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Question Number 35416 by abdo mathsup 649 cc last updated on 18/May/18
let −1≤x≤1 simplify A=sin{ arcsinx +2arcsin)(1−x)}
$${let}\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1}\:{simplify} \\ $$$$\left.{A}={sin}\left\{\:{arcsinx}\:\:+\mathrm{2}{arcsin}\right)\left(\mathrm{1}−{x}\right)\right\} \\ $$
Commented by abdo mathsup 649 cc last updated on 20/May/18
we have A = sin(arcsinx)cos(2arcsin(1−x)) +cos(arcsinx) sin(2arcsin(1−x)) = x ( 1−2sin^2 (arcsin(1−x))) +cos(arcsinx) sin(arcsin(1−x)cos(arcsin(1−x)) =x( 1−2 (1−x)^2 ) +(1−x)cos(arcsinx)cos(arcsin(1−x))...
$${we}\:{have}\: \\ $$$${A}\:=\:{sin}\left({arcsinx}\right){cos}\left(\mathrm{2}{arcsin}\left(\mathrm{1}−{x}\right)\right) \\ $$$$+{cos}\left({arcsinx}\right)\:{sin}\left(\mathrm{2}{arcsin}\left(\mathrm{1}−{x}\right)\right) \\ $$$$=\:{x}\:\left(\:\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left({arcsin}\left(\mathrm{1}−{x}\right)\right)\right)\:+{cos}\left({arcsinx}\right)\:{sin}\left({arcsin}\left(\mathrm{1}−{x}\right){cos}\left({arcsin}\left(\mathrm{1}−{x}\right)\right)\right. \\ $$$$={x}\left(\:\mathrm{1}−\mathrm{2}\:\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \right)\:+\left(\mathrm{1}−{x}\right){cos}\left({arcsinx}\right){cos}\left({arcsin}\left(\mathrm{1}−{x}\right)\right)… \\ $$
Commented by abdo mathsup 649 cc last updated on 20/May/18
but cos(arcsinx)=ξ(√(1−sin^2 (arcsinx))) =ξ(√(1−x^2 )) and cos(arcsin(1−x))=ξ^′ (√(1−(1−x)^2 )) so A=x{ 1−2(1−x)^2 }+ ξξ^′ (1−x)(√(1−x^2 )) (√(1−(1−x)^2 )) with ξ^2 =1 and ξ^′^2 =1
$${but}\:{cos}\left({arcsinx}\right)=\xi\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \left({arcsinx}\right)} \\ $$$$=\xi\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}\:\:\:{and}\:\:{cos}\left({arcsin}\left(\mathrm{1}−{x}\right)\right)=\xi^{'} \sqrt{\mathrm{1}−\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$${so}\: \\ $$$${A}={x}\left\{\:\mathrm{1}−\mathrm{2}\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \right\}+\:\xi\xi^{'} \left(\mathrm{1}−{x}\right)\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\sqrt{\mathrm{1}−\left(\mathrm{1}−{x}\right)^{\mathrm{2}} } \\ $$$${with}\:\xi^{\mathrm{2}} \:=\mathrm{1}\:{and}\:\xi^{'^{\mathrm{2}} } =\mathrm{1} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 18/May/18
sin(sin^(−1) x)cos{2sin^(−1) (1−x)}+cos(sin^(−1) x) ×sin{2sin^− (1−x)} x×(1−2sin^2 θ)+(√(1−x^2 )) ×2sinθ.(√(1−sin^2 θ)) sinθ=1−x =x×{1−2(1−2x+x^2 )+(√(1−x^2 )) ×2(1−x)× (√(1−(1−2x+x^2 ))) =x×(1−2+4x−2x^2 )+(√(1−x^2 )) ×(2−2x)× (√(1−1+2x−x^2 )) =x×(4x−2x^2 −1)+(√(1−x^2 )) ×(2−2x)(√(2x−2x^2 )) pls check
$${sin}\left({sin}^{−\mathrm{1}} {x}\right){cos}\left\{\mathrm{2}{sin}^{−\mathrm{1}} \left(\mathrm{1}−{x}\right)\right\}+{cos}\left({sin}^{−\mathrm{1}} {x}\right) \\ $$$$×{sin}\left\{\mathrm{2}{sin}^{−} \left(\mathrm{1}−{x}\right)\right\} \\ $$$${x}×\left(\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \theta\right)+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:\:}\:×\mathrm{2}{sin}\theta.\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \theta}\: \\ $$$${sin}\theta=\mathrm{1}−{x} \\ $$$$={x}×\left\{\mathrm{1}−\mathrm{2}\left(\mathrm{1}−\mathrm{2}{x}+{x}^{\mathrm{2}} \right)+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}\:×\mathrm{2}\left(\mathrm{1}−{x}\right)×\right. \\ $$$$\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{2}{x}+{x}^{\mathrm{2}} \right)} \\ $$$$={x}×\left(\mathrm{1}−\mathrm{2}+\mathrm{4}{x}−\mathrm{2}{x}^{\mathrm{2}} \right)+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:×\left(\mathrm{2}−\mathrm{2}{x}\right)× \\ $$$$\sqrt{\mathrm{1}−\mathrm{1}+\mathrm{2}{x}−{x}^{\mathrm{2}} \:} \\ $$$$={x}×\left(\mathrm{4}{x}−\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} \:}\:×\left(\mathrm{2}−\mathrm{2}{x}\right)\sqrt{\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{2}} } \\ $$$${pls}\:{check} \\ $$

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