Want to see correct answers?
Login or join for free!
Question Group Info

This question group is public and is used in 2 tests.

Author: nsharp1
No. Questions: 4
Created: Aug 20, 2020
Last Modified: 4 years ago

Periodicity of Trig Functions

View group questions.

To print this group, add it to a test.

The following questions will investigate the period of the sine, cosine, and tangent functions. The definition of a periodic function is that, for some constant T0, f(x+T)=f(x) for all x in the domain of f.

For the following questions, let θ be an angle in standard position. Let the intersection of the terminal arm of θ and the unit circle centered at the origin be (a,b).
Grade 11 Trigonometry CCSS: HSF-TF.A.4
C.
The period of a periodic function is the smallest value of T such that f(x+T)=f(x) for all values of x in the domain of f. Using the information in the previous questions, which of the following gives the best reasoning as to why the period of the sine function is 2π?
  1. Since values of T less than 2π weren't valid, 2π must work.
  2. Using T=2π and trying a few values of θ, such as 3π7 and 8π11, the equation sin(θ+2π)=sin(θ) is valid. Therefore, it is true for all values of θ.
  3. For T=2π, θ+T and θ are coterminal angles for any value of θ. Therefore, referring to the unit circle, sin(θ)=b and sin(θ+2π)=b, and therefore sin(θ+2π)=sin(θ).
  4. There are no other values of T such that sin(x+T)=sin(x).
Grade 11 Trigonometry CCSS: HSF-TF.A.4
D.
The period of the cosine function is also 2π, and showing this is true is similar to the process used for the sine function. However, the period of the tangent function is π. Which of the following equations correctly shows why the period of tangent is π?
  1. tan(θ+π)=sin(θ+π)cos(θ+π)=-sin(θ)-cos(θ)=sin(θ)cos(θ)=tan(θ)
  2. tan(θ+π)=sin(θ+π)cos(θ+π)=sin(θ)+πcos(θ)+π=sin(θ)cos(θ)=tan(θ)
  3. tan(θ+π)=sin(θ+π)cos(θ+π)=sin(θ)+sin(π)cos(θ)+cos(π)=sin(θ)+0cos(θ)+0=tan(θ)
  4. tan(θ+π)=tan(θ)+tan(π)=tan(θ)