I'm having trouble understanding why I was taught the way I was today when finding an answer to this problem is so much easier. Can someone thoroughly explain to me why?
I have to evaluate
cos(2pi / 3)
Basically what I do is, I look at a unit circle and see what x value is for (2pi/3) and I find (-1/2). Done that is the answer according to my book.
The method my teacher told us we should do seems like its a few extra steps only to arrive at the answer I get just from directly looking at a Unit Circle chart.
He tells us we need to first find the reference number for (2pi / 3) which is (pi / 3). Then since (2pi / 3) is in quadrant II, the cos(2pi / 3) is gonna be negative. From here we use the reference number and the fact it's negative to achieve...
-cos (pi / 3) = (-1/2)
It is the same answer as the method I do it which is simple. I mean I don't mind doing it the way he taught us except i'm tryin got understand what is the significance of doing it his way if i can do it my way and still get the same answer every single time.
Oh yea that's another question i've been meaning to ask you guys. Say I don't have a Unit Circle chart in front of me. How do I go about figuring out the point of say (pi / 3)? So far all he's told us is, refer to the unit circle. I know from looking at the unit circle, the point is (1 / 2 , sqrt(3) / 2). So how do I get to know the point without referring to the chart? Thanks
There are some basic angles you are going to simply need to memorize. These angles are . If you know the sine, cosine, and tangent of these angles you are doing well.
There is no nice formula to figure out trig functions. If you have any other angle besides the above you are going to need a calculator. (Well, for right now anyway. You'll learn double and half angle formulas at some point.)
-Dan
Hi ok one more question pertaining to evaluating. I'm doing the problems according to how the professor taught us as explained above.
Now i'm having difficulty with this problem
sec(11pi / 3)
Now from here i know that the t-bar or reference number is gonna(pi / 3)
Now I know that (11pi / 3) falls in the second quadrant so x will be negative and y will be positive.
I'm asked to find the sec(11pi / 3) which will be
-sec(pi / 3) = 1 / (-1 / 2).
This should equal -2 but the book is telling me the answer is not negative but a positive 2. Where did I go wrong? When finding secant, cosecant and cotangent, does it not matter whether they x and y values are positive and negative? If not why is that?
I just tried to do another problem with csc and it is still coming out to the sign opposite of what it is supposed to be. I'm totally clueless as to what is going on right now. When doing sin, cos, and tangent it works out perfectly fine. MmMm....
sec(11pi / 3)
and
csc(11pi / 3)