Question

if A,B and C are interior angles of a triangle ABC,then show that
sin(B+C/2)=cosA/2​

Answer 1

Answer:

Given △ABC

We know that sum of three angles of a triangle is 180

Hence ∠A+∠B+∠C=180

o

or A+B+C=180

o

B+C=180

o

−A

Multiply both sides by

2

1

2

1

(B+C)=

2

1

(180

o

−A)

2

1

(B+C)=90

o

−

2

A

...(1)

Now

2

1

(B+C)

Taking sine of this angle

sin(

2

B+C

)[

2

B+C

=90

o

−

2

A

]

sin(90

o

−

2

A

)

cos

2

A

[sin(90

o

−θ)=cosθ]

Hence sin(

2

B+C

)=cos

2

A

proved

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Given:

A, B and C are interior angles of the triangle ABC.

To prove:

$sin(\frac{B+C}{2})=cos(\frac{A}{2} )$

Solution:

Sum of all angles of the triangle ABC = 180°

A + B + C = 180°

$\frac{A+B+C}{2}=\frac{180}{2}$

$\frac{A+B+C}{2}=90$

$\frac{B+C}{2} =90-\frac{A}{2}$

$sin(\frac{B+C}{2}) =sin(90-\frac{A}{2})$

$sin(\frac{B+C}{2}) =cos(\frac{A}{2})$ $($∵ $cos\theta=sin(90-\theta))$

Hence proved.