sin^8X+cos^8X+4sin^2Xcos^2x-2sin^4xcos^4X(化简）

来源：学生作业帮助网编辑：作业帮时间：2024/06/17 05:03:07

x��)�+�̋��N�/Q& �Q�gT�k�T�x&O{�=_��~O�MR�>Y��lȷ�i��{�m!�QUb5��J]�� iqF�FCUAT�T��B��ucь]!�1F�2��V��M9��HC�m��n�<_;��';vB�/��?_��~�,X�b \T1L�*��#`� �S��0�F�dH6�b�Q8�`F��a��/�W<�8�鞆�8P�� Fi�-�j�%�ҌB�F�Q�pm��k~x6eۋ�ۑd%��

sin^8X+cos^8X+4sin^2Xcos^2x-2sin^4xcos^4X(化简）
sin^8X+cos^8X+4sin^2Xcos^2x-2sin^4xcos^4X(化简）

sin^8X+cos^8X+4sin^2Xcos^2x-2sin^4xcos^4X(化简）
原式=sin^8X-2sin^4xcos^4X+cos^8X+4sin^2Xcos^2x
=（sin^4x-cos^4x）^2+4sin^2Xcos^2x
=[（sin^2x-cos^2x）（sin^2x+cos^2x）]^2+4sin^2Xcos^2x
=（sin^2x-cos^2x）^2+4sin^2Xcos^2x
=sin^4x-2sin^2Xcos^2x+cos^4x+4sin^2Xcos^2x
=sin^4x+2sin^2Xcos^2x+cos^4x
=（sin^2x+cos^2x）^2
=1

答案为1
化简过程：=sin^8X-sin^4xcos^4X+cos^8X-sin^4xcos^4X+4sin^2Xcos^2x
=sin^4X（sin^4X-cos^4X）+cos^4X（cos^4X-sin^4x）+4sin^2Xcos^2x
=sin^4X（sin^2X-cos^2X））+cos^4X...

全部展开

答案为1
化简过程：=sin^8X-sin^4xcos^4X+cos^8X-sin^4xcos^4X+4sin^2Xcos^2x
=sin^4X（sin^4X-cos^4X）+cos^4X（cos^4X-sin^4x）+4sin^2Xcos^2x
=sin^4X（sin^2X-cos^2X））+cos^4X（cos^2X-sin^2x）+4sin^2Xcos^2x
=（sin^4X-cos^4X）*（sin^2x-cos^2X）+4sin^2Xcos^2x
=（sin^2x-cos^2X）*（sin^2x-cos^2X）+4sin^2Xcos^2x
=（sin^2x=cos^2X）^2
=1

收起

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^8X+cos^8X+4sin^2Xcos^2x-2sin^4xcos^4X(化简） d/dx sin^8(cos(4x)) 化简sin^4x-sin^2x+cos^2x 化简:sin^4 x-sin^2 x+cos^2 x 三角化简sin(x-5兀)cos[(-兀/2)-x]cos(8兀-x)/sin[x-(3兀/2)]sin(-x-4兀) 化简sin^4x+cos^2x 化简(sin^2x-cos^4x+cos^2x-sin^4x)/(sin^2x-cos^6x+cos^2x-sin^6x) 化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x 求证 cos^8(x)-sin^8(x)=cos2x【1-1/2sin^2（2x)】 (1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x 化简[1-(sin^4 x-sin^2 xcos^2 x+cos^4 x)]/(sin^2 x)+3sin^2 x 求证（cos^2 x-sin^2 x)(cos^4 x+sin^4 x)+1/4 sin 2x sin 4x=cos 2x 证明sin^4x-cos^4x=sin^2x-cos^2x 求证sin^4x-cos^4x=sin^2x-cos^2x 证明sin^2(x)+cos^2(x+30)+sin(x)cos(x+30)=3/4 用cos x表示sin^4 x -sin^2 x +cos^2 x 用cos x表示sin^4x-sin^2 x+cos^2 x