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imperfectlyperfect
@imperfectlyperfect
June 2021
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SVP c'est urgent je ne comprends pas l'ex 4a !
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slyz007
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sin(2π/3+x)=sin(π-π/3+x)=-sin(-π/3+x)=sin(π/3-x)=sin(π/3)cosx-cos(π/3)sinx
sin(4π/3+x)=sin(π+π/3+x)=-sin(π/3+x)=-sin(π/3)cosx-cos(π/3)sinx
Donc
sinx+sin(2π/3+x)+sin(4π/3+x)=sinx+sinxcos(π/3)-cos(π/3)sinx-sinxcos(π/3)-cos(π/3)sinx
sinx+sin(2π/3+x)+sin(4π/3+x))sinx-1/2*sinx-1/2*sinx=sinx-sinx=0
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Verified answer
sin(2π/3+x)=sin(π-π/3+x)=-sin(-π/3+x)=sin(π/3-x)=sin(π/3)cosx-cos(π/3)sinxsin(4π/3+x)=sin(π+π/3+x)=-sin(π/3+x)=-sin(π/3)cosx-cos(π/3)sinx
Donc
sinx+sin(2π/3+x)+sin(4π/3+x)=sinx+sinxcos(π/3)-cos(π/3)sinx-sinxcos(π/3)-cos(π/3)sinx
sinx+sin(2π/3+x)+sin(4π/3+x))sinx-1/2*sinx-1/2*sinx=sinx-sinx=0