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Question Number 90629 by Josephbaraka@gmail.com last updated on 25/Apr/20

If sin^(−1) (1−x)−2 sin^(−1) x = (π/2), then x=

$$\mathrm{If}\:\:\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{1}−{x}\right)−\mathrm{2}\:\mathrm{sin}^{−\mathrm{1}} {x}\:=\:\frac{\pi}{\mathrm{2}},\:\mathrm{then}\:{x}= \\ $$

Commented by jagoll last updated on 25/Apr/20

let sin^(−1) (1−x) = y 1−x = sin y ⇒x = 1−sin y 2sin^(−1) (x) = p ⇒x = sin ((p/2)) ⇒cos (y−p) = 0 cos y cos p + sin y sin p = 0 (√(2x−x^2 )) (1−2x^2 ) = (x−1)(2x(√(1−x^2 )))

$${let}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{1}−{x}\right)\:=\:{y} \\ $$$$\mathrm{1}−{x}\:=\:\mathrm{sin}\:{y}\:\Rightarrow{x}\:=\:\mathrm{1}−\mathrm{sin}\:{y} \\ $$$$\mathrm{2sin}^{−\mathrm{1}} \left({x}\right)\:=\:{p}\:\Rightarrow{x}\:=\:\mathrm{sin}\:\left(\frac{{p}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{cos}\:\left({y}−{p}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{cos}\:{y}\:\mathrm{cos}\:{p}\:+\:\mathrm{sin}\:{y}\:\mathrm{sin}\:{p}\:=\:\mathrm{0} \\ $$$$\sqrt{\mathrm{2}{x}−{x}^{\mathrm{2}} }\:\left(\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} \right)\:=\:\left({x}−\mathrm{1}\right)\left(\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right) \\ $$$$ \\ $$

Answered by TANMAY PANACEA. last updated on 25/Apr/20

sin^(−1) (1−x)=(π/2)+2sin^(−1) x sin(sin^(−1) (1−x))=sin((π/2)+2sin^(−1) x) 1−x=cos(2sin^(−1) x) sinθ=x so cos2θ=2sin^2 θ−1=2x^2 −1 1−x=2x^2 −1 2x^2 +x−2=0 x=((−1±(√(1+16)))/4)=((−1±(√(17)))/4)

$${sin}^{−\mathrm{1}} \left(\mathrm{1}−{x}\right)=\frac{\pi}{\mathrm{2}}+\mathrm{2}{sin}^{−\mathrm{1}} {x} \\ $$$${sin}\left({sin}^{−\mathrm{1}} \left(\mathrm{1}−{x}\right)\right)={sin}\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}{sin}^{−\mathrm{1}} {x}\right) \\ $$$$\mathrm{1}−{x}={cos}\left(\mathrm{2}{sin}^{−\mathrm{1}} {x}\right) \\ $$$${sin}\theta={x}\:\:\:{so}\:\:\:\:{cos}\mathrm{2}\theta=\mathrm{2}{sin}^{\mathrm{2}} \theta−\mathrm{1}=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{1}−{x}=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +{x}−\mathrm{2}=\mathrm{0} \\ $$$${x}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{16}}}{\mathrm{4}}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{17}}}{\mathrm{4}} \\ $$

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