# How is my text book getting the components of a vector?

1. Dec 31, 2012

### student34

1. The problem statement, all variables and given/known data
My text book shows a vector B 130° from the positive x-axis. B has a value of 5; it asks for the x and y components for B.

But they show a way other than SOH CAH TOA. They got the x component by multiplying 5(cos130°) and the y component by multiplying 5(sin130°). What is this method?

2. Relevant equations
All I know is SOH CAH TOA, but that obviously will not work in this case.

3. The attempt at a solution
I can easily figure it out by putting B into a right triangle to get the x and y components. But I am interested in learning this technique too which I apparently missed some where along the way.

2. Dec 31, 2012

### lurflurf

That is the usual method. sin and cos are defined for angles greater than 90, they can also be reduced
5(cos130°)=-5(cos50°)
5(sin130°)=5(sin50°)

3. Dec 31, 2012

### BruceW

(@student34)
I'm guessing the way you did it was to consider the triangle between the line and the y axis (or x axis), then relate this to the x and y components with respect to the origin? This way is fine, but yeah, their way is faster.

Do you know how to use double-angle formula? You can use this to show why their method and your method both give the same result. hint: 130=90+40 and use this in the double-angle formula.

Fundamentally, their method is using a rotation matrix to rotate the line counter-clockwise. are you familiar with rotation matrix? You can sort of justify their method by drawing a couple of different positions of the line, and seeing how the angle changes and how that changes the x and y components. (try this out, I think it helps give a feel for what is going on)