Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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

montrer que sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)

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

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

LHS=2sin ((A+B)/2)cos ((A−B)/2)+sin C =2sin ((π−C)/2)cos ((A−B)/2)+sin C =2cos (C/2)cos ((A−B)/2)+2sin (C/2)cos (C/2) =2cos (C/2)(cos ((A−B)/2)+sin (C/2)) =2cos (C/2)(cos ((A−B)/2)+cos ((A+B)/2)) =2cos (C/2)(2cos ((A−B+A+B)/4)cos ((A−B−A−B)/4)) =4cos(A/2)cos(B/2)cos(C/2) =RHS

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

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

thank you

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com