# 1 - tan(theta)

1. Jun 5, 2013

This problem is:

$\frac{sin(45)cos(\theta) - cos(45)sin(\theta)}{x} = 1 - tan(\theta)$

I've been trying to find x mostly just by multiplying things by 1-tan(theta), but so far nothing I've tried yields the numerator.

Also, $sin(45)cos(\theta) - cos(45)sin(\theta) = sin(45 - \theta)$, it that helps...

2. Jun 5, 2013

### rock.freak667

If you are to solve for 'x' why can't you just cross-multiply by 'x' and then divide by '1-tanθ '?

Do you need to simplify or something like that?

3. Jun 5, 2013

### Staff: Mentor

Is your goal to solve for x? To reduce confusion, that's why we ask posters to use the template and enter the complete problem description.

If you're trying to solve for x, start by multiplying both sides of the equation by x.

4. Jun 5, 2013

### Curious3141

Note that $\sin(45)\cos(\theta) - \cos(45)\sin(\theta) = \sin(45 - \theta)$ and that $\tan(45 - \theta) = \frac{1 - \tan\theta}{1 + \tan\theta}$.

5. Jun 10, 2013

Wow, that seems really obvious to me now that you brought that up, haha. I'm having a pretty bad time doing the long division though... Check it out ->

$x = \frac{sin(45)cos(\theta) - cos(45)sin(\theta)}{-tan(\theta) + 1}$

I'm doing the leading terms in polynomial long division and I end up with...

$\frac{-sin(45)cos(\theta)}{tan(\theta}$

Ignoring the sin(45) term, I'm getting $\frac{cos(\theta)}{tan(\theta)}$ = $\frac{adjacent}{hypotenuse}$ * $\frac{adjacent}{opposite}$ = $\frac{adjacent^2}{hypotenuse*opposite}$

The $\frac{adjacent^2}{hypotenuse*opposite}$ term means nothing to me and doesn't really help solve. It's pretty difficult to figure this one out.

6. Jun 10, 2013

### Mentallic

Which is exactly why that approach won't help you in this problem.

Take a look at Curious3141's post and see what you can do with that.

7. Jun 11, 2013

Oh I gotcha... because $\frac{sine(x)}{cosine(x)} = tan(x)$... So I can just divide by cos(45 - $\theta$) to end up with tan(45 - $\theta$) which is equal to $\frac{1 - tan(\theta)}{1 + tan(\theta)}$