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TS EAMCET 6-Aug-2021 Shift 1 Question Paper

Section: Mathematics

Question : 22 of 160

Marks: +1, -0

\cos A+\cos B+\cos C

a+b\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}

a+b

Solution:

A, B, C

are angles of a triangle.

\therefore A+B+C=180^{\circ} \Rightarrow A+B=180^{\circ}-C

\Rightarrow \;\; \;\frac{A+B}{2}=90^{\circ}-\;\frac{C}{2}

Now,

\cos A+\cos B+\cos C

\;\; =2 \cos \left(\;\frac{A+B}{2}\right) \cdot \cos \left(\;\frac{A-B}{2}\right)+\left(1-2 \sin^{2} \;\frac{C}{2}\right)

\;\; =2 \cos \left(90^{\circ}-\;\frac{C}{2}\right) \cdot \cos \left(\;\frac{A-B}{2}\right)+1-2 \sin^{2} \;\frac{C}{2}

\;\; =2 \sin \;\frac{C}{2} \cdot \cos \left(\;\frac{A-B}{2}\right)-2 \sin^{2} \;\frac{C}{2}+1

\;\; =2 \sin \;\frac{C}{2}\left[\cos \left(\;\frac{A-B}{2}\right)-\sin \;\frac{C}{2}\right]+1

\;\; =2 \sin \;\frac{C}{2}\left[\cos \left(\;\frac{A-B}{2}\right)-\cos \left(\;\frac{A+B}{2}\right)\right]+1

\;\; =2 \sin \;\frac{C}{2}\left[2 \sin \;\frac{A}{2} \cdot \sin \;\frac{B}{2}\right]+1

\Rightarrow \cos A+\cos B+\cos C=1+4 \sin \;\frac{A}{2} \sin \;\frac{B}{2} \cdot \sin \;\frac{C}{2}

On comparing, we get

\;a=1, b=4

\therefore \;\; a+b=5

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