Question 1: Is this mainly saying that when the angle is greater than 90 degrees use the reference angle to evaluate a trig function?
Question 2: Is this then saying that if you choose to evaluate a trig function without using a reference angle, you have to go through a lot of work, such as using these trig identities:
sin(180 + θ) = -sin(θ) or cos(90 + θ)= -sin(θ)
and doing so allows you to see that this following equation is true:
sin(180 + θ) + cos(270 - θ) = 0 ?
If I have the basic idea of what this text is saying in general, then i can move on to say that I have no idea what this text is saying in specifics.
I understand the first part. I use a reference angle by convention because I know of no other way to evaluate a trig function with an angle greater than 90. Reference angles allows me to see the angle as right triangle and thus is possible to evaluate.
However, when the text goes on to say that if you choose not to use the reference angle, you'll have to evaluate using the given trig identities. I have two questions about that:
1. I don't understand how this equation is possible: cos(90 + θ)= -sin(θ)
For example: If I have angle 2∏/3 and I find that sin(2∏/3) = √3/2
and then I put angle 2∏/3 into cos(∏/2+2∏/3) and find that cos(7∏/6) = (-√3/2)
Then if I divide both sides of cos(7∏/6) = (-√3/2) by -1, I get -cos(7∏/6) = √3/2
Then I could say -cos(7∏/6) = sin(2∏/3), then if i divide that by -1, I get
cos(7∏/6) = -sin(2∏/3) or cos(∏/2+2∏/3) = -sin(2∏/3)
So is this logic/evaluating correct? If so, then this is my next question:
2) How does that have anything to do with finding the components of a vector?
The Attempt at a Solution
Please help, I feel lost on this.