Question

Solve no. 5 fasttttt...........

Answer 1

Ya, Answer Is Here!!!!!!!

◆Question: cosA+cosB+cosC=1+4sin(A/2)sin(B/2)sinc(C/2)

Now,

L.H.S,

cosA+cosB+cosC

=(cosA+cosB)+cosC

=[2×cos{ (A+B)/2 × cos (A-B)/2 }+ cosC

=[2cos { (π/2)-(C/2)] × [cos (A-B)/2] +cosC

=[ 2sin C/2 × cos (A-B)/2 ] + [1-2×sin^2 (C/2)]

=[1+2sin (C/2) × cos (A-B)/2 ] - sin (C/2)

=[1+2sin (C/2) × cos (A-B)/2 ] - cos (A+B)/2]

=[1+2sin (C/2) × 2sin (A/2) × sin (B/2)]

=[1+4 sin (A/2) × Sin(B/2) × sin(C/2)]

=RHS

So, We Can Say cosA+cosB+cosC=1+4sin(A/2)sin(B/2)sinc(C/2).

Hope My Answer Helped☺️