Question

If A, B and C are interior angles of a triangle ABC, then show that

    sin (B+C/2) = cos A/2

​

Answer 1

Step-by-step explanation:

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

Answer 2

Proof:

We know that, for a given triangle, sum of all the interior angles of a triangle is equal to 180°

⇒A + B + C = 180° ….(1)

To find the value of (B+ C)/2, simplify the equation (1)

⇒ B + C = 180° – A

⇒ (B+C)/2 = (180°-A)/2

⇒ (B+C)/2 = (90°-A/2)

Now, multiply both sides by sin functions, we get

⇒ sin (B+C)/2 = sin (90°-A/2)

Since sin (90°-A/2) = = cos A/2, the above equation is equal to

⇒sin (B+C)/2 = cos A/2

Hence proved!!