Question

IF A, B and C are the interior angles of a triangle ABC, show that -
cos((a+b)/2)= sin (c/2)​

Answer 1

Answer:

IN ΔABC

∠A+∠B+∠C=180

∠A+∠B=180-∠C                (1)

NOW, cos(a+b/2)

          sin(90- (a+b)/2)                   (cosθ=sin(90-θ))

          sin(180-(180-c)/2)                (using eq 1)

           sin(c/2)=RHS

Step-by-step explanation:

Answer 2

Answer:

proved

Step-by-step explanation:

show that -

cos((a+b)/2)= sin (c/2)​

a + b + c = 180  

a + b = 180 - c

(a + b )/2 = (180 - c)/2 = (90 - c/2)

cos((a+b)/2) = cos (90 - c/2)

cos((a+b)/2) = sin ( c/2 )