431(1,3) большое спасибо!!!!!!!!!

Решите задачу:

$1)\; \; \left \{ {{x=5cost} \atop {y=5sint}} \right. \; \left \{ {{cost=\frac{x}{5}} \atop {y=5sint}} \right. \; \left \{ {{t=arccos\frac{x}{5}} \atop {y=5sin (arccos\frac{x}{5})}} \right. \\\\sin(arccos\frac{x}{5})=\sqrt{1-cos^2(arccos\frac{x}{5})}=\sqrt{1-(\frac{x}{5})^2}=\frac{\sqrt{25-x^2}}{5}\; \; \Rightarrow \\\\ y=\sqrt{25-x^2}$

$3)\; \; \left \{ {{x=sint+cost} \atop {y=sint\cdot cost}} \right. \; \left \{ {{x=sint+cost} \atop {y=\frac{1}{2}sin2t}} \right. \; \left \{ {{x=sint+cost} \atop {sin2t=2y}} \right. \; \left \{ {{x=sint+cost} \atop {2t=arcsin(2y)}} \right. \\\\ \left \{ {{x=sint+cost} \atop {t=\frac{1}{2}arcsin(2y)}} \right. \\\\x=sint+cost=sin(\frac{arcsin2y}{2})+cos(\frac{arcsin2y}{2})$

$\star \; \; sin\frac{ \alpha }{2}=\pm \sqrt{\frac{1-cos\alpha }{2}}\; ,\; \; cos\frac{ \alpha }{2}=\pm \sqrt{\frac{1+cos\alpha }{2}$

$x=sin(\frac{arcsin2y}{2})+cos(\frac{arcsin2y}{2})=\\\\x=\sqrt{\frac{1-cos(arcsin2y)}{2}}+\sqrt{\frac{1+cos(arcsin2y)}{2}}\\\\cos(arcsin2y)=\sqrt{1-sin^2(arcsin2y)}=\sqrt{1-(2y)^2}=\sqrt{1-4y^2}\\\\x=\sqrt{\frac{1-\sqrt{1-4y^2}}{2}}+\sqrt{\frac{1+\sqrt{1-4y^2}}{2}}$