# Evaluation trig functions

1. Mar 16, 2009

### Nyasha

1. The problem statement, all variables and given/known data
$$cos(ArcSec(-\sqrt2+\frac{\pi}{4})$$

3. The attempt at a solution

$$cos(ArcSec(-\sqrt2+\frac{\sqrt2}{2})$$

In order to solve this problem how do l deal with the $$\frac{\pi}{4}$$. Is it correct to substitute $$\frac{\pi}{4}$$ with $$\frac{\sqrt2}{2}$$

2. Mar 16, 2009

### yyat

Hi Nyasha!

Why do you think that $$\frac{\pi}{4}\approx .785$$ is the same as $$\frac{\sqrt{2}}{2}\approx .707$$?

You can however simplify cos(arcsec(...)).

3. Mar 16, 2009

### Nyasha

So is this correct:

$$cos(ArcSec(-\sqrt2+(0.785))$$

4. Mar 16, 2009

### yyat

Well, you don't need to use decimal notation (and probably should not in this case), I was just trying to demonstrate that the two numbers are not the same.

What I was hinting at before: sec(x)=1/cos(x), so arcsec(x)=arccos(1/x) and also cos(arccos(x))=x by definition. Use this to simplify your expression.

5. Mar 16, 2009

### Nyasha

I can evaluate this thing without the $$\frac{\pi}{4}$$ by the assistance of of the principal range of arcsec and a diagram. I just am just getting confused with the $$\frac{\pi}{4}$$.

6. Mar 16, 2009

### Nyasha

Thanks, l have solved the question.

$\text{Using pricinpal values: }\:\text{arcsec}\left(\text{-}\sqrt{2}\right) \:=\:\frac{3\pi}{4}$

And then add the $\frac{3\pi}{4}$ to the$\frac{\pi}{4}$

7. Mar 17, 2009

### yyat

So it really was $$\text{cos}(\text{arcsec}(-\sqrt{2})+\frac{\pi}{4})$$? Otherwise, it is not correct.

8. Mar 17, 2009

### Nyasha

Yes it was :

$$\{cos}(\text{arcsec}(-\sqrt{2})+\frac{\pi}{4})$$

It seemed as if l had copied the thing wrongly.

However,it seems as if l have solved it.