1. Oct 18, 2013

### TheKracken

1. The problem statement, all variables and given/known data

Simplify cot(θ)sin(-θ) so that it is a single trig function of positive θ

2. Relevant equations

3. The attempt at a solution
I changed cot(θ) to
cos(θ)/sin(θ) X Sin(-θ)/1

I stated that sin(-θ) and sin(θ) were the same and then cancled out the sin(θ) and sin (-θ) to get cos(θ)/1 or Cos(θ)

I feel like this is right but I only got half the points I was supposed to get, not quite sure what I did wrong.

2. Oct 18, 2013

### Staff: Mentor

sin(-θ) and sin(θ) are not quite the same. What is different about them?

3. Oct 18, 2013

### TheKracken

The degree angle is negative in one of them, which would give you a value between 0 and negative one? I just cant seem to grasp how to simplify this.

4. Oct 18, 2013

### Staff: Mentor

Is the sin() function even or odd? Look at a plot of sin(x) over -360 degrees to +360 degrees. What does it do around x=0?

5. Oct 18, 2013

### TheKracken

Wait this would then give me
cosθ/sinθ X -sinθ/1
yeah?
Or...maybe I am on the wrong track...

6. Oct 18, 2013

### Staff: Mentor

So do you see why you only got partial points?

7. Oct 18, 2013

### TheKracken

Sin is a odd function, though I do not understand why. Or what that quite means, I just found it in my notes. except maybe it is opposite of it such as it is the same value except with a negative sign on it.

8. Oct 18, 2013

### TheKracken

Ok well if this is right then the sinθ would cancel out and wouldn't the negative still be there? As in it would then simplify to -cosθ which is not what it asked for. It asked that "so that it is a single trig function of positive θ "

9. Oct 18, 2013

### Staff: Mentor

For odd functions, f(-x) = -f(x). For even functions, f(-x) = f(x). Can you see how sin() is an odd function, and cos() is an even function?

The question is asking for the answer to be expressed in terms of positive θ. -cos(θ) is a function of positive θ.

10. Oct 18, 2013

### TheKracken

OH!!!! Thank you so very much, yes that makes lots of sense to me. Alright, thank you.