sin^6a+cos^6a-2sin^4a-cos^4a+sin^2a=0我想知道,这道题为什么等于零,
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/30 20:22:16
x)+̋3KN/RF I.k
%tO^i6QKD9jjkL5vRqF`-`!!:j@(t!FU@
l`@]FB!a= VֆO.6m%jBUhDV"m`3УHA [5
sin^6a+cos^6a-2sin^4a-cos^4a+sin^2a=0我想知道,这道题为什么等于零,
sin^6a+cos^6a-2sin^4a-cos^4a+sin^2a=0
我想知道,这道题为什么等于零,
sin^6a+cos^6a-2sin^4a-cos^4a+sin^2a=0我想知道,这道题为什么等于零,
原式左边=cos^4A*(cos^2A-1)+(sin^3A-sinA)^2
=-sin^2Acos^4A+sin^2A(sin^2A-1)^2
=-sin^2Acos^4A+sin^2A(-cos^2A)^2
=-sin^2Acos^4A+sin^2Acos^4A
=0
左边=(sin^2a+cos^2a)(sin^4a-sin^2acos^2a+cos^4a)-2sin^4a-cos^4a+sin^2a
=sin^4a-sin^2acos^2a+cos^4a-2sin^4a-cos^4a+sin^2a
=-sin^4a-sin^2acos^2a+sin^2a
=sin^2a(1-sin^2a)-sin^2acos^2a
=sin^2acos^2a-sin^2acos^2a
=0=右边
(1-sin^6a-cos^6a)/(sin^2a-sin^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)
由cos(a+b)=cos a cos b-sin a sin b cos(a-b)=cos a cos b+sin a sin b解题设a为锐角,证:1、2分之根3乘cos a + 2分之1乘sin a=cos(6分之π-a)2、cos a-sin a=根号2cos(4分之π+a)
已知(4sin a-2cos a)/(5cos a +3sin a)=6/11,求cos^4 a-sin^4 a
化简(1-sin^6 a-cos^6 a)/(cos^2 a-cos^4 a)==
(1-sin^6 a-cos^6 a)/(sin^2 a-sin^4 a)的化简结果,
sin^6 a+cos^6 a+3sin^2 acos^2a=(sin^2 a+cos^2 a)(sin^4 a+cos^4 a-sin^2 acos^2 a)是如何计算出的
2(sin^6 α +cos^6α)-3(sin^4a+cos^4α)mingbaidian
2(sin^6 α +cos^6α)-3(sin^4a+cos^4α)
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^6a-cos^6a)/(1-sin^4a-cos^4a)=?
(1-sin^4a-cos^4a)/(1-sin^6a-cos^6a)
sin^6 a+cos^6 a+3×sin^2 a×cos^2 a=?
sin^4+sin^2cos^2+cos^2a=
sin(-19π/6)=?已知[3Sin(π+a)+Cos(π-a)]/4Sin(-a)-Sin(5π/2+a)=2,求tan a
化简:(sin^6a+cos^6a-1) / (sin^4ª+cos^4a-1)
sin^6a+cos^6a-2sin^4a-cos^4a+sin^2a=0我想知道,这道题为什么等于零,