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Question-211732

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Question Number 211732 by BaliramKumar last updated on 19/Sep/24
Answered by TonyCWX08 last updated on 19/Sep/24
Using quadratic formula x=((sin^2 θ±(√(sin^4 θ + 4cos^2 θ)))/2) p=((sin^2 θ+(√(sin^4 θ + 4cos^2 θ)))/2) p^2 =(((sin^2 θ+(√(sin^4 θ + 4cos^2 θ)))/2))^2 p^2 = (((sin^2 θ+(√(sin^4 θ + 4cos^2 θ)))^2 )/4) q=((sin^2 θ−(√(sin^4 θ + 4cos^2 θ)))/2) q^2 =(((sin^2 θ−(√(sin^4 θ + 4cos^2 θ)))^2 )/4) p^2 +q^2 =((sin^4 θ + 2sin^2 θ(√(sin^4 θ+4cos^2 θ ))+ sin^4 θ + 4cos^2 θ +sin^4 θ −2sin^2 θ(√(sin^4 θ+4cos^2 θ))+sin^4 θ+4cos^2 θ)/4) p^2 +q^2 =((4sin^4 θ + 8cos^2 θ )/4) p^2 +q^2 =sin^4 θ+2cos^2 θ Minimum value = 1
$${Using}\:{quadratic}\:{formula} \\ $$$${x}=\frac{{sin}^{\mathrm{2}} \theta\pm\sqrt{{sin}^{\mathrm{4}} \theta\:+\:\mathrm{4}{cos}^{\mathrm{2}} \theta}}{\mathrm{2}} \\ $$$$ \\ $$$${p}=\frac{{sin}^{\mathrm{2}} \theta+\sqrt{{sin}^{\mathrm{4}} \theta\:+\:\mathrm{4}{cos}^{\mathrm{2}} \theta}}{\mathrm{2}} \\ $$$${p}^{\mathrm{2}} =\left(\frac{{sin}^{\mathrm{2}} \theta+\sqrt{{sin}^{\mathrm{4}} \theta\:+\:\mathrm{4}{cos}^{\mathrm{2}} \theta}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${p}^{\mathrm{2}} =\:\frac{\left({sin}^{\mathrm{2}} \theta+\sqrt{{sin}^{\mathrm{4}} \theta\:+\:\mathrm{4}{cos}^{\mathrm{2}} \theta}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$ \\ $$$${q}=\frac{{sin}^{\mathrm{2}} \theta−\sqrt{{sin}^{\mathrm{4}} \theta\:+\:\mathrm{4}{cos}^{\mathrm{2}} \theta}}{\mathrm{2}} \\ $$$${q}^{\mathrm{2}} =\frac{\left({sin}^{\mathrm{2}} \theta−\sqrt{{sin}^{\mathrm{4}} \theta\:+\:\mathrm{4}{cos}^{\mathrm{2}} \theta}\right)^{\mathrm{2}} }{\mathrm{4}} \\ $$$$ \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\frac{{sin}^{\mathrm{4}} \theta\:+\:\mathrm{2}{sin}^{\mathrm{2}} \theta\sqrt{{sin}^{\mathrm{4}} \theta+\mathrm{4}{cos}^{\mathrm{2}} \theta\:}+\:{sin}^{\mathrm{4}} \theta\:+\:\mathrm{4}{cos}^{\mathrm{2}} \theta\:+{sin}^{\mathrm{4}} \theta\:−\mathrm{2}{sin}^{\mathrm{2}} \theta\sqrt{{sin}^{\mathrm{4}} \theta+\mathrm{4}{cos}^{\mathrm{2}} \theta}+{sin}^{\mathrm{4}} \theta+\mathrm{4}{cos}^{\mathrm{2}} \theta}{\mathrm{4}} \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\frac{\mathrm{4}{sin}^{\mathrm{4}} \theta\:+\:\mathrm{8}{cos}^{\mathrm{2}} \theta\:}{\mathrm{4}} \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} ={sin}^{\mathrm{4}} \theta+\mathrm{2}{cos}^{\mathrm{2}} \theta \\ $$$${Minimum}\:{value}\:=\:\mathrm{1} \\ $$
Commented by BaliramKumar last updated on 19/Sep/24
first step 4cos^2 θ
$${first}\:{step}\:\mathrm{4}{cos}^{\mathrm{2}} \theta \\ $$
Commented by TonyCWX08 last updated on 19/Sep/24
Edited
$${Edited} \\ $$
Answered by BHOOPENDRA last updated on 19/Sep/24
p+q=((−b)/a) p.q=c/a (p+q)^2 =p^2 +q^2 +2pq p^2 +q^2 =(p+q)^2 −2pq p^2 +q^2 = sin^4 θ +2cos^2 θ =(1−cos^2 θ)^2 +2cos^2 θ p^2 +q^2 =(1+cos^4 θ) = 0≤cos^4 θ≤1 = 0+1≤cos^4 θ +1≤1+1 = 1≤cos^4 θ +1≤2 p^2 +q^2 =1 (minimum value)
$${p}+{q}=\frac{−{b}}{{a}} \\ $$$${p}.{q}={c}/{a} \\ $$$$\left({p}+{q}\right)^{\mathrm{2}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} +\mathrm{2}{pq} \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\left({p}+{q}\right)^{\mathrm{2}} −\mathrm{2}{pq} \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\:{sin}^{\mathrm{4}} \theta\:+\mathrm{2}{cos}^{\mathrm{2}} \theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{1}−{cos}^{\mathrm{2}} \theta\right)^{\mathrm{2}} +\mathrm{2}{cos}^{\mathrm{2}} \theta \\ $$$$\:{p}^{\mathrm{2}} +{q}^{\mathrm{2}} \:=\left(\mathrm{1}+{cos}^{\mathrm{4}} \theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\:\:\mathrm{0}\leq{cos}^{\mathrm{4}} \theta\leq\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\:\:\:\mathrm{0}+\mathrm{1}\leq{cos}^{\mathrm{4}} \theta\:+\mathrm{1}\leq\mathrm{1}+\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{1}\leq{cos}^{\mathrm{4}} \theta\:+\mathrm{1}\leq\mathrm{2} \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\mathrm{1}\:\left({minimum}\:{value}\right) \\ $$$$ \\ $$

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