Question

if sin^-1x+sin^-1y+sin^-1z=π

Answer 1

sin⁻¹x + sin⁻¹y + sin⁻¹z = π 

　　　　　sin⁻¹x = A......x = sinA 
　　　　　sin⁻¹y = B......y = sinB 
　　　　　sin⁻¹z = C......z = sinC 


then A + B + C = π 

　　　　　x[√(1 - x²)]+ y[√(1 - y²)]+z[√(1 - z²)] = 2xyz 
　　　　　　　sinAcosA + sinBcosB +sinCcosC = 2sinAsinBsinC 
　　sin2A + sin2B + sin2C = 4sinAsinBsinC.....（１） 


so prove （１） 

　sin2A + sin2B + sin2C 

= 2sin(A + B)cos(A - B) + sin[2π- 2A - 2B] 

= 2sin(A + B)cos(A - B) - sin(2A + 2B) 

= 2sin(A + B)cos(A - B) - 2sin(A + B)cos(A + B) 

= 2sin(A + B)[cos(A - B) - cos(A + B)] 

= 2sin(A + B)・2[(-1)(-1) sinAsinB] 

= 4sin(π - C)・sinAsinB 

= 4sinAsinBsinC