Question

prove that cosA-cosB+cosC+1=4cosA/2.sinB/2.cosC/2

Answer 1

Answer:

CosA - CosB + CosC + 1 = 4 Cos(A/2)Sin(B/2)Cos(C/2)

Step-by-step explanation:

cosA-cosB+cosC+1=4cosA/2.sinB/2.cosC/2

Complete Question : A , B & C are angles of a triangles

=> A + B + C = 180°

LHS

= CosA - CosB + CosC + 1

= (CosA + CosC) - CosB + 1

= (2Cos((A+C)/2)Cos((A-C)/2)) - CosB + 1

= 2Cos((π - B)/2)Cos((A-C)/2)) - CosB + 1

= 2Sin(B/2)Cos((A-C)/2)  - (1-2Sin²B/2) + 1

= 2Sin(B/2)Cos((A-C)/2)  - 1 + 2Sin²B/2 + 1

= 2(Sin(B/2)Cos((A-C)/2)  + 2Sin²B/2

= 2(Sin(B/2) (Cos((A-C)/2)  + Sin(B/2))

= 2Sin(B/2)(Cos((A-C)/2) +Sin((π - (A+C)/2))

= 2Sin(B/2)(Cos((A-C)/2) + Cos(A+C)/2))

= 2Sin(B/2)(Cos((A-C)/2) + Cos(A+C)/2))

=  2Sin(B/2) 2Cos(A/2)Cos(-C/2)

= 2Sin(B/2) 2Cos(A/2)Cos(C/2)

= 4 Cos(A/2)Sin(B/2)Cos(C/2)

QED

Proved

CosA - CosB + CosC + 1 = 4 Cos(A/2)Sin(B/2)Cos(C/2)