=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)
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=1-(sin^6a+cos^6a)------(1)=1-(sin^a+cos^a)(sin^4a-sin^acos6a+cos^4a)--------(2)由(1)怎么能到(2)呢=1-(sin^
=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)
=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)
=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)
1-(sin^2a+cos^2a)(sin^4 a-sin^2acos^2a+cos^4a)
=1-(sin^4 a+2sin^2acos^2a+cos^4a-3sin^2acos^2a)
=3sin^2acos^2a
=3sin^2(2a)/4
解释为:
收起
=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)
(1-sin^6a-cos^6a)/(1-sin^4a-cos^4a)=?
化简(1-sin^6 a-cos^6 a)/(cos^2 a-cos^4 a)==
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Sina+sin^2a=1,cos^2a+cos^s4a+cos^6a
当sinˆa+sin^2a=1求cos^+cos^6的值
cosα+cos^2α=1求sin^2a+sin^6a+sin^8a=?
(1-sin(a)^6-cos(a)^6)/(1-sin(a)^4-cos(a)^4) (1-sin(a)^6-cos(a)^6)/(1-sin(a)^4-cos(a)^4)
化简1-sin^4a-cos^4a/1-sin^6a-cos^6a
1-sin^6a-cos^6a分之1-sin^4a-cos^4a 化简
(1-sin^4a-cos^4a)/(1-sin^6a-cos^6a)
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=?
证明sin(4A)sin(2A)(1-cos(2A)) cos(4A)cos(2A)(1 cos(2A))=cos(2A)(1 cos(6A))应该是sin4Asin2A(1-cos2A) + cos4Acos2A(1+cos2A)=cos2A(1+cos6A)
求证1+sin a+cos a+2sin acos a/1+sin a+cos a=sin a+cos a
sinA+sin^2A=1,cos^2A+cos^6A=?
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
已知sina+sin^2a=1,则cos^2a+cos^6a+cos^8a=