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Question Number 172408 by infinityaction last updated on 26/Jun/22
Answered by mr W last updated on 26/Jun/22
Commented by mr W last updated on 26/Jun/22
AD=DC=1 CP=1 sin θ=PQ DQ=1 cos θ+1 sin θ=sin θ+cos θ AQ^2 =1^2 +(sin θ+cos θ)^2 −2×1×(sin θ+cos θ) cos ((π/2)+θ) =2+sin 2θ+2(sin θ+cos θ) sin θ =3+2 sin 2θ−cos 2θ =3+(√5)((2/( (√5))) sin 2θ−(1/( (√5))) cos 2θ) =3+(√5)(cos α sin 2θ−sin α cos 2θ) =3+(√5) sin (2θ−α) (with α=sin^(−1) (1/( (√5)))=cos^(−1) (2/( (√5)))=tan^(−1) (1/2)) =3+(√5) sin (2θ−tan^(−1) (1/2)) (AQ^2 )_(max) =3+(√5) AQ_(max) =(√(3+(√5)))≈2.288 at 2θ−tan^(−1) (1/2)=(π/2) or θ=(π/4)+(1/2)tan^(−1) (1/2)≈58.3°
$${AD}={DC}=\mathrm{1} \\ $$$${CP}=\mathrm{1}\:\mathrm{sin}\:\theta={PQ} \\ $$$${DQ}=\mathrm{1}\:\mathrm{cos}\:\theta+\mathrm{1}\:\mathrm{sin}\:\theta=\mathrm{sin}\:\theta+\mathrm{cos}\:\theta \\ $$$${AQ}^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\left(\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\right)^{\mathrm{2}} −\mathrm{2}×\mathrm{1}×\left(\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\right)\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}+\theta\right) \\ $$$$\:\:=\mathrm{2}+\mathrm{sin}\:\mathrm{2}\theta+\mathrm{2}\left(\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\right)\:\mathrm{sin}\:\theta \\ $$$$\:\:=\mathrm{3}+\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\theta−\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\:\:=\mathrm{3}+\sqrt{\mathrm{5}}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\mathrm{sin}\:\mathrm{2}\theta−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:\mathrm{cos}\:\mathrm{2}\theta\right) \\ $$$$\:\:=\mathrm{3}+\sqrt{\mathrm{5}}\left(\mathrm{cos}\:\alpha\:\mathrm{sin}\:\mathrm{2}\theta−\mathrm{sin}\:\alpha\:\mathrm{cos}\:\mathrm{2}\theta\right) \\ $$$$\:\:=\mathrm{3}+\sqrt{\mathrm{5}}\:\mathrm{sin}\:\left(\mathrm{2}\theta−\alpha\right)\:\: \\ $$$$\left({with}\:\alpha=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:=\mathrm{3}+\sqrt{\mathrm{5}}\:\mathrm{sin}\:\left(\mathrm{2}\theta−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\left({AQ}^{\mathrm{2}} \right)_{{max}} =\mathrm{3}+\sqrt{\mathrm{5}} \\ $$$${AQ}_{{max}} =\sqrt{\mathrm{3}+\sqrt{\mathrm{5}}}\approx\mathrm{2}.\mathrm{288} \\ $$$${at}\:\mathrm{2}\theta−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}=\frac{\pi}{\mathrm{2}}\: \\ $$$${or}\:\theta=\frac{\pi}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\approx\mathrm{58}.\mathrm{3}° \\ $$
Commented by infinityaction last updated on 26/Jun/22
explain sir 3+(√5)sin (2θ−tan^(−1) (1/2))
$${explain}\:{sir} \\ $$$$\mathrm{3}+\sqrt{\mathrm{5}}\mathrm{sin}\:\left(\mathrm{2}\theta−\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$
Commented by mr W last updated on 26/Jun/22
i have added some steps more.
$${i}\:{have}\:{added}\:{some}\:{steps}\:{more}. \\ $$
Commented by infinityaction last updated on 26/Jun/22
AD=DC=1 CP=1 sin θ=PQ DQ=1 cos θ+1 sin θ=sin θ+cos θ AQ^2 =1^2 +(sin 𝛉+cos 𝛉)^2 −2(sin 𝛉+cos 𝛉) cos ((𝛑/2)+𝛉) AQ^2 =2+sin 2θ+2(sin θ+cos θ) sin θ AQ^2 =3+2 sin 2θ−cos 2θ let p = 2sin2θ − cos2θ (dp/dθ) = 4cos2θ + 2sin2θ = 0 tan2θ = −2 sin2θ = (2/( (√5))) and cos2θ = −(1/( (√5))) AQ^2 = 3+(√5) AQ = (√(3+(√5)))
$$\:\:\:\:{AD}={DC}=\mathrm{1} \\ $$$$\:\:\:\:\:{CP}=\mathrm{1}\:\mathrm{sin}\:\theta={PQ} \\ $$$$\:\:\:{DQ}=\mathrm{1}\:\mathrm{cos}\:\theta+\mathrm{1}\:\mathrm{sin}\:\theta=\mathrm{sin}\:\theta+\mathrm{cos}\:\theta \\ $$$$\boldsymbol{{AQ}}^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\left(\boldsymbol{\mathrm{sin}}\:\boldsymbol{\theta}+\boldsymbol{\mathrm{cos}}\:\boldsymbol{\theta}\right)^{\mathrm{2}} −\mathrm{2}\left(\boldsymbol{\mathrm{sin}}\:\boldsymbol{\theta}+\boldsymbol{\mathrm{cos}}\:\boldsymbol{\theta}\right)\:\boldsymbol{\mathrm{cos}}\:\left(\frac{\boldsymbol{\pi}}{\mathrm{2}}+\boldsymbol{\theta}\right) \\ $$$$\:\:{AQ}^{\mathrm{2}} \:\:=\mathrm{2}+\mathrm{sin}\:\mathrm{2}\theta+\mathrm{2}\left(\mathrm{sin}\:\theta+\mathrm{cos}\:\theta\right)\:\mathrm{sin}\:\theta \\ $$$$\:\:\:\:\:{AQ}^{\mathrm{2}} \:\:=\mathrm{3}+\mathrm{2}\:\mathrm{sin}\:\mathrm{2}\theta−\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\:\:\:\:\:\:\:\:{let}\:\:\:\:\:{p}\:\:=\:\:\mathrm{2sin2}\theta\:−\:\mathrm{cos2}\theta \\ $$$$\:\:\:\:\:\:\:\:\frac{{dp}}{{d}\theta}\:\:=\:\:\mathrm{4cos2}\theta\:+\:\mathrm{2sin2}\theta\:\:=\:\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{tan2}\theta\:\:\:=\:\:−\mathrm{2} \\ $$$$\:\:\:\mathrm{sin2}\theta\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\:\:{and}\:\:\:\:\mathrm{cos2}\theta\:\:\:=\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\: \\ $$$$\:\:\:\:{AQ}^{\mathrm{2}} \:\:=\:\:\mathrm{3}+\sqrt{\mathrm{5}} \\ $$$$\:\:\:{AQ}\:\:\:=\:\:\sqrt{\mathrm{3}+\sqrt{\mathrm{5}}}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by infinityaction last updated on 26/Jun/22
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by Tawa11 last updated on 26/Jun/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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