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Q.In △ABC\triangle ABC, if sin⁡θ=ab+c\sin\theta = \dfrac{a}{b+c}, then show that cos⁡θ=2bcb+ccos⁡A2\cos\theta = \dfrac{2\sqrt{bc}}{b+c} \cos \dfrac{A}{2}.

Telangana TsbieTelangana Board of Intermediate Education (Intermediate 1st Year) 2022Subjective· 4mImportance★★★★★
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Square the given relation, factor (b+c)2−a2(b+c)^2-a^2 in terms of the semi-perimeter ss, then bring in the half-angle formula cos⁡2A2=s(s−a)bc\cos^2\frac{A}{2}=\frac{s(s-a)}{bc} to finish.

Given: sin⁡θ=ab+c\sin\theta = \dfrac{a}{b+c}.

Step 1 — square and use sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1:

cos⁡2θ=1−sin⁡2θ=1−a2(b+c)2=(b+c)2−a2(b+c)2=(b+c−a)(b+c+a)(b+c)2\cos^2\theta = 1-\sin^2\theta = 1-\frac{a^2}{(b+c)^2} = \frac{(b+c)^2-a^2}{(b+c)^2} = \frac{(b+c-a)(b+c+a)}{(b+c)^2}

Step 2 — bring in the semi-perimeter s=a+b+c2s=\dfrac{a+b+c}{2}, so a+b+c=2sa+b+c=2s and b+c−a=2s−2a=2(s−a)b+c-a = 2s-2a = 2(s-a):

cos⁡2θ=2(s−a)⋅2s(b+c)2=4s(s−a)(b+c)2\cos^2\theta = \frac{2(s-a)\cdot 2s}{(b+c)^2} = \frac{4s(s-a)}{(b+c)^2}

Step 3 — use the standard half-angle formula cos⁡2A2=s(s−a)bc\cos^2\dfrac{A}{2} = \dfrac{s(s-a)}{bc}, i.e. s(s−a)=bccos⁡2A2s(s-a) = bc\cos^2\dfrac{A}{2}: …

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