Question

In triangle ABC prove that cos^2A/2+cos^2b/2-cos^2c/2=2cosA/2cosB/2sinc/2

Answer 1

We have to prove that cos²A/2 + cos²B/2 - cos²C/2 = 2cosA/2 cosB/2 sinC/2 for a triangle ABC.

proof : LHS = cos²A/2 + cos²B/2 - cos²C/2

we know, cos²θ = (1 + cos2θ)/2

so, cos²A/2 = (1 + cosA)/2

cos²B/2 = (1 + cosB)/2

cos²C/2 = (1 - cosC)/2

= (1 + cosA)/2 + (1 + cosB)/2- (1 + cosC)/2

= 1/2 + (cosA + cosB - cosC)/2

= 1/2 + 1/2 [2cos(A + B)/2 cos(A - B)/2 - cosC]

we know, A + B + C = π

so, A + B = π - C.....(1)

= 1/2 + 1/2 [2 cos(π - C)/2 cos(A - B)/2 - cosC]

= 1/2 + 1/2 [ 2sinC/2 cos(A - B)/2 - 1 + 2sin²C/2 ]

We know, cos2x = 1 - 2sin²x

= sinC/2[cos(A - B)/2 + cosC/2]

= SinC/2 [ 2cos(A - B + C)/2 cos(A - B - C)/2]

= sinC/2 [ 2cos(π - B - B)/2 cos(A - π + A)/2 ]

= sinC/2 [ 2cosB/2 cosA/2 ]

= 2cosA/2 cosB/2 sinC/2 = RHS

Answer 2

Answer:-

We have to prove that cos²A/2 + cos²B/2 - cos²C/2 = 2cosA/2 cosB/2 sinC/2 for a triangle ABC.

proof:

LHS = cos²A/2 + cos²B/2 - cos²C/2

we know, cos²θ = (1 + cos2θ)/2

so,

cos²A/2 = (1 + cosA)/2

cos²B/2 = (1 + cosB)/2

cos²C/2 = (1 - cosC)/2

= (1 + cosA)/2 + (1 + cosB)/2- (1 + cosC)/2

= 1/2 + (cosA + cosB - cosC)/2

= 1/2 + 1/2 [2cos(A + B)/2 cos(A - B)/2 - cosC]

we know,

A + B + C = π

so,

A + B = π - C.....(1)

= 1/2 + 1/2 [2 cos(π - C)/2 cos(A - B)/2 - cosC]

= 1/2 + 1/2 [ 2sinC/2 cos(A - B)/2 - 1 + 2sin²C/2 ]

We know,

cos2x = 1 - 2sin²x

= sinC/2[cos(A - B)/2 + cosC/2]

= SinC/2 [ 2cos(A - B + C)/2 cos(A - B - C)/2]

= sinC/2 [ 2cos(π - B - B)/2 cos(A - π + A)/2 ]

= sinC/2 [ 2cosB/2 cosA/2 ]

= 2cosA/2 cosB/2 sinC/2 = RHS

Hope it's help you❤️