Question

28. If A+B+C =180°, prove that : Sin² A + Sin²B - Sin²C = 2 Sin A Sin B CosC​

Answer 1

Step-by-step explanation:

sin^2A+sin^2B+sin^2C = 2+2cosA.cosB.cosC

I can write,

sin^2A as,

sin^2A = (1-cos(2A))/2

Therefore,

LHS =

= (1-cos(2A))/2+(1-cos(2B))/2+(1-cos(2C))/2

=3/2 - (cos(2A)+cos(2B)+cos(2C))

= 1/2(3 - (2cos(A+B)cos(A-B)+cos(2C)))

C = 180 - (A+B)

cos(C) = cos(180 - (A+B))

cos(C) = -cos(A+B)

Therefore,

LHS

=3/2 - (-2cos(C)cos(A-B)+cos(2C))

cos(2C) = 2cos^2(C) - 1

LHS

=1/2(3 - (-2cos(C)cos(A-B)+2cos^2(C)-1))

= 1/2(4 - (2cos(C)(cos(C)-cos(A-B)))

= 1/2(4 - 2cos(C)(-cos(A+B)-cos(A-B)))

= 1/2(4+2cos(C)(cos(A+B)cos(A-B)))

= 1/2(4+2cos(C)xx2cos(A)cos(B))

= 2 + 2cos(A)cos(B)cos(C)

Therefore, if A+B+C = 180,

Answer 2

Answer:

this is the answer

hope it will help u