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 8503 by MNG last updated on 13/Oct/16

Q. 1 sinA+sinB+sinC = 4cos(A/2) cos (B/2) cos (C/2) . Q.2 cosA cosB − cosC = 4cos(A/2) cos(B/2) cos(C/2) −1 Q.3 ((sin2A+sin2B+sin2C)/(sinA+sinB+sinC)) =8sin (A/2) sin(B/2) sin(C/2)

$${Q}.\:\mathrm{1}\:\:{sinA}+{sinB}+{sinC}\:=\:\mathrm{4}{cos}\frac{{A}}{\mathrm{2}}\:{cos} \\ $$$$\frac{{B}}{\mathrm{2}}\:{cos}\:\frac{{C}}{\mathrm{2}}\:. \\ $$$${Q}.\mathrm{2}\:\:{cosA}\:{cosB}\:−\:{cosC}\:=\:\mathrm{4}{cos}\frac{{A}}{\mathrm{2}}\:{cos}\frac{{B}}{\mathrm{2}} \\ $$$${cos}\frac{{C}}{\mathrm{2}}\:−\mathrm{1} \\ $$$$ \\ $$$${Q}.\mathrm{3}\:\:\frac{{sin}\mathrm{2}{A}+{sin}\mathrm{2}{B}+{sin}\mathrm{2}{C}}{{sinA}+{sinB}+{sinC}}\:=\mathrm{8}{sin}\:\frac{{A}}{\mathrm{2}} \\ $$$${sin}\frac{{B}}{\mathrm{2}}\:{sin}\frac{{C}}{\mathrm{2}} \\ $$$$ \\ $$

Commented by sandy_suhendra last updated on 13/Oct/16

Is A+B+C=180° ?

$$\mathrm{Is}\:\mathrm{A}+\mathrm{B}+\mathrm{C}=\mathrm{180}°\:? \\ $$

Commented by MNG last updated on 14/Oct/16

yes

$${yes} \\ $$

Answered by prakash jain last updated on 14/Oct/16

Q1 sin A+sin B+sin C =2sin ((A+B)/2)cos ((A−B)/2)+sin C A+B+C=π⇒((A+B)/2)=(π/2)−(C/2) =2sin ((π/2)−(C/2))cos ((A−B)/2)+2sin (C/2)cos (C/2) =2cos (C/2)cos ((A−B)/2)+2sin ((π/2)−((A+B)/2))cos (C/2) =2cos (C/2){cos ((A−B)/2)+cos ((A+B)/2)} =2cos (C/2)cos (A/2)cos (B/2) formula sin A+sin B=2sin ((A+B)/2)cos ((A−B)/2) cos A+cos B=2cos ((A+B)/2)cos ((A−B)/2)

$$\mathrm{Q1} \\ $$$$\mathrm{sin}\:{A}+\mathrm{sin}\:{B}+\mathrm{sin}\:{C} \\ $$$$=\mathrm{2sin}\:\frac{{A}+{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}}+\mathrm{sin}\:{C} \\ $$$${A}+{B}+{C}=\pi\Rightarrow\frac{{A}+{B}}{\mathrm{2}}=\frac{\pi}{\mathrm{2}}−\frac{{C}}{\mathrm{2}} \\ $$$$=\mathrm{2sin}\:\left(\frac{\pi}{\mathrm{2}}−\frac{{C}}{\mathrm{2}}\right)\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}}+\mathrm{2sin}\:\frac{{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{C}}{\mathrm{2}} \\ $$$$=\mathrm{2cos}\:\frac{{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}}+\mathrm{2sin}\:\left(\frac{\pi}{\mathrm{2}}−\frac{{A}+{B}}{\mathrm{2}}\right)\mathrm{cos}\:\frac{{C}}{\mathrm{2}} \\ $$$$=\mathrm{2cos}\:\frac{{C}}{\mathrm{2}}\left\{\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}}+\mathrm{cos}\:\frac{{A}+{B}}{\mathrm{2}}\right\} \\ $$$$=\mathrm{2cos}\:\frac{{C}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{2}} \\ $$$${formula} \\ $$$$\mathrm{sin}\:{A}+\mathrm{sin}\:{B}=\mathrm{2sin}\:\frac{{A}+{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:{A}+\mathrm{cos}\:{B}=\mathrm{2cos}\:\frac{{A}+{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}−{B}}{\mathrm{2}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com