Question

if a+b+c = π then prove that









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Answer 1

Answer:

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Answer 2

Step-by-step explanation:

A+BC=π (given)

⇒,C=π−(A+B)−(i)

Now,

sin

2

A+sin

2

B+sin

2

C=

2

1−cos2A

+

2

1−cos2B

+

2

1−cos2C

[As cos2x=1−2sin

2

x ]

=

2

3

−

2

1

[cos2A+cos2B+cos2C]

=[Using cosC+cos0=2cos

2

C+D

cos

2

C−D

]

=

2

3

−

2

1

[2cos(A+B)cos(A−B)+cos2C]

=

2

3

−

2

1

[2cos(π−C)cos(A−B)+cos2C] (From eq (i))

=

2

3

−

2

1

[−2cosCcos(A−B)+2cos

2

C−1] (cos(π−x)=−cosx)

=

2

3

+cosCcos(A−B)−cos

2

C+

2

1

[multiplying

2

1

inside bracket]

=2+cosC[cos(A−B)−cosC]

=2+cosC[cos(A−B)−cos(π−(A+B))] [From eq (i)]

=2+cosC[cos(A−B)+cos(A+B)]

=2+cosC×[2cosAcosB]

=2+2cosAcosBcosC