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Question Number 99539 by Ar Brandon last updated on 21/Jun/20

$$\mathrm{If}\mathrm{A}+\mathrm{B}+\mathrm{C}=180°\mathrm{Prove}$$$$\mathrm{sinA}+\mathrm{sinB}+\mathrm{sinC}=4\cos \frac{1}{2}\mathrm{Acos}\frac{1}{2}\mathrm{Bcos}\frac{1}{2}\mathrm{C}$$

Answered by Dwaipayan Shikari last updated on 22/Jun/20

$$sinA+sinB+sinC=2sin\frac{A}{2}cos\frac{A}{2}+2sin\frac{B+C}{2}cos\frac{B-C}{2}$$$$=2cos\frac{A}{2}(sin\frac{A}{2}+cos\frac{B-C}{2})$$$$=4cos\frac{A}{2}\left(cos\frac{\frac{\pi }{2}-\frac{A+C}{2}+\frac{B}{2}}{2}cos\frac{\frac{\pi }{2}-\frac{A+B}{2}+\frac{C}{2}}{2}\right)$$$$=4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}\left[Proved\right]\{Asyoucansee(\frac{\pi }{2}-\frac{A+C}{2})=\frac{B}{2}$$$$\{And(\frac{\pi }{2}-\frac{A+B}{2})=\frac{C}{2},sin\frac{A}{2}=cos(\frac{\pi }{2}-\frac{A}{2})$$$$\{sin\frac{C+B}{2}=cos\frac{A}{2}$$

Commented by Ar Brandon last updated on 22/Jun/20

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