Question Number 202295 by liuxinnan last updated on 24/Dec/23
$${psin}\theta{con}^{\mathrm{2}} \theta={a} \\ $$$${pcos}\theta{sin}^{\mathrm{2}} \theta={b} \\ $$$${p}\neq\mathrm{0}\:\:\:\:\theta\in\left(\mathrm{0}\:,\frac{\pi}{\mathrm{2}}\right) \\ $$$${prove}\:{p}=\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{{ab}} \\ $$
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)) .](Q202312.png)
$$ \\ $$$$\:\:\:\underline{\boldsymbol{{Exercice}}}\:: \\ $$$$\left.\:\:\:{p}\:{sin}\theta\:{cos}^{\mathrm{2}} \theta\:=\:{a}\:\:\:\:{et}\:\:\:\:{p}\:{cos}\theta\:{sin}^{\mathrm{2}} \:\theta\:=\:{b}\:{o}\grave {{u}}\:\theta\in\right]\mathrm{0};\frac{\pi}{\mathrm{2}}\left[.\right. \\ $$$$\:\:{Montrons}\:{que}\:\:{p}\:=\:\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{{ab}}\:. \\ $$$$\:\:\:{ab}\:=\:{p}^{\mathrm{2}} {sin}^{\mathrm{3}} \theta\:{cos}^{\mathrm{3}} \theta\:\:\:{et}\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:=\:{p}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta\:{cos}^{\mathrm{2}} \theta\:\left({cos}^{\mathrm{2}} \theta+{sin}^{\mathrm{2}} \theta\right) \\ $$$$\:\:\Rightarrow\:{abp}\:=\:\left({p}\:{sin}\theta\:{cos}\theta\right)^{\mathrm{3}} \:\:{et}\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:=\:\left({p}\:{sin}\theta\:{cos}\theta\right)^{\mathrm{2}} \\ $$$$\:\:\:\Rightarrow\:{p}\:{sin}\theta\:{cos}\theta\:=\:\left({abp}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\:\:{et}\:\:\:{p}\:{sin}\theta\:{cos}\theta\:=\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\:\:\:\Rightarrow\:{abp}\:=\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:\:\Rightarrow\:{p}\:=\:\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{{ab}}\:. \\ $$
Commented by MathematicalUser2357 last updated on 26/Dec/23
$$\mathrm{no}\:\mathrm{french} \\ $$