# Trig problem

1. May 11, 2006

### jacy

I have to calculate this without using the calculator.
cos (arctan 5/12)

So far i draw a triangle and i have the opposite side to be 5, adjacent to be 12, and hypotenuse to be 13. Please suggest me some hint, thanks.

Last edited: May 11, 2006
2. May 11, 2006

### Integral

Staff Emeritus
Once you have drawn the triangle it is easy. Mark the angle represented by ArcTan $\frac 5 {12}$

then use the definition of the cos of that angle to get your answer.

3. May 11, 2006

### jacy

am getting 12/13, am i correct

4. May 11, 2006

### Curious3141

Correct. BTW, if you're allowed access to a calculator, you can use that to verify your answer (even if you're not allowed to use the calc to derive the answer).

5. May 12, 2006

### jacy

Thanks, should the unit be the length of the sides, since 12/13 is not an angle.

6. May 12, 2006

### Curious3141

No, trig ratios have no unit. You're dividing a length by a length, so they're dimensionless.

7. May 13, 2006

### arunbg

Just for another take on this problem (I find the construction of a rt triangle a bit cumbersome) we could use basic trig. results.

For eg, if you have to do something like cos(arctan(x))
we can proceed by taking arctan(x)=y
so x=tan(y)
$$\frac{1}{\sqrt{1+x^2}} = \frac{1}{\sqrt{1+tan^2y}} = cos(y) or y=arccos(\frac{1}{\sqrt{1+x^2}})$$
So cos(arctan(x)) = cos(y) = $\frac{1}{\sqrt{1+x^2}})$
which gives the answer.This approach works for all such problems.
No messy triangles.

Arun

8. May 13, 2006

### jacy

Thanks everyone for helping me out.