sin^6+COS^6===sin^6+cos^6====
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sin^6+COS^6===sin^6+cos^6====sin^6+COS^6===sin^6+cos^6====sin^6+COS^6===sin^6+cos^6====sin(x)^6+cos(
sin^6+COS^6===sin^6+cos^6====
sin^6+COS^6===
sin^6+cos^6====
sin^6+COS^6===sin^6+cos^6====
sin(x)^6+cos(x)^6 ＝[sin(x)^2+cos(x)^2]*[sin(x)^4-sin(x)^2*cos(x)^2+cos(x)^4] ＝sin(x)^4+2*sin(x)^2*cos(x)^2+cos(x)^4-3*sin(x)^2*cos(x)^2 ＝[sin(x)^2+cos(x)^2]^2-3*sin(x)^2*cos(x)^2 =1-3*sin(x)^2*cos(x)^2 =5/8+3/8*cos(4*x)
原式=(sin^2+cos^2)(sin^4-sin^2cos^2+cos^4)=sin^4-sin^2cos^2+cos^4=(sin^2+cos^2)^2-3sin^2cos^2 =1-3 X (1-cos2)/2 X (1+cos2)/2=1-3/4 X sin2^2=1-3/4 X (1-cos4)/2=5/8+3/8 cos4
6sin^2α-sinαcosα-cos^2α=0,求cosα+sinα的值
cos=-6/11,sin=?
cosπ/6=(),sinπ=()
证明一个sin&cos的等式证明 1+sin^2(x)+sin^4(x)+sin^6(x)=[1-sin^8(x)]/[cos^2(x)]
sin^6+COS^6===sin^6+cos^6====
cosα+cos^2α=1求sin^2a+sin^6a+sin^8a=?
已知cosθ+cosθ^2=1,则sinθ^2+sinθ^6+sinθ^8=
已知cosθ+cosθ=1,则sinθ+sin∧6θ+sin∧8θ=
若sinα+cosα=1,求证:(sinα)^6+(cosα)^6=1
已知sinα+cosα=1 证明(sinα)^6+(cosα)^6=1
（1-sin^6a-cos^6a)/(1-sin^4a-cos^4a)=?
[1-（sinα）^4-(cosα)^4]/[1-(sinα)^6-(cos)^6)]=?
化简：sin^6α+cos^6α+3sin^2α*cos^2α=?
cos(6+a)cos(a-54)+sin(6+a)sin(a-54)=
sin^6 a＋cos^6 a＋3×sin^2 a×cos^2 a=?
当sinˆa+sin^2a=1求cos^+cos^6的值
设角a=-35π/6,则2sin(π+α)cos(π-α)-cos(π+α)/ 1+sin^2α+sin(π-α)-cos^2(π+α)sina=sin(-35π/6)=sin(-6π+π/6)=1/2 cosα=根号3/22sin(π＋α)cos(π-α)-cos(π+α)/ [1+sin^2α+sin(π-α)-cos^2(π+α)]=2（-sinα)(-cosα）+cosα/[1+sin²
=1-（sin^6a+cos^6a） ------（1）=1-（sin^a+cos^a)(sin^4 a-sin^acos6a+cos^4 a) --------（2）由（1）怎么能到（2）呢=1-（sin^6a+cos^6a）=1-（sin^2a+cos^2a)(sin^4 a-sin^2acos^2a+cos^4a)