1,计算简化（sinβ）的四次+（cosβ）的四次 2,sin10°×cos20°×cos40°的解题过程

（sinβ)^4+（cosβ）^4=[（sinβ）^2+（cosβ)^2]^2-2(sinβ）^2*（cosβ)^2
=1-1/2 (2sinβcosβ)^2
=1-1/2 (sin2β)^2
sin10°×cos20°×cos40°
=sin10°×cos10°×cos20°×cos40°/cos10°
=1/2 sin20°×cos20°×cos40°/cos10°
=1/4sin40°×cos40°/cos10°
=1/8 sin80°/cos10°
=1/8 cos10°/cos10°
=1/8

1.=（sin^2+cos^2)^2-2sin2*cos2=1-1/2*sin^2 2β=
2.=8*cos10* */8cos10=sin80/8cos10=1/8

一,sin^4a+cos^4a
=(sin²a+cos²a)²-2sin²acos²a
=1-(1/2)(2sinacosa)²
=1-(1/2)sin²2a
二：sin10°*cos20°*cos40°
=cos10°sin10°*cos20°*cos40°/cos10°
=(1/2)sin20°*cos20°*cos40°/cos10°
=(1/4)sin40°*cos40°/cos10°
=(1/8)sin80°/cos10°
=(1/8)cos10°/cos10°
=1/8

1.原式=[(sinb)^2+(cosb)^2]^2-2(sinb)^2(cosb)^2
=1-[(sin2b)^2]/2
2.原式=[sin20*sin140*sin100/(8cos10*cos50*cos70)
=1/8