If A + B + C = π/3 , then sin ≤ft(π-6A/6) + sin ≤ft(π-6B/6) + sin C =

Correct Answer is : 4 cos (π−6A12)≤ft(π-6A/12)(12π−6A) cos (π−6B12)≤ft(π-6B/12)(12π−6B) sin C2C/22C

- 1 + 4 cos ≤ft(π-6A/12) cos ≤ft(π-6B/12) /2

4 sin ≤ft(π+6A/12) sin ≤ft(π+6B/12) cos C/2

1 - 4 cos ≤ft(π-6A/12) cos ≤ft(π-6B/12) cos π-6C/12

4 cos ≤ft(π-6A/12) cos ≤ft(π-6B/12) sin C/2

Correct Answer is : 4 cos (π−6A12)≤ft(π-6A/12)(12π−6A) cos (π−6B12)≤ft(π-6B/12)(12π−6B) sin C2C/22C

Test Index

TS EAMCET 04 May 2019 Shift 2 Solved Paper

Section: Mathematics

Question : 22 of 160

Marks: +1, -0

If A + B + C =

\frac{\pi}{3}

\left(\frac{\pi-6A}{6}\right)

\left(\frac{\pi-6B}{6}\right)

+ sin C =

- 1 + 4 cos $\left(\frac{\pi-6A}{12}\right)$ cos $\left(\frac{\pi-6B}{12}\right) \sin\frac{C}{2}$
4 sin $\left(\frac{\pi+6A}{12}\right)$ sin $\left(\frac{\pi+6B}{12}\right)$ cos $\frac{C}{2}$
1 - 4 cos $\left(\frac{\pi-6A}{12}\right)$ cos $\left(\frac{\pi-6B}{12}\right)$ cos $\frac{\pi-6C}{12}$
4 cos $\left(\frac{\pi-6A}{12}\right)$ cos $\left(\frac{\pi-6B}{12}\right)$ sin $\frac{C}{2}$

Solution:

It is given that

A + B + C = \frac{\pi}{3}

\text{So}

Y = \sin\left(\frac{\pi-6A}{6}\right) + \sin\left(\frac{\pi-6B}{6}\right) + \sin C

= 2 \sin\left(\frac{\pi-6A+\pi+6B}{12}\right) \cos\left(\frac{\pi-6A-\pi+6B}{12}\right) + \sin C

= 2 \sin\left(\frac{\pi}{6} - \left(\frac{A+B}{2}\right)\right) \cos\left(\frac{\pi}{6} - \left(\frac{A+B}{2}\right)\right) + \sin C

= 2 \sin\left(\frac{\pi}{6} - \frac{\pi}{6} + \frac{C}{2}\right) \cos\left(\frac{A-B}{2}\right) + 2 \sin\frac{C}{2} \cos\frac{C}{2}

Solve further

Y = 2 \sin\frac{C}{2} \left( \cos\frac{A-B}{2} + \cos\frac{C}{2} \right)

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

= 4 \sin\frac{C}{2} \left( \cos\frac{A+C-B}{4} \right) \cos\left( \frac{A-(B+C)}{4} \right)

= 4 \left( \cos\frac{\pi-6B}{12} \right) \left( \cos\frac{\pi-6A}{12} \right) \sin\frac{C}{2}

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