Question

in triangle ABC ,prove that cos (B+C/2)=sin A/2 ​

Answer 1

Step-by-step explanation:

In Triangle ABC

A°+B°+C°=180°

A+B+C/2=180/2

A+B/2=90-C/2

MULTIPLY BY Sin Both Side

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

We Know That Sin(90-A)=CosA

So,Sin(A+B/2)=CosC/2

Answer 2

$cos(\dfrac{B+C}{2} )= sin \dfrac{A}{2}$   is the required result proved below

Step-by-step explanation:

To prove: $cos(\dfrac{B+C}{2} )= sin \dfrac{A}{2}$

Taking left hand side

$cos (\dfrac{B+C}{2} )$   -----(i)

In triangle ABC

By triangle sum property : The sum of all the angles of a triangle is 180°

A+B+C= 180°

⇒B+C= 180°-A

Substitute the value in (i) we get

$cos(\dfrac{180^\circ-A}{2}) = cos(90^\circ- \dfrac{A}{2} )$

Now as we know

cos(90°-∅)= sin∅

therefore $cos (90^\circ -\dfrac{A}{2} )= sin\dfrac{A}{2}$

Hence proved

#Learn more

Sin(90-theta)=cos(90-theta) then find theta

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