ifcosX+cosY=a and sinX+sinY=b than prove thatsin2X+sin2Y=2ab(1-2/(a^2+b^2)

Answer 1

ab = sinx cosx + sinx cosy + cosx siny + siny cosy
= sin(x+y) + sinx cosx + siny cosy
using your sum/difference formulas,
= sin(x+y) + 1/2 sin2x + 1/2 sin2y
= sin(x+y) + sin(x+y)cos(x-y)
= sin(x+y)(1+cos(x-y))

a^2+b^2 = cos^2x + 2cosx cosy + cos^2y + sin^2x + 2sinx siny + sin^2y
= 2+2cos(x-y)
2/(a^2+b^2) = 1/(1+cos(x-y))
1-2/(a^2+b^2) = cos(x-y)/(1+cos(x-y))

so, using your sum/difference formulas as above,

2ab(1-2/(a^2+b^2)) = 2sin(x+y)cos(x-y)
= sin2x + sin2y

*whew*

Answer 2

not well

Answer 3

not well this problem

Answer 4

cosx+cosy=a sinx+siny=b find tha value of sin2x+sin2y=?

Answer 5

2+3