# Thread: Trigonometric Functions

1. ## Trigonometric Functions

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.

2. Originally Posted by JonathanEyoon
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.
There is value in knowing both methods. And consider that at some point you might not have that unit circle chart handy and might need to do it by hand.

-Dan

3. Will you always have a unit circle to refer to?

4. Originally Posted by topsquark
There is value in knowing both methods. And consider that at some point you might not have that unit circle chart handy and might need to do it by hand.

-Dan

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

5. Originally Posted by JonathanEyoon
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 $0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \text{ and } \frac{\pi}{2}$. 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

6. 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?

7. 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)

8. Originally Posted by JonathanEyoon
sec(11pi / 3)
and csc(11pi / 3)
No, $\frac{{11\pi }}{3} \in IV$ that is it is in the quadrant four.
Go for the nearest even multiple of $2\pi = \frac{{12\pi }}{3}$.
Thus the equivalent angle is $\frac{{ - \pi }}{3}$

The equivalent angle of $\frac{{ 19 \pi }}{4}$ is $\frac{{ 3 \pi }}{4}$, that is closest to $\frac{{16\pi }}{4} \equiv 2\pi$.

9. Originally Posted by Plato
No, $\frac{{11\pi }}{3} \in IV$ that is it is in the quadrant four.
Go for the nearest even multiple of $2\pi = \frac{{12\pi }}{3}$.
Thus the equivalent angle is $\frac{{ - \pi }}{3}$

The equivalent angle of $\frac{{ 19 \pi }}{4}$ is $\frac{{ 3 \pi }}{4}$, that is closest to $\frac{{16\pi }}{4} \equiv 2\pi$.

I forgot that a full circle was 2pi. I was thinking 11pi / 3 as making 3 full revolutions and then 2/3 or another which kept me in the second quadrant. Ok i'm gonna try it out with your advice.