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Question Number 124501 by mnjuly1970 last updated on 03/Dec/20

          ....nice  calculus..     in AB^Δ C : sin^2 (A)+sin^2 (B)+sin^2 (C)=2  prove that: AB^Δ C is right triangle                 Good luck.

$$\:\:\:\:\:\:\:\:\:\:....{nice}\:\:{calculus}.. \\ $$$$\:\:\:{in}\:{A}\overset{\Delta} {{B}C}\::\:{sin}^{\mathrm{2}} \left({A}\right)+{sin}^{\mathrm{2}} \left({B}\right)+{sin}^{\mathrm{2}} \left({C}\right)=\mathrm{2} \\ $$$${prove}\:{that}:\:{A}\overset{\Delta} {{B}C}\:{is}\:{right}\:{triangle} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathscr{G}{ood}\:{luck}. \\ $$

Answered by mindispower last updated on 03/Dec/20

sin(c)=sin(a+b)  ⇔sin^2 (A)+sin^2 (B)+(sin(A+B))^2 )=2  sin^2 (A+B)=sin^2 (A)cos^2 (B)+cos^2 (A)sin^2 (B)  +(1/2)sin(2A)sin(2B)  sin^2 (A)−1+sin^2 (B)−1+((sin(2A)sin(2B))/2)+  sin^2 (A)cos^2 (B)+sin^2 (B)cos^2 (A)=0  ⇔−cos^2 (A)+cos^2 (A)sin^2 (B)−cos^2 (B)+cos^2 (B)sin^2 (A)  +((sin(2A)sin(2B))/2)=0  ⇔−2cos^2 (A)cos^2 (B)+2sin(A)sin(B)cos(A)cos(B)=0  ⇔2cos(A)cos(B)(sin(A)sin(B)−cos(A)cos(B))=0  ⇒cos(A)cos(B)cos(A+B)=0  ⇒A=(π/2),B=(π/2),or A+B=(π/2)⇒C=(π/2)

$${sin}\left({c}\right)={sin}\left({a}+{b}\right) \\ $$$$\left.\Leftrightarrow{sin}^{\mathrm{2}} \left({A}\right)+{sin}^{\mathrm{2}} \left({B}\right)+\left({sin}\left({A}+{B}\right)\right)^{\mathrm{2}} \right)=\mathrm{2} \\ $$$${sin}^{\mathrm{2}} \left({A}+{B}\right)={sin}^{\mathrm{2}} \left({A}\right){cos}^{\mathrm{2}} \left({B}\right)+{cos}^{\mathrm{2}} \left({A}\right){sin}^{\mathrm{2}} \left({B}\right) \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{A}\right){sin}\left(\mathrm{2}{B}\right) \\ $$$${sin}^{\mathrm{2}} \left({A}\right)−\mathrm{1}+{sin}^{\mathrm{2}} \left({B}\right)−\mathrm{1}+\frac{{sin}\left(\mathrm{2}{A}\right){sin}\left(\mathrm{2}{B}\right)}{\mathrm{2}}+ \\ $$$${sin}^{\mathrm{2}} \left({A}\right){cos}^{\mathrm{2}} \left({B}\right)+{sin}^{\mathrm{2}} \left({B}\right){cos}^{\mathrm{2}} \left({A}\right)=\mathrm{0} \\ $$$$\Leftrightarrow−{cos}^{\mathrm{2}} \left({A}\right)+{cos}^{\mathrm{2}} \left({A}\right){sin}^{\mathrm{2}} \left({B}\right)−{cos}^{\mathrm{2}} \left({B}\right)+{cos}^{\mathrm{2}} \left({B}\right){sin}^{\mathrm{2}} \left({A}\right) \\ $$$$+\frac{{sin}\left(\mathrm{2}{A}\right){sin}\left(\mathrm{2}{B}\right)}{\mathrm{2}}=\mathrm{0} \\ $$$$\Leftrightarrow−\mathrm{2}{cos}^{\mathrm{2}} \left({A}\right){cos}^{\mathrm{2}} \left({B}\right)+\mathrm{2}{sin}\left({A}\right){sin}\left({B}\right){cos}\left({A}\right){cos}\left({B}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{2}{cos}\left({A}\right){cos}\left({B}\right)\left({sin}\left({A}\right){sin}\left({B}\right)−{cos}\left({A}\right){cos}\left({B}\right)\right)=\mathrm{0} \\ $$$$\Rightarrow{cos}\left({A}\right){cos}\left({B}\right){cos}\left({A}+{B}\right)=\mathrm{0} \\ $$$$\Rightarrow{A}=\frac{\pi}{\mathrm{2}},{B}=\frac{\pi}{\mathrm{2}},{or}\:{A}+{B}=\frac{\pi}{\mathrm{2}}\Rightarrow{C}=\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 03/Dec/20

thank you so much  mr mindspower  nice as always...

$${thank}\:{you}\:{so}\:{much} \\ $$$${mr}\:{mindspower} \\ $$$${nice}\:{as}\:{always}... \\ $$

Answered by Dwaipayan Shikari last updated on 03/Dec/20

2sin^2 A+2sin^2 B+2sin^2 C=4  1−cos2A+1−cos2B+1−sin2C=4  cos2A+cos2B+cos2C=−1  2cos(A+B)cos(A−B)+cos2C=−1  2cos(A+B)cos(A−B)+2cos^2 C=0  cos(A+B)cos(A−B)+cos^2 C=0  cosC cos(A−B)=cos^2 C  cosC(cosC−cos(A−B))=0  cosC=0  C=(π/2)      or  cosC=cos(A−B) ⇒C=A−B⇒C+B=A

$$\mathrm{2}{sin}^{\mathrm{2}} {A}+\mathrm{2}{sin}^{\mathrm{2}} {B}+\mathrm{2}{sin}^{\mathrm{2}} {C}=\mathrm{4} \\ $$$$\mathrm{1}−{cos}\mathrm{2}{A}+\mathrm{1}−{cos}\mathrm{2}{B}+\mathrm{1}−{sin}\mathrm{2}{C}=\mathrm{4} \\ $$$${cos}\mathrm{2}{A}+{cos}\mathrm{2}{B}+{cos}\mathrm{2}{C}=−\mathrm{1} \\ $$$$\mathrm{2}{cos}\left({A}+{B}\right){cos}\left({A}−{B}\right)+{cos}\mathrm{2}{C}=−\mathrm{1} \\ $$$$\mathrm{2}{cos}\left({A}+{B}\right){cos}\left({A}−{B}\right)+\mathrm{2}{cos}^{\mathrm{2}} {C}=\mathrm{0} \\ $$$${cos}\left({A}+{B}\right){cos}\left({A}−{B}\right)+{cos}^{\mathrm{2}} {C}=\mathrm{0} \\ $$$${cosC}\:{cos}\left({A}−{B}\right)={cos}^{\mathrm{2}} {C} \\ $$$${cosC}\left({cosC}−{cos}\left({A}−{B}\right)\right)=\mathrm{0} \\ $$$${cosC}=\mathrm{0}\:\:{C}=\frac{\pi}{\mathrm{2}}\:\:\:\:\:\:{or}\:\:{cosC}={cos}\left({A}−{B}\right)\:\Rightarrow{C}={A}−{B}\Rightarrow{C}+{B}={A} \\ $$

Commented by mnjuly1970 last updated on 03/Dec/20

thank you so much  extraordinary...

$${thank}\:{you}\:{so}\:{much} \\ $$$${extraordinary}... \\ $$$$ \\ $$

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