Question

in triangle A + B + C = π show that sin2A+sin2B+sin2C=2sinA sinB cosC​

Answer 1

Proof :

L.H.S. = sin2A + sin2B + sin2C

= 2 sin{(2A + 2B)/2} cos{(2A - 2B)/2} + sin2C

= 2 sin(A + B) cos(A - B) + sin2C

= 2 sin(π - C) cos(A - B) + sin2C,

since A + B + C = π

= 2 sinC cos(A - B) + sin2C

= 2 sinC cos(A - B) + 2 sinC cosC,

since sin2A = 2 sinA cosA

= 2 sinC {cos(A - B) + cosC}

= 2 sinC [2 cos{(A - B + C)/2} cos{(A - B - C)/2}]

= 4 sinC cos{(π - 2B)/2} cos{(π - 2A)/2},

since A + B + C = π & cos(- A) = cosA

= 4 sinC cos(π/2 - B) cos(π/2 - A)

= 4 sinC sinB sinA

= 4 sinA sinB sinC

= R.H.S.

Hence, proved.