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psin-con-2-a-pcos-sin-2-b-p-0-0-pi-2-prove-p-a-2-b-2-3-2-ab-




Question Number 202295 by liuxinnan last updated on 24/Dec/23
psinθcon^2 θ=a  pcosθsin^2 θ=b  p≠0    θ∈(0 ,(π/2))  prove p=(((a^2 +b^2 )^(3/2) )/(ab))
$${psin}\theta{con}^{\mathrm{2}} \theta={a} \\ $$$${pcos}\theta{sin}^{\mathrm{2}} \theta={b} \\ $$$${p}\neq\mathrm{0}\:\:\:\:\theta\in\left(\mathrm{0}\:,\frac{\pi}{\mathrm{2}}\right) \\ $$$${prove}\:{p}=\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{{ab}} \\ $$
Answered by 1990mbodji last updated on 24/Dec/23
     Exercice :     p sinθ cos^2 θ = a    et    p cosθ sin^2  θ = b ou^�  θ∈]0;(π/2)[.    Montrons que  p = (((a^2 +b^2 )^(3/2) )/(ab)) .     ab = p^2 sin^3 θ cos^3 θ   et  a^2 +b^2  = p^2 sin^2 θ cos^2 θ (cos^2 θ+sin^2 θ)    ⇒ abp = (p sinθ cosθ)^3   et  a^2 +b^2  = (p sinθ cosθ)^2      ⇒ p sinθ cosθ = (abp)^(1/3)    et   p sinθ cosθ = (a^2 +b^2 )^(1/2)      ⇒ abp = (a^2 +b^2 )^(3/2)   ⇒ p = (((a^2 +b^2 )^(3/2) )/(ab)) .
$$ \\ $$$$\:\:\:\underline{\boldsymbol{{Exercice}}}\:: \\ $$$$\left.\:\:\:{p}\:{sin}\theta\:{cos}^{\mathrm{2}} \theta\:=\:{a}\:\:\:\:{et}\:\:\:\:{p}\:{cos}\theta\:{sin}^{\mathrm{2}} \:\theta\:=\:{b}\:{o}\grave {{u}}\:\theta\in\right]\mathrm{0};\frac{\pi}{\mathrm{2}}\left[.\right. \\ $$$$\:\:{Montrons}\:{que}\:\:{p}\:=\:\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{{ab}}\:. \\ $$$$\:\:\:{ab}\:=\:{p}^{\mathrm{2}} {sin}^{\mathrm{3}} \theta\:{cos}^{\mathrm{3}} \theta\:\:\:{et}\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:=\:{p}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\:{cos}^{\mathrm{2}} \theta\:\left({cos}^{\mathrm{2}} \theta+{sin}^{\mathrm{2}} \theta\right) \\ $$$$\:\:\Rightarrow\:{abp}\:=\:\left({p}\:{sin}\theta\:{cos}\theta\right)^{\mathrm{3}} \:\:{et}\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:=\:\left({p}\:{sin}\theta\:{cos}\theta\right)^{\mathrm{2}} \\ $$$$\:\:\:\Rightarrow\:{p}\:{sin}\theta\:{cos}\theta\:=\:\left({abp}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\:\:{et}\:\:\:{p}\:{sin}\theta\:{cos}\theta\:=\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\:\:\:\Rightarrow\:{abp}\:=\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:\:\Rightarrow\:{p}\:=\:\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{{ab}}\:. \\ $$
Commented by MathematicalUser2357 last updated on 26/Dec/23
no french
$$\mathrm{no}\:\mathrm{french} \\ $$

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