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Question Number 81884 by rajesh4661kumar@gmail.com last updated on 16/Feb/20

Commented by mind is power last updated on 16/Feb/20

applie  we can find triangle   withe   sin^−  x,sin^− y,sin^− z angle  withe oriantation  a+b+c=π  ⇒sin(a)cos(a)+sin(b)cos(b)+sin(c)cos(c)=2sin(a)sin(b)sin(c)..1?  ⇒sin(2a)+sin(2b)+sin(2c)=4sin(a)sin(b)sin(c)..2  sin(2a)+sin(2b)=2cos(a−b)sin(a+b)  sin(2c)=sin(2π−2a−2b)=−sin(2a+2b)=−2sin(a+b)cos(a+b)  sin(2a)+sin(2b)+sin(2c)  =2cos(a−b)sin(a+b)−2sin(a+b)cos(a+b)  =2sin(a+b)(cos(a−b)−cos(a+b))  =4sin(a)sin(b)sin(π−c)=4sin(a)sin(b)sin(c)  2..proved  ⇔1 True  for a=sin^− (x),b=sin^− (y),sin^− (z)=c  sin(a)=x,cos(a)=(√(1−x^2 ))......⇒  1⇔x(√(1−x^2 ))+y(√(1−y^2 ))+z(√(1−z^2 ))=2xyz

appliewecanfindtrianglewithesinx,siny,sinzanglewitheoriantationa+b+c=πsin(a)cos(a)+sin(b)cos(b)+sin(c)cos(c)=2sin(a)sin(b)sin(c)..1?sin(2a)+sin(2b)+sin(2c)=4sin(a)sin(b)sin(c)..2sin(2a)+sin(2b)=2cos(ab)sin(a+b)sin(2c)=sin(2π2a2b)=sin(2a+2b)=2sin(a+b)cos(a+b)sin(2a)+sin(2b)+sin(2c)=2cos(ab)sin(a+b)2sin(a+b)cos(a+b)=2sin(a+b)(cos(ab)cos(a+b))=4sin(a)sin(b)sin(πc)=4sin(a)sin(b)sin(c)2..proved1Truefora=sin(x),b=sin(y),sin(z)=csin(a)=x,cos(a)=1x2......1x1x2+y1y2+z1z2=2xyz

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