Question

cosA +cosB - cosC = -1 + 4cosA/2 cosB/2 sinC/2

Answer 1

Step-by-step explanation:

We have to prove that :

cosa+cosb−cosc=4cos(a/2)cos(b/2)sin(c/2)−1

Here, we make an assumption that a+b+c=π

i.e, the sum of the angles a, b and c is 180°

L.H.S=(cosa+cosb)−cosc

=2cosa+b2cosa−b2−cosc

=2cos(π−c2)cos(a−b2)−cosc

=2cos(π2−c2)cos(a−b2)−cosc

=2sinc2cos(a−b2)−(1−2sin2c2)

=2sinc2(cosa−b2+sinc2)−1

=2sinc2(cosa−b2+sinπ−(a+b)2)−1

=2sinc2(cosa−b2+sin[π2−(a+b)2])−1

=2sinc2(cosa−b2+cos[(a+b)2])−1

=2sinc2(cos2a4+cos[(−2b)4])−1

=2sinc2(cosa2cos[−b2])−1

=4cos(a2)(cosb2)sin(c2)−1

=R.H.S

Happy math!