A right triangle with an angle of has a hypotenuse of . Calculate the side of the triangle opposite to the angle (y), and the side adjacent to the angle (x). Round your answer to 3 decimal places.
Table of contents
- 0. Review of College Algebra4h 45m
- 1. Measuring Angles40m
- 2. Trigonometric Functions on Right Triangles2h 5m
- 3. Unit Circle1h 19m
- 4. Graphing Trigonometric Functions1h 19m
- 5. Inverse Trigonometric Functions and Basic Trigonometric Equations1h 41m
- 6. Trigonometric Identities and More Equations2h 34m
- 7. Non-Right Triangles1h 38m
- 8. Vectors2h 25m
- 9. Polar Equations2h 5m
- 10. Parametric Equations1h 6m
- 11. Graphing Complex Numbers1h 7m
2. Trigonometric Functions on Right Triangles
Solving Right Triangles
Problem 3
Textbook Question
Graph one period of each function. y = 2 tan x/2
Verified step by step guidance1
Identify the given function: \(y = 2 \tan \left( \frac{x}{2} \right)\).
Recall the basic properties of the tangent function \(\tan x\): it has vertical asymptotes where its argument equals \(\frac{\pi}{2} + k\pi\), for any integer \(k\), and its period is \(\pi\).
Determine the period of the transformed function \(\tan \left( \frac{x}{2} \right)\) using the formula for the period of \(\tan(bx)\), which is \(\frac{\pi}{|b|}\). Here, \(b = \frac{1}{2}\), so the period is \(\pi \div \frac{1}{2} = 2\pi\).
Find the vertical asymptotes by solving \(\frac{x}{2} = \frac{\pi}{2} + k\pi\), which gives \(x = \pi + 2k\pi\). For one period, consider \(k=0\) to find asymptotes at \(x = \pi\) and \(x = -\pi\).
Plot the graph between the vertical asymptotes \(x = -\pi\) and \(x = \pi\), noting that the function is scaled vertically by a factor of 2, so the values of \(\tan \left( \frac{x}{2} \right)\) are multiplied by 2.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Period of the Tangent Function
The standard tangent function, y = tan(x), has a period of π, meaning it repeats every π units. When the function is modified to y = tan(bx), the period changes to π/|b|. For y = 2 tan(x/2), the coefficient inside the tangent is 1/2, so the period is π divided by 1/2, which equals 2π.
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Amplitude and Vertical Stretch
Although tangent functions do not have a maximum or minimum value, the coefficient outside the function, here 2, vertically stretches the graph. This means the output values are multiplied by 2, making the graph steeper and the values grow faster, but it does not affect the period or asymptotes.
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Vertical Asymptotes of Tangent Functions
Tangent functions have vertical asymptotes where the function is undefined, occurring at points where the cosine in the denominator is zero. For y = tan(bx), asymptotes occur at x = (π/2 + kπ)/b, where k is any integer. For y = 2 tan(x/2), asymptotes are at x = π + 2kπ, guiding the graph's shape and domain restrictions.
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