Question

xsin³p+ ycos³p = sinpcosp, xsinp= ycosp, prove that x²+y²=1

Answer 1

Answer:

${x}^{2} + {y}^{2} = 1$

Step-by-step explanation:

$x { \sin(p) }^{3} + y { \cos(p) }^{3} = \sin(p) \cos(p) \\ \\ x \sin(p) = y \cos(p) \\ \\ from \: these \: equations \: \: \\ \\ y \cos(p) { \sin(p) }^{2} + y { \cos(p) }^{3} = \sin(p) \cos(p) \\ \\ y \cos(p) ( { \sin(p) }^{2} + { \cos(p) }^{2} ) = \sin(p) \cos(p) \\ \\ y \cos(p) = \sin(p) \cos(p) \\ \\ y = \sin(p) \\ \\ therefore \: \: \: \: x = \cos(p) \\ \\ \\ then \: \: {x}^{2} + {y}^{2} = \cos(p) ^2 + { \sin(p) }^{2} \\ \\ = 1 \\ \\ therefore \: \: \: {x}^{2} + {y}^{2} = 1$