as \(1/\cos 90^{\circ},\) is also undefined.
Sine and Cosine Functions
The sine and cosine functions play central roles in the unit circle, directly relating to its coordinates. The sine function outputs the y-coordinate, while the cosine corresponds to the x-coordinate of any point on the circle.
For example, at \(90^{\circ}\) or \(-270^{\circ}\):
For example, at \(90^{\circ}\) or \(-270^{\circ}\):
- \(\sin 90^{\circ} = 1\), meaning the point is at the top of the circle, with maximum height.
- \(\cos 90^{\circ} = 0\), indicating the point lies directly above the center on the circle's edge.
Other exercises in this chapter
Problem 3
Fill in the blanks with the appropriate short answers. Do not use a calculator. The least positive angle coterminal with \(-180^{\circ}\) has degree measure___.
View solution Problem 4
Refer to the equations in the definition of simple harmonic motion in this section. and consider the following equation. \(s(t)=5 \cos 2 t, \quad\) where \(t\)
View solution Problem 4
Determine what fraction of the circumference of the unit circle each value of s represents. For example, \(s=\pi\) represents \(\frac{1}{2}\) of the circumferen
View solution Problem 4
Fill in the blanks with the appropriate short answers. Do not use a calculator. The complement of a \(40^{\circ}\) angle is ___ and the supplement of a \(40^{\c
View solution