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Question Number 70394 by Cmr 237 last updated on 04/Oct/19

$$\mathrm{montrer}\mathrm{que}$$$$\mathrm{sinA}+\mathrm{sinB}+\mathrm{sinC}=4\cos \frac{\mathrm{A}}{2}\cos \frac{\mathrm{B}}{2}\cos \frac{\mathrm{C}}{2}$$

Answered by $@ty@m123 last updated on 05/Oct/19

$$LHS=2\sin \frac{A+B}{2}\cos \frac{A-B}{2}+\sin C$$$$=2\sin \frac{\pi -C}{2}\cos \frac{A-B}{2}+\sin C$$$$=2\cos \frac{C}{2}\cos \frac{A-B}{2}+2\sin \frac{C}{2}\cos \frac{C}{2}$$$$=2\cos \frac{C}{2}(\cos \frac{A-B}{2}+\sin \frac{C}{2})$$$$=2\cos \frac{C}{2}(\cos \frac{A-B}{2}+\cos \frac{A+B}{2})$$$$=2\cos \frac{C}{2}\left(2\cos \frac{A-B+A+B}{4}\cos \frac{A-B-A-B}{4}\right)$$$$=4\cos \frac{\mathrm{A}}{2}\cos \frac{\mathrm{B}}{2}\cos \frac{\mathrm{C}}{2}$$$$=RHS$$

Commented by Cmr 237 last updated on 04/Oct/19

$$\mathrm{thank}\mathrm{you}$$

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