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Question Number 140477 by ajfour last updated on 08/May/21

If sin θ+sin^2 θ+sin^3 θ+.... = cos θ and 0<θ<(π/2) then find θ.

$${If}\:\:\mathrm{sin}\:\theta+\mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{sin}\:^{\mathrm{3}} \theta+.... \\ $$$$\:\:\:\:\:\:=\:\mathrm{cos}\:\theta\:\:\:\:\:{and}\:\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}}\:{then} \\ $$$${find}\:\theta. \\ $$

Answered by benjo_mathlover last updated on 08/May/21

⇒((sin θ)/(1−sin θ)) = cos θ let sin θ = x ⇒cos θ =(√(1−x^2 )) ⇒ (x/(1−x)) = (√(1−x^2 )) ⇒ x ≈ 0.468989 ⇒ θ = sin^(−1) (0.468989) sin^(−1) (0.468989)= 0.488146 rad

$$\:\Rightarrow\frac{\mathrm{sin}\:\theta}{\mathrm{1}−\mathrm{sin}\:\theta}\:=\:\mathrm{cos}\:\theta \\ $$$$\mathrm{let}\:\mathrm{sin}\:\theta\:=\:\mathrm{x}\:\Rightarrow\mathrm{cos}\:\theta\:=\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\frac{\mathrm{x}}{\mathrm{1}−\mathrm{x}}\:=\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\mathrm{x}\:\approx\:\mathrm{0}.\mathrm{468989} \\ $$$$\Rightarrow\:\theta\:=\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{0}.\mathrm{468989}\right) \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{0}.\mathrm{468989}\right)=\:\mathrm{0}.\mathrm{488146}\:\mathrm{rad} \\ $$$$ \\ $$

Commented by ajfour last updated on 08/May/21

Thanks sir, i also tried.

Answered by ajfour last updated on 08/May/21

((sin θ)/(1−sin θ))=cos θ p^2 +q^2 =1 (p/(1−p))=q pq=t t^2 +q^4 =q^2 t=q^2 −qt q^2 =qt+t (q^2 t^2 +t^2 +2qt^2 )=qt+t−t^2 qt^3 +t^3 +t^2 +2qt^2 =qt+t−t^2 ⇒ q(t^3 +2t^2 −t)+(t^3 +2t^2 −t)=0 let t^2 +2t−1=0 ⇒ (t+1)^2 =2 t=−1+(√2) ⇒ sin θcos θ=(√2)−1 sin 2θ=2(√2)−2 θ=(1/2)sin^(−1) (2(√2)−2) θ ≈ 0.488146802 rad

$$\:\frac{\mathrm{sin}\:\theta}{\mathrm{1}−\mathrm{sin}\:\theta}=\mathrm{cos}\:\theta \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\mathrm{1} \\ $$$$\frac{{p}}{\mathrm{1}−{p}}={q} \\ $$$${pq}={t} \\ $$$${t}^{\mathrm{2}} +{q}^{\mathrm{4}} ={q}^{\mathrm{2}} \\ $$$${t}={q}^{\mathrm{2}} −{qt} \\ $$$${q}^{\mathrm{2}} ={qt}+{t} \\ $$$$\left({q}^{\mathrm{2}} {t}^{\mathrm{2}} +{t}^{\mathrm{2}} +\mathrm{2}{qt}^{\mathrm{2}} \right)={qt}+{t}−{t}^{\mathrm{2}} \\ $$$${qt}^{\mathrm{3}} +{t}^{\mathrm{3}} +{t}^{\mathrm{2}} +\mathrm{2}{qt}^{\mathrm{2}} ={qt}+{t}−{t}^{\mathrm{2}} \\ $$$$\Rightarrow\:{q}\left({t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} −{t}\right)+\left({t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} −{t}\right)=\mathrm{0} \\ $$$${let}\:\:\:\:{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\left({t}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2} \\ $$$$\:\:\:{t}=−\mathrm{1}+\sqrt{\mathrm{2}}\:\:\: \\ $$$$\:\:\Rightarrow\:\:\mathrm{sin}\:\theta\mathrm{cos}\:\theta=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\mathrm{sin}\:\mathrm{2}\theta=\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2} \\ $$$$\:\:\:\theta=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}\right) \\ $$$$\:\:\:\theta\:\approx\:\mathrm{0}.\mathrm{488146802}\:\:{rad} \\ $$

Answered by mr W last updated on 08/May/21

sin θ+sin^2 θ+sin^3 θ+...=cos θ sin θ+sin^2 θ+sin^3 θ+... =sin θ(1+sin θ+sin^2 θ+sin^3 θ+...) =sin θ×(1+cos θ) sin θ+sin^2 θ+sin^3 θ+... =sin θ×(1/(1−sin θ)) ⇒1+cos θ=(1/(1−sin θ)) ⇒cos θ−sin θ=sin θ sin θ ⇒1−2sin θcos θ=(sin θ sin θ)^2 ⇒(sin θcos θ)^2 +2sin θcos θ−1=0 ⇒sin θ cos θ=(√2)−1 ⇒2sin θ cos θ=2((√2)−1) ⇒sin 2θ=2((√2)−1) ⇒θ=((sin^(−1) 2((√2)−1))/2)

$$\mathrm{sin}\:\theta+\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{sin}^{\mathrm{3}} \:\theta+...=\mathrm{cos}\:\theta \\ $$$$ \\ $$$$\mathrm{sin}\:\theta+\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{sin}^{\mathrm{3}} \:\theta+... \\ $$$$=\mathrm{sin}\:\theta\left(\mathrm{1}+\mathrm{sin}\:\theta+\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{sin}^{\mathrm{3}} \:\theta+...\right) \\ $$$$=\mathrm{sin}\:\theta×\left(\mathrm{1}+\mathrm{cos}\:\theta\right) \\ $$$$ \\ $$$$\mathrm{sin}\:\theta+\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{sin}^{\mathrm{3}} \:\theta+... \\ $$$$=\mathrm{sin}\:\theta×\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}\:\theta} \\ $$$$ \\ $$$$\Rightarrow\mathrm{1}+\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{sin}\:\theta} \\ $$$$\Rightarrow\mathrm{cos}\:\theta−\mathrm{sin}\:\theta=\mathrm{sin}\:\theta\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\mathrm{1}−\mathrm{2sin}\:\theta\mathrm{cos}\:\theta=\left(\mathrm{sin}\:\theta\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left(\mathrm{sin}\:\theta\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\mathrm{2sin}\:\theta\mathrm{cos}\:\theta−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\Rightarrow\mathrm{2sin}\:\theta\:\mathrm{cos}\:\theta=\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{2}\theta=\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$\Rightarrow\theta=\frac{\mathrm{sin}^{−\mathrm{1}} \mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2}} \\ $$

Commented by ajfour last updated on 08/May/21

nice sir, you got it straight!

$${nice}\:{sir},\:{you}\:{got}\:{it}\:{straight}! \\ $$

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