# Some help needed

1. Sep 10, 2006

### mubashirmansoor

(z-x)/g = tan(0.5(arctan((z-y)/g)+arctan((y-x)/g)))

I'll be thankfull...

2. Sep 10, 2006

### quasar987

Sometimes it's not possible to make 'z' the subject of a formula.

3. Sep 11, 2006

### Robokapp

ok let's begin by writing that more mathematically.

$$\frac{z-x} {g} = tan{\frac{tan^{-1}(\frac{z-y} {g})+tan^{-1}(\frac{y-x} {g})} {2}$$

IS this what you mean? 6am...I might have done it wrong.

tip: maybe trying to get both sides to look messy might help...i have a way to solve this in mind and if it works that way, this is a beautiful problem. if it doesn't... i dont like this problem.

Last edited: Sep 11, 2006
4. Sep 11, 2006

### Robokapp

Oh I hate you now, it awakened my curiousity and...50 minutes later I think I have it. One Z on the process decided to turn squared on me and instead of a nice common factor i had to use quadratic formula. indication that i did something awfully wrong but...i checked and it looks ok.

EDIT: Ok I re-read the stickies, this isnt the homework section so i can post the work I did. Basically i got rid of that nasty 1/2 by...well this way: (sorry for the notation, i wrote this is a .txt file.

Last edited: Sep 11, 2006
5. Sep 11, 2006

### mubashirmansoor

thanks robokapp, I'll be thankfull if you do the whole thing (applying the formulas) what is z equal to????

I don't really know which formulas you are talking about...

thanks once again

6. Sep 11, 2006

### Robokapp

Okay...sure. After all I do am curious if I did any horrible mistakes (i might have)

I disliked the 1/2 part. I knew there are formulas for different half-angle, duble angle, angle plus angle etc so i went to wikipedia and typed in 'trigonometry.' the formulas are on bottom.

I knew the "tan" and "arctan" will cancel eachother out so I picked formulas so that I only work in terms of tan. This is my full work:

(z-x)/g = tan(0.5(arctan((z-y)/g)+arctan((y-x)/g)))

2arctan[(z-x)/g] = arctan((z-y)/g)+arctan((y-x)/g)

tan{2arctan[(z-x)/g]} = tan{arctan((z-y)/g)+arctan((y-x)/g)}

Using the double angle formula in first part and angle plus angle formula in second part...

2tan{arctan[(z-x)/g]}
----------------------------------------------- =
1 - tan{arctan[(z-x)/g]} * tan{arctan[(z-x)/g]}

tan{arctan((z-y)/g)}+tan{arctan((y-x)/g)}
= -------------------------------------------- ====>
1+tan{arctan((z-y)/g)}*tan{arctan((y-x)/g)}

2(z-x)/g.............(z-y)/g+(y-x)/g
-------------- = ------------------- ====>
1-(z-x)^2/g^2.....1+(z-y)/g * (y-x)/g

2(z-x).................(z-y)+(y-x)
-------------- = ---------------- =====>
1-(z-x)^2/g.......1+(z-y) * (y-x)

2gz-2gx ...............z-x
------------- = ---------------- ======> flip it.
g-z^2-2zx+x^2......1+zy-zx-y^2+xy

g-z^2-2zx+x^2....1+zy-zx-y^2+xy
------------- = -------------- =======>
2gz-2gx...............z-x

g-z^2-2zx+x^2...... 1+zy-zx-y^2+xy
------------- = -------------- =======> multiply second by 2g/2g
2g(z-x).................. z-x

g-z^2-2zx+x^2....2g(1+zy-zx-y^2+xy)
------------- = ------------------- =======>
2g(z-x)..................2g(z-x)

g-z^2-2zx+x^2 = 2g+2zgy-2zgx-2gy^2+2gxy

-z^2-2zx-2zgy-2zgx = 2g+2gxy-2gy^2-g-x^2

-z^2-2zx-2zgy-2zgx = g+2gxy-2gy^2-x^2

-z^2+z(2x-2gy-2gx) - g-2gxy+2gy^2+x^2 = 0

Calling

a= -1
b= 2x-2gy-2gx
c= -g-2gxy+2gy^2+x^2

z=0.5{-(2x-2gy-2gx) +/- sqrt[(2x-2gy-2gx)^2+4(-g-2gxy+2gy^2+x^2)]}

-------------------------------------------

Now...I dont think it should have 2 answers, I dont seem to have any way around that z^2 however.

The ......... takes place of..balnk so numbers dont run into eachother.

plz someone check my work...

Last edited: Sep 11, 2006
7. Sep 12, 2006

### mubashirmansoor

Thanks Robokapp, I had actually done a sily mistake in writing the question, It should had been:

(z-x)/2g = tan(0.5(arctan((z-y)/g)+arctan((y-x)/g)))

But I used the formulas you provided and got the following results:

z = 2x - y +/- sqrt(3y^2-6yx+3x^2-4g^2)

Thankyou you were a great help

8. Sep 12, 2006

### mubashirmansoor

but chk it out, Did I do it correctly? becase I'm little puzzeled It's not working as I wanted...