montrer-que-sinA-sinB-sinC-4cos-A-2-cos-B-2-cos-C-2-

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

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

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

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

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

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

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