Question

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

Answer 1

Answer:

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Given:

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

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To show:

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sin{(B + C)/2} = cosA/2

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Solution:

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$\longrightarrow$ (A + B + C )/2 = 90

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$\longrightarrow$ B/2 + C/2 = 90 – A/2

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$\longrightarrow$ sin(B + 2)/2 = sin(90 – A/2)

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$\longrightarrow$ sin(B + C)/2 = cosA/2

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L.H.S = R.H.S

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Hence proved.

Answer 2

Given 

$△ABC$

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

Hence 

$∠A+∠B+∠C=180$

$or A+B+C=180o$

$B+C=180o−A$

Multiply both sides

by 21

$21(B+C)=21(180o−A)$

$21(B+C)=90o−2A...(1)$

Now 

$21(B+C)$

Taking sine of this angle

$sin(2B+C) \\ [2B+C=90o−2A]$

$sin(90o−2A)$

$cos2A[sin(90o−θ)=cosθ]$

Hence 

$sin(2B+C)=cos2A proved$