# Need help with sec, sin, etc.

1. Feb 14, 2008

### Nexus555

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

4sec180° - 2sin2270°

3. The attempt at a solution

sec(180 is -1.

So we have 4(-1) which is -4

sin2(270 is -1
So we have 2(-1)

This now reads -4 - -2

Is this correct or do I suck at this? =(

2. Feb 14, 2008

### arildno

What sign must the square of a real number always have?

3. Feb 14, 2008

### Nexus555

Positive?

4. Feb 14, 2008

### HallsofIvy

Staff Emeritus
Yes. sin(270)= -1 so sin2(270)= (-1)2= (-1)(-1)= 1.

5. Feb 14, 2008

### Nexus555

Ok let me try this again:

The attempt at a solution

sec(180 is -1.

So we have 4(-1) which is -4

sin2(270 is (-1)2 which is 1

So we have 2(1)

This now reads -4 - 2

Is this right?

Last edited: Feb 14, 2008
6. Feb 14, 2008

### Nexus555

Ok here is another question, a bit off topic. I know that sec(180 = -1. But this is the question I have.

-When looking at 180° in standard position, we realize that y is 0 and x is any given number along the x axis.
-sec is understood to be r/x.

- So let's say I choose (5,0)
-I need to find r, so we do square root of (5)2 + (0)2
-Which leads to the square root of 25
-Squared is 5

Therefore 5/5 = 1 and sec=1

My question is why in the calculator it says sec(180 is -1? Is it because 180° lies in both the negative x quadrants (II and III) and the postive x quadrants (I and IV?) Since it lies in both a negative and positive x axis, it must be -1 instead of 1? This confuses me a little bit, I know I probably sound like a moron for even asking such a question.

Last edited: Feb 14, 2008
7. Feb 15, 2008

### Gib Z

Remember $$\sec x = \frac{1}{\cos x}$$ by definition. Do you remember the definitions of the trig functions on the unit circle? If you do, Cos 180° should be quite easy seen to be 1 =]

8. Feb 15, 2008

### TVP45

Plot your (5,0) point on a graph. Draw an arrow from the origin to the point. Which direction (degrees) does the arrow point?