Question

If A, B and C are interior angles of a triangle ABC, then show that sin [(B + C)/2] = cos A/2.​

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Answer 1

Answer:

Step-by-step explanation:

For this you need to remember the trigonometric identity (complementary):-

So now, you have to find the relation in sin and cos.

$\rm{sin \theta = 90 - cos \theta }$

Assuming that you are introduced to this formula, I will solve the question with the interior angle theorem as well.

Now, since, A, B and C are interior angles,

$\rm{A + B + C = 180}$

$\rm{B + C = 180 - A}$ Why we did this is because we want the value of sin(B+C/2)

Replace the value of B+C now,

$\rm{sin\bigg{(\dfrac{B+C}{2}\bigg) \implies sin\bigg(\dfrac{180-A}{2}\bigg)}$

It can be written as :-

$\rm{ sin\bigg(\dfrac{180}{2} - \dfrac{A}{2}\bigg)}$

$\rm{sin \theta = 90 - cos \theta }$ $\implies$ See the formula. We already go the solution now.

$\rm{ sin\bigg(90 - \dfrac{A}{2}\bigg)}$ $\implies$ $\sf{cos \bigg(\dfrac{A}{2}\bigg)$

Hence, Proved.

Answer 2

Given :-

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

To Prove :-

sin [(B + C)/2] = cos A/2.​

Solution :-

According to the angle sum property

A + B + C = 180

B + C = 180 - A

Now Multiplying both side 1/2

B + C × 1/2 = 180 - A× 1/2

B + C/2 = 180 - A/2

B + C/2 = 180/2 - A/2

B + C/2 = 90 - A/2

sin(B + C/2) = sin(90 - A/2)

As sin 90 - Ф = cos Ф

sin(B + C/2) = cos A/2