How Do You Make 'z' the Subject in This Trigonometric Formula?

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The discussion focuses on manipulating a trigonometric formula to isolate 'z'. The original equation involves a tangent and arctangent relationship, which participants attempt to simplify using trigonometric identities. After several transformations, one user arrives at a quadratic equation for 'z', suggesting that there may be two potential solutions. A correction is made to the initial equation, leading to a final expression for 'z' that includes a square root term, but the user expresses confusion about the validity of their result. The conversation highlights the complexities of solving trigonometric equations and the importance of careful manipulation of formulas.
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Can you please help me for making 'z' the subject of formula;

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

I'll be thankfull...
 
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Sometimes it's not possible to make 'z' the subject of a formula.
 
mubashirmansoor said:
Can you please help me for making 'z' the subject of formula;

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

I'll be thankfull...


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 don't like this problem.
 
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Oh I hate you now, it awakened my curiosity 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 isn't 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.

Robokapp's work said:
(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...
 
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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
 
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 each other 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 don't think it should have 2 answers, I don't seem to have any way around that z^2 however.


The ... takes place of..balnk so numbers don't run into each other.

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