If A+B+C=90 degress, prove that
sin2A+sin2B+sin2C = 4cosAcosBcosC

• Since A + B + C = 90, sinA = cos(B+C), sinB = cos(A+C), sinC = cos(A+B)...
  
  Sin2A + sin2B + sin2C
  • 2sin(A+B)cos(A-B)+2sinCcosC
  • 2sin(A+B)cos(A-B)+2cos(A+B)sin(A+B)
  • 2sin(A+B) [cos(A-B) + cos(A+B)]
  • 2sin(A+B) [cosAcosB + sinAsinB + cosAcosB – sinAsinB]
  • 4sin(A+B)cosAcosB
  • 4cosCcosAcoB
  • 4cosAcosBcosC (proven)
  
  Hope that helps XD
  
• 9 years agoHelpfull: Yes(1) No(1)