四道三角函数的题,1.用cosα表示(sinα)^4-(sinα)^2+(cosα)^22.化简:(根号下(1-2sin
6个回答

(一)

(sinα)^4-(sinα)^2+(cosα)^2

=(1-cosα^2)^2-(1-cosα^2)+(cosα)^2

=1-2cosα^2+cosα^4-1+cosα^2+cosα^2

=cosα^4

(二)化简分子

=√[sin^2(10°)+cos^2(10°)-2sin10°cos10°]

=√(sin10°-cos10°)^2=sin10°-cos10°;

化简分母

=cos10°-√[1-cos^2(180°-10°)]

=cos10°-√[1-cos^2(10°)]

=cos10°-sin10°

原式=(sin10°-cos10°)/(cos10°-sin10°)=-1

(三)第一项化简

=cosα√{[(1-sinα)(1-sinα)]/[(1+sinα)(1-sinα)]}

=cosα√[(1-sinα)^2]/[1-sin^2(α)]

=cosα√[(1-sinα)^2]/cos^2(α)

=cos(1-sinα)/(-cosα) =sinα-1

因为α在第二象限,所以cosα是负数.开方出来要加负号.

第二项化简

=sinα√{[(1-cosα)(1-cosα)]/[(1+cosα)(1-cosα)]}

=sinα√[(1-cosα)^2]/[1-cos^2(α)]

=sinα√[(1-cosα)^2]/sin^2(α)

=sinα(1-cosα)/sinα =1-cosα

所以原式=sinα-1+1-cosα=sinα-cosα

(四)化简分子

=(sinx)^2+(cosx)^2-(sinx)^4(sinx)^2-(cosx)^4(cosx)^2

=(sinx)^2[1-(sinx)^4]+(cosx)^2[1-(cosx)^4]

=(sinx)^2{[1+(sinx)^2][1-(sinx)^2]}+(cosx)^2{[(1+(cosx)^2)][(1-(cosx)^2]}

=(sinx)^2[1+(sinx)^2](cosx)^2+(cosx)^2{[(1+(cosx)^2)](sinx)^2

=(sinx)^2(cosx)^2{1+(sin)^2+1+(cosx)^2)}

=3(sinx)^2(cosx)^2

化简分母

=(sinx)^2+(cosx)^2-(sinx)^2(sinx)^2-(cosx)^2(cosx)^2

=(sinx)^2[1-(sinx)^2]+(cosx)^2[1-(cosx)^2]

=(sinx)^2(cosx)^2+(cosx)^2(sinx)^2

=2(sinx)^2(cosx)^2

原式=[3(sinx)^2(cosx)^2]/[2(sinx)^2(cosx)^2]=3/2