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 125389 by mnjuly1970 last updated on 10/Dec/20

... nice calculus ... in AB^Δ C prove : sin^2 (B)+sin^2 (C)−sin^2 (A)=2cos(A)sin(B)sin(C)

$$\:\:\:\:\:\:\:\:\:\:\:\:\:...\:{nice}\:\:{calculus}\:... \\ $$$$\:\:\:\:\:\:\:\:{in}\:\:{A}\overset{\Delta} {{B}C}\:\:{prove}\:: \\ $$$$\:\:\:\:\:\:{sin}^{\mathrm{2}} \left({B}\right)+{sin}^{\mathrm{2}} \left({C}\right)−{sin}^{\mathrm{2}} \left({A}\right)=\mathrm{2}{cos}\left({A}\right){sin}\left({B}\right){sin}\left({C}\right) \\ $$$$ \\ $$

Answered by mnjuly1970 last updated on 10/Dec/20

l.h.s ::sin^2 (B)+sin^2 (C)−sin^2 (A)=((b^2 +c^2 −a^2 )/(4R^2 ))=((2bccos(A))/(4R^2 )) =((2(2Rsin(B).2Rsin(C)cos(A))/(4R^2 )) =2sin(B)sin(C)cos(A) ✓

$${l}.{h}.{s}\:::{sin}^{\mathrm{2}} \left({B}\right)+{sin}^{\mathrm{2}} \left({C}\right)−{sin}^{\mathrm{2}} \left({A}\right)=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{4}{R}^{\mathrm{2}} }=\frac{\mathrm{2}{bccos}\left({A}\right)}{\mathrm{4}{R}^{\mathrm{2}} } \\ $$$$\:\:=\frac{\mathrm{2}\left(\mathrm{2}{Rsin}\left({B}\right).\mathrm{2}{Rsin}\left({C}\right){cos}\left({A}\right)\right.}{\mathrm{4}{R}^{\mathrm{2}} } \\ $$$$\:=\mathrm{2}{sin}\left({B}\right){sin}\left({C}\right){cos}\left({A}\right)\:\checkmark \\ $$

Answered by Dwaipayan Shikari last updated on 10/Dec/20

2sin^2 (B)+2sin^2 (C)−2sin^2 (A)=2y 1−cos2B+1−cos2C−1+cos2A=2y −2cos(B+C)cos(B−C)+cos2A=2y−1 2cos(A)cos(B−C)+2cos^2 A=2y y=cosA(cosA+cos(B−C))=2cosAcos(((A+B−C)/2))cos(((A−B+C)/2)) =2cosAcos(((π−2C)/2))cos(((π−2B)/2))=2cosAsinBsinC

$$\mathrm{2}{sin}^{\mathrm{2}} \left({B}\right)+\mathrm{2}{sin}^{\mathrm{2}} \left({C}\right)−\mathrm{2}\boldsymbol{{sin}}^{\mathrm{2}} \left(\boldsymbol{{A}}\right)=\mathrm{2}\boldsymbol{{y}} \\ $$$$\mathrm{1}−\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{B}}+\mathrm{1}−\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{C}}−\mathrm{1}+\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{A}}=\mathrm{2}\boldsymbol{{y}} \\ $$$$−\mathrm{2}\boldsymbol{{cos}}\left({B}+{C}\right){cos}\left({B}−{C}\right)+{cos}\mathrm{2}{A}=\mathrm{2}{y}−\mathrm{1} \\ $$$$\mathrm{2}{cos}\left({A}\right){cos}\left({B}−{C}\right)+\mathrm{2}{cos}^{\mathrm{2}} {A}=\mathrm{2}{y} \\ $$$${y}={cosA}\left({cosA}+{cos}\left({B}−{C}\right)\right)=\mathrm{2}{cosAcos}\left(\frac{{A}+{B}−{C}}{\mathrm{2}}\right){cos}\left(\frac{{A}−{B}+{C}}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}{cosAcos}\left(\frac{\pi−\mathrm{2}{C}}{\mathrm{2}}\right){cos}\left(\frac{\pi−\mathrm{2}{B}}{\mathrm{2}}\right)=\mathrm{2}{cosAsinBsinC} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 10/Dec/20

thank you so much hard−working and powerful ....grateful..

$${thank}\:{you}\:{so}\:{much} \\ $$$$\:{hard}−{working}\:{and}\:\:{powerful} \\ $$$$....{grateful}.. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com