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Subject: Re: Equations depicting 2 DOF
Date: Mon, 12 Apr 2010 19:16:05 +0000 (UTC)
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"Imran " <imran.shafi@gmail.com> wrote in message <hpvhko$ri4$1@fred.mathworks.com>...
> I am sorry to bother you again but Roger
> but once i tried to solve the two equations mutually i get a little different outcome, please correct me where i am wrong
> L1 Cos (theta1)+ L2 Cos (theta2)=x
> L1 Sin (theta1) + L2 Sin (theta2)=y
> squaring both sides and adding give me
> cos (theta1-theta2)=(L3^2 - L1^2 - L2^2)/(2* L1 *L2), here i am stuck up
> IN FACT how
> theta1 = theta3 +/- acos((L3^2+L1^2-L2^2)/(2*L1*L3)); please if you can explain
> Imran
-------------------
  You almost have it, Imran!  With the steps

 L3 = sqrt(x^2+y^2); 
 theta3 = atan2(y,x);

the equations become

 L1*cos(theta1)+L2*cos(theta2) = L3*cos(theta3)
 L1*sin(theta1)+L2*sin(theta2) = L3*sin(theta3)

as you may have realized.  On the next step instead of squaring these as they stand and adding, try transposing the two terms containing 'L2' to the right sides of the equations and then squaring and adding.  What do you get?  As I say, you're almost there.

Roger Stafford