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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)}
let1x1simplifyA=sin{arcsinx+2arcsin)(1x)}
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))...
wehaveA=sin(arcsinx)cos(2arcsin(1x))+cos(arcsinx)sin(2arcsin(1x))=x(12sin2(arcsin(1x)))+cos(arcsinx)sin(arcsin(1x)cos(arcsin(1x))=x(12(1x)2)+(1x)cos(arcsinx)cos(arcsin(1x))
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
butcos(arcsinx)=ξ1sin2(arcsinx)=ξ1x2andcos(arcsin(1x))=ξ1(1x)2soA=x{12(1x)2}+ξξ(1x)1x21(1x)2withξ2=1andξ2=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(sin1x)cos{2sin1(1x)}+cos(sin1x)×sin{2sin(1x)}x×(12sin2θ)+1x2×2sinθ.1sin2θsinθ=1x=x×{12(12x+x2)+1x2×2(1x)×1(12x+x2)=x×(12+4x2x2)+1x2×(22x)×11+2xx2=x×(4x2x21)+1x2×(22x)2x2x2plscheck

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