If $\sin q\ne \cos q$ and $x,y,z$ satisfy the equations $x\cos p-y \sin p+z=\cos q+1$, $x\sin p+y\cos p+z=1-\sin q$, $x\cos(p+q)-y\sin(p+q)+z=2$

Question

If $\sin q\ne \cos q$ and $x,y,z$ satisfy the equations $$ x\cos p-y \sin p+z=\cos q+1\\ x\sin p+y\cos p+z=1-\sin q\\ x\cos(p+q)-y\sin(p+q)+z=2 $$ then find the value of $x^2+y^2+z^2$

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I multiplied the first equation by $\cos p$ and the second equation by $\sin p$ and added to get $x+z(\cos p+\sin p)=\cos p\cos q+\cos p+\sin p-\sin p\sin q$

and then I multiplied the first equation by $\sin p$ and the second equation by $\cos p$ and subtracted to get $y+z(\cos p-\sin p)=\cos p-\cos p\sin q-\sin p\cos q-\sin p$
Now i am stuck.I dont know how to solve further.

Answer 1

Rather, multiply the first equation by $\cos q$ and the second by $-\sin q$ and add these scaled equations to obtain $$x\cos(p+q)-y\sin(p+q)+z(\cos q -\sin q)=\cos q -\sin q+1.$$ Since from the last of the original triple we know that $x\cos(p+q)-y\sin(p+q)=2-z,$ we can substitute in the above sum and letting $K=\cos q -\sin q,$ we obtain $$2-z+Kz=K+1,$$ which you may now substitute in any pair of the original system to eliminate $z,$ so that you now have a pair in $x$ and $y$ alone. I believe you can find the way from there.

Answer 2

Your equations are linear in all variables, solve the first equation for $z$ and plug this in equation two and three, solve one of this equations for $y$ and you will get an equation only in $x$. By this method i got

$$x={\frac { \left( -\cos \left( q \right) -\sin \left( q \right) \right) \sin \left( p+q \right) + \left( -\cos \left( q \right) +1 \right) \cos \left( p \right) +\sin \left( p \right) \left( \sin \left( q \right) +1 \right) }{ \left( \sin \left( p \right) -\cos \left( p \right) \right) \sin \left( p+q \right) -1+ \left( \cos \left( p \right) +\sin \left( p \right) \right) \cos \left( p+q \right) }} $$

Answer 3

From the first two equations, put z on the right-hand side.
1. square and sum to get an expression for $x^2+y^2$ in terms of $q$ and $z$.
2. Combine the two to find another expression for $x\cos(p+q)-y\sin(p+q)$