# Simple trig functions

1. Jan 6, 2010

### tomwilliam

1. The problem statement, all variables and given/known data
(In radians I assume) Using the exact values of sin, cos, tan of 1/3 pi and 1/6 pi, and the symmetry of the graphs of sin, cos and tan, find the exact values of sin(-1/6pi), cos (5/3pi) and tan(4/3pi).

2. Relevant equations
cos(x)=sin(x+pi/2)
sin(x) = sin(x+2pi)
sin^2(x)+cos^2(x)=1

3. The attempt at a solution

I have a little difficulty understanding what is required here. I have a list of trig functions but these don't seem to be here. I imagine the answer is some square root with a coefficient, but I'd like to understand it better. Can someone point me in the right direction without just giving me the answer? Thanks

2. Jan 6, 2010

### rock.freak667

Try the numbers as combination of π, for example to find tan(5π/4)=tan(π+π/4). Like that.

3. Jan 6, 2010

4. Jan 6, 2010

### vela

Staff Emeritus
The idea is to look at the graphs of functions to figure out properties that would be useful in solving the problems. For example, if you graph $$\sin x$$, you'll see that it's an odd function; in other words, it has the property $$\sin(x) = -\sin(-x)$$. If $$x=-\pi/6$$, you can say $$\sin(-\pi/6) = -\sin[-(-\pi/6)] = -\sin(\pi/6)$$.

5. Jan 6, 2010

### tomwilliam

Thanks guys. I guess I'm being slow, but I've had a few tries now and can't equate, e.g. sin(-pi/6) with either sin, cos, or tan of 1/3 or 1/6 pi.
For example:

sin(-pi/6) = -sin(pi/6)
fine.
cos(pi/6 + pi/2) = -sin(pi/6)
so
cos(4/6 pi) = sin(-pi/6)

but what I really want to equate the LHS to is
-cos(1/3 pi).
I know I'm being really slow, but I can't seem to get there.

6. Jan 6, 2010

### tomwilliam

The graph does help a lot.
I can see that sin(-1/6 pi) is equal to cos exactly half pi earlier in the graph. So is there an identity that states sin(x - pi/2)=cos(x)?
If so, I've cracked it.

7. Jan 6, 2010

### l'Hôpital

Sorta but not quite.

Notice if you put x = 0, you'd get
sin (-pi/2) = cos 0 = 1 which is wrong. The right identity.

It'd be
$$\sin (\frac{\pi}{2} - x ) = \cos x$$

8. Jan 6, 2010

### rock.freak667

should be sin(x-π/2)= -cos(x)

9. Jan 6, 2010

### vela

Staff Emeritus
Actually, I think Tom discovered the identity

$$sin(x+\frac{\pi}{2}) = cos x$$

which just says that the graph of cosine is the sine shifted by $$\pi/2$$. It's easy to screw up the sign of the phase shift because, to me, it seems like it should be the opposite sign intuitively. I always have to double-check to make sure I get it right.

10. Jan 6, 2010

### vela

Staff Emeritus
You probably can't because that's actually kind of hard to do. :)

The problem wants you to figure out what $$\cos(5\pi/3)$$ equals in terms of $$\cos(\pi/6)$$ or $$\cos(\pi/3)$$. Same with $$\tan(4\pi/3)$$ and $$\tan(\pi/6)$$ or $$\tan(\pi/3)$$.

11. Jan 6, 2010

### tomwilliam

Thanks all.
Well I think I'm getting there.
I think for the first one I can just say

sin(pi/6)=0.5
sin (-x) = -sin(x), therefore
sin(-pi/6)=-sin(pi/6)=-0.5

Simple.

For the second one, I'm guessing I have to use:
cos(pi/3)=0.5
and somehow show that it is equal to:
cos(5pi/3).

I can see the graph very clearly, and understand why this is the case. But I'm not sure how I can show it. I know that as the angle goes from 0 to pi, the cosine of the angle goes from 1 to -1, and then as the angle goes from pi to 2pi, the cosine goes back round from -1 to 1. So of course, the cosine at 1/3 of the way round to pi will be the same as it is at 2pi - 1/3pi. But how to show that mathematically?

Thanks again, and sorry for being slow-witted.

12. Jan 6, 2010

### rock.freak667

5π/3 = 2π- π/3

Use this.

13. Jan 6, 2010

### vela

Staff Emeritus
So essentially what you've figured out is that $$\cos x = \cos(2\pi - x)$$. This another one of the many trig identities. You can just say

$$\cos 5\pi/3 = \cos(2\pi-\pi/3) = \cos \pi/3 = ...$$