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

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