Question

Hi solve trignometry question :-<br />if A, B, and C r interior angles of a triangle ABC then show that Sin B+C/2 = Cos A/2​

Answer 1

Given:

• a triangle ABC

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

To Prove:

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

Solution:

we know that:

sum of all angles of triangle = 180° [ASP]

so, in ∆ABC,

$\sf{A +B+C=180°}$

$⟹$$\sf{B+C=180°-A}$

$\sf{multiplying\:both\:side\:by\: \frac{1}{2}}$

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

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

$\sf { \frac{B + C}{2} = 90 \degree - \frac{A}{2} \: \: \: \: \: \: .....(1)}$

Taking L.H.S,

$\sf \sin( \frac{B + C}{2} )$

$\implies \sf \sin(90 \degree - \frac{A}{2} )$

$\implies \sf \cos \frac{A}{2} \: \: \: \: \: \: ( \sin(90 - \theta) = \cos( \theta) )$

$\implies\sf{R.H.S}$

Hence Proved

Answer 2

Given :

• A , B and C are interior angles of a ∆ABC.

To Prove :

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

Solution:

We know that:

Sum of all Angles the in a triangle is 180°

Then , In ∆ ABC

Sum of the angles = 180°

$\sf\angle\:A+\angle\:B+\angle\:C=180\degree$

$\sf\implies\angle\:B+\angle\:C=180\degree-\angle\:A$

Now Divide both sides by 2

$\sf\implies\dfrac{\angle\:B+\angle\:C}{2}=\dfrac{180\degree-\angle\:A}{2}$

$\sf\implies\dfrac{\angle\:B+\angle\:C}{2}=\dfrac{180\degree}{2}-\dfrac{\angle\:A}{2}$

$\sf\implies\dfrac{\angle\:B+\angle\:C}{2}=90\degree-\dfrac{\angle\:A}{2}$

Taking sine of the Angle both sides , Then

$\sf\implies\sin(\dfrac{\angle\:B+\angle\:C}{2})=\sin(90\degree-\dfrac{\angle\:A}{2})$

$\sf\implies\sin(\dfrac{\angle\:B+\angle\:C}{2})=\cos(\dfrac{\angle\:A}{2})$

Hence , Proved

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Formula's Used :

$\sf1)\sin(90-x)=\cos\:x$

$\sf2)\cos(90-x)=\sin\:x$

$\sf3)\tan(90-x)=\cot\:x$

$\sf4)\sec(90-x)=\csc\:x$

$\sf5)\cot(90-x)=\tan\:x$

$\sf6)\csc(90-x)=\sec\:x$