Question-193874

Question Number 193874 by cortano12 last updated on 22/Jun/23

Answered by Subhi last updated on 22/Jun/23

cos(y)=1−2sin^2 ((y/2)) 1−cos(cx^2 +bx+a)=2sin^2 (((cx^2 +bx+a)/2)) lim_(x→(1/α)) ((2sin^2 (((cx^2 +bx+a)/2)))/((1−αx)^2 )) ax^2 +bx+c ⇛ x = ((−b±(√(b^2 −4ac)))/(2a)) suppose: α = ((−b+(√(b^2 −4ac)))/(2a)) , β = ((−b−(√(b^2 −4ac)))/(2a)) (ax^2 +bx+c) ⇛αβ = (c/a) (cx^2 +bx+a) ⇛ δγ = (a/c) (a/c) = ((c/a))^(−1) ∴ (1/α)=(1/δ) , (1/β)=(1/γ) (ax^2 +bx+c) = a(x−α)(x−β) (cx^2 +bx+a)= c(x−(1/α))(x−(1/β))=(c/(αβ))(1−αx)(1−βx)=a(1−αx)(1−βx) lim_(x→(1/α)) ((2sin^2 (a(((1−βx))/2)).(1−αx)))/((1−αx)^2 ))=((2a^2 (1−βx)^2 )/4)=((a^2 (1−βx)^2 )/2)=((a^2 (1−(β/α))^2 )/2) ((a^2 (1−(β/α))^2 )/2)=((a^2 (1−((−b−(√(b^2 −4ac)))/(−b+(√(b^2 −4ac)))))^2 )/2)=((2a^2 (b^2 −4ac))/((−b+(√(b^2 −4ac)))^2 ))=(((b^2 −4ac))/(2α^2 ))

$$ \\ $$$${cos}\left({y}\right)=\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{y}}{\mathrm{2}}\right) \\ $$$$\mathrm{1}−{cos}\left({cx}^{\mathrm{2}} +{bx}+{a}\right)=\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{cx}^{\mathrm{2}} +{bx}+{a}}{\mathrm{2}}\right) \\ $$$${lim}_{{x}\rightarrow\frac{\mathrm{1}}{\alpha}} \:\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{cx}^{\mathrm{2}} +{bx}+{a}}{\mathrm{2}}\right)}{\left(\mathrm{1}−\alpha{x}\right)^{\mathrm{2}} } \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}\:\Rrightarrow\:{x}\:=\:\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$${suppose}:\:\alpha\:=\:\frac{−{b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}}\:\:,\:\beta\:=\:\frac{−{b}−\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{\mathrm{2}{a}} \\ $$$$\left({ax}^{\mathrm{2}} +{bx}+{c}\right)\:\Rrightarrow\alpha\beta\:=\:\frac{{c}}{{a}} \\ $$$$\left({cx}^{\mathrm{2}} +{bx}+{a}\right)\:\Rrightarrow\:\delta\gamma\:=\:\frac{{a}}{{c}} \\ $$$$\frac{{a}}{{c}}\:=\:\left(\frac{{c}}{{a}}\right)^{−\mathrm{1}} \\ $$$$\therefore\:\:\frac{\mathrm{1}}{\alpha}=\frac{\mathrm{1}}{\delta}\:,\:\frac{\mathrm{1}}{\beta}=\frac{\mathrm{1}}{\gamma} \\ $$$$\left({ax}^{\mathrm{2}} +{bx}+{c}\right)\:=\:{a}\left({x}−\alpha\right)\left({x}−\beta\right) \\ $$$$\left({cx}^{\mathrm{2}} +{bx}+{a}\right)=\:{c}\left({x}−\frac{\mathrm{1}}{\alpha}\right)\left({x}−\frac{\mathrm{1}}{\beta}\right)=\frac{{c}}{\alpha\beta}\left(\mathrm{1}−\alpha{x}\right)\left(\mathrm{1}−\beta{x}\right)={a}\left(\mathrm{1}−\alpha{x}\right)\left(\mathrm{1}−\beta{x}\right) \\ $$$${lim}_{{x}\rightarrow\frac{\mathrm{1}}{\alpha}} \frac{\mathrm{2}{sin}^{\mathrm{2}} \left({a}\left(\frac{\left.\mathrm{1}−\beta{x}\right)}{\mathrm{2}}\right).\left(\mathrm{1}−\alpha{x}\right)\right)}{\left(\mathrm{1}−\alpha{x}\right)^{\mathrm{2}} }=\frac{\mathrm{2}{a}^{\mathrm{2}} \left(\mathrm{1}−\beta{x}\right)^{\mathrm{2}} }{\mathrm{4}}=\frac{{a}^{\mathrm{2}} \left(\mathrm{1}−\beta{x}\right)^{\mathrm{2}} }{\mathrm{2}}=\frac{{a}^{\mathrm{2}} \left(\mathrm{1}−\frac{\beta}{\alpha}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\frac{{a}^{\mathrm{2}} \left(\mathrm{1}−\frac{\beta}{\alpha}\right)^{\mathrm{2}} }{\mathrm{2}}=\frac{{a}^{\mathrm{2}} \left(\mathrm{1}−\frac{−{b}−\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}{−{b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}}\right)^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{2}{a}^{\mathrm{2}} \left({b}^{\mathrm{2}} −\mathrm{4}{ac}\right)}{\left(−{b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\right)^{\mathrm{2}} }=\frac{\left({b}^{\mathrm{2}} −\mathrm{4}{ac}\right)}{\mathrm{2}\alpha^{\mathrm{2}} } \\ $$

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