Chapter 12.1, Problem 19WE

To determine

To sketch given angle in standard positionwhen $n=0,n=1,n=2,n=-1$

Answer to Problem 19WE

The angles are ${45}^{\circ },{405}^{\circ },{765}^{\circ },-{315}^{\circ }$

Explanation of Solution

Given information :

  ${45}^{\circ }+n\cdot {360}^{\circ }$

According to the given conditions the position of the angle shall remain same when $n$ is a positive integer on the graph.

Therefore the standard position of angle remain same when $n=0,n=1,n=2$

However the angle shall become,

Graph for ${45}^{\circ }+0={45}^{\circ }$ is,

  

graph of ${45}^{\circ }+{360}^{\circ }={405}^{\circ }$ is,

  

Graph for ${45}^{\circ }+2\left(360\right)={765}^{\circ }$ is,

  

Graph for ${45}^{\circ }+\left(-{360}^{\circ }\right)=-{315}^{\circ }$ is,

  

Hence the angles are ${45}^{\circ },{405}^{\circ },{765}^{\circ },-{315}^{\circ }$