How to solve advanced algebra equation with multiple variables?

ohaiyo88
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Homework Statement


x= cosθ1p3(cosθ23)+cosθ1(cosθ2p2-a)-sinθ1p1
y= sinθ1p3(cosθ23)+sinθ1(cosθ2p2-a)+cosθ1p1
z= sinθ23p3-sinθ2p3

Homework Equations


cosθ23 = cos(θ23)
sinθ23 = sin(θ23)

Finding θ123. Solve in terms of x,y,z,p
Pls help to solve this algebra equation by eliminating the variable. Thanks in advance for the helping..im doing my final year project and came across to this unsolvable problem..
 
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What is it that you want to do? Do you want to express l_1,l_2,l_3 in terms of x,y,z and \theta_1,\theta_2,\theta_3?

Your notation is kind of hard to follow. This post explains how you can make your math look nice.
 
Find θ1,θ2,θ3 in terms of x,y,z and p. Sorry for the inconvenience.
 
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Uuuh, perhaps you should try a numerical solution...

If you don't want that, then you can always try to use

\cos(x)=\frac{e^{ix}+e^{-ix}}{2}~\text{and}~\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}

This turns the geometric problem into an exponential problem which might be easier to solve.

Another thing you should consider are the t-formula's:

\sin(\theta_i)=\frac{2t_i}{1+t_i^2},~\cos(\theta_i)=\frac{1-t_i^2}{1+t_i^2}

with t_i=\tan(\theta_i/2)

This puts the sines and cosines in the same variable. This might be easier to solve. I guarantee nothing however...
 
but i tried both methods, it just make the question more complicated than ever. I used to squaring up x and y, and add both together, I am sucessfully eliminate θ1, but others seems to have trouble.
 
Anyone here can lend a golden hand?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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