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Assmue the $m,n,u,v$ be real numbers,and such $$m^2+n^2=1,u^2+v^2=1,nv>0,m>0,u>0$$ and $$5mu=3(1-nv)$$

show that $$mv-3nu=m-3u$$

Following is My methods:

let $$m=\cos{x},n=\sin{x},u=\cos{y},v=\sin{y}$$ and $$5mu=3(1-nv)\Longrightarrow 5\cos{x}\cos{y}=3(1-\sin{x}\sin{y})$$ $$\Longrightarrow 5\cos{x}\cos{y}+3\sin{x}\sin{y}=3$$ $$4\cos{(x-y)}+\cos{(x+y)}=3\Longrightarrow -8\sin^2{\dfrac{x-y}{2}}+4+2\cos^2{\dfrac{x+y}{2}}-1=3$$ $$\Longrightarrow \cos{\dfrac{x+y}{2}}=2\sin{\dfrac{x-y}{2}}$$ $$\Longrightarrow 1-\tan{\dfrac{x}{2}}\tan{\dfrac{y}{2}}=2\tan{\dfrac{x}{2}}-2\tan{\dfrac{y}{2}}$$ then we must show that $$\Longleftrightarrow \cos{x}\sin{y}-3\sin{x}\cos{y}=\cos{x}-3\cos{y}$$ $$\Longleftrightarrow \cos{x}(1-\sin{y})=3\cos{y}(1-\sin{x})$$ $$\Longleftrightarrow\dfrac{\cos{x}}{1-\sin{x}}=3\dfrac{\cos{y}}{1-\sin{y}}\Longleftrightarrow \dfrac{1+\tan{\frac{x}{2}}}{1-\tan{\frac{x}{2}}}=3\dfrac{1+\tan{\frac{y}{2}}}{1-\tan{\frac{y}{2}}}\tag{2}$$ $$\Longleftrightarrow 1-\tan{\dfrac{x}{2}}\tan{\dfrac{y}{2}}=2\tan{\dfrac{x}{2}}-2\tan{\dfrac{y}{2}}$$ so it is clear

My Question:have without this trigonometric methods?such this famous Lagrange's identity

$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$$ since $m^2+n^2=1,u^2+v^2=1$ $$(5mu+3nv)=3\Longrightarrow (mv+3u)=3nu+m$$

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It follows also directly by substituting the equations. Let $u= - 3(nv - 1)/(5m)$. Then the desired equation is given by $$ 9(nv - 1)(n - 1) + 5(v - 1)m^2=0. $$ Now substitute $m^2:=1-n^2$. Then we need to show $$ (4nv + 5n - 5v - 4)(n-1)=0. $$ Since $n-1\neq 0$ because of $m>0$ we need to show $4nv + 5n - 5v - 4=0$. This follows from $u^2+v^2=1$, which is equivalent to $$ (4nv + 5n - 5v - 4)(4nv-5n+5v-4)=0. $$ Here the second factor cannot be zero. Assume it is zero, i.e., $v=(5n+4)/(4n+5)$. Then $u=-3(n-1)(n+1)/(4n+5)m)<0$, a contradiction. It follows that the first factor is zero, i.e., $$ 4nv + 5n - 5v - 4=0. $$

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  • $\begingroup$ so Now you must show $4nv+5n-5v-4=0$ then you want prove $(4nv+5n-5v-4)(4nv-5n+5v-4)=0$,then you can must show that $$(4nv-4)^2-(5n-5v)^2=0$$.. $\endgroup$ – china math Jan 12 '15 at 9:51
  • $\begingroup$ Hello,It's not me downvote, becasue you idea following must prove $4nv+5n-5v-4=0$ is true $\endgroup$ – china math Jan 12 '15 at 9:59
  • $\begingroup$ Yes, you are right. I have edited the post. $\endgroup$ – Dietrich Burde Jan 12 '15 at 12:03

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