Homework Help: 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

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