Sine Function Solutions and Transformations

Sine Function Solutions and Transformations

Assessment

Interactive Video

Mathematics

9th - 10th Grade

Hard

Created by

Liam Anderson

FREE Resource

The video tutorial addresses solving a sine equation with a given domain, focusing on graph transformations and domain modifications. It explains the effects of 2x and pi/3 on the graph, including amplitude changes and horizontal shifts. The instructor demonstrates modifying the domain to align with the equation's argument, solving a simplified version of the equation, and identifying valid solutions within the modified domain. The tutorial concludes with finalizing the solution and verifying it using graph analysis.

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the main challenge in solving the equation sine(2x + π/3) = 1/2?

Understanding the horizontal shift and dilation

Calculating the amplitude

Identifying the vertical shift

Finding the period of the sine function

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How does the transformation 2x affect the sine graph?

It stretches the graph vertically

It increases the amplitude

It shifts the graph vertically

It compresses the graph horizontally

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the purpose of modifying the domain in this problem?

To match the range of the sine function

To simplify the equation

To align with the transformed function

To eliminate complex numbers

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which angle is a solution to sine(θ) = 1/2 in radians?

π/4

π/3

π/6

5π/6

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Why is 5π/6 a valid solution in the modified domain?

It is less than π

It is greater than 2π

It fits within the modified domain limits

It is a common angle for sine

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the period of the sine function?

π

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

How do you find additional solutions for sine(θ) = 1/2?

By adding π to the initial solution

By subtracting 2π from the initial solution

By adding 2π to the initial solution

By subtracting π from the initial solution

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