y=sin4xcos4x ÇóT

sin4xcos4x=1/2 * sin8x ËùÒÔT=2pi/8=pi/4


ÒÑÖªy=sin4x+cos4x,Çóy(n)

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f(x)=(sinx)^4+(cosx)^4f(x)=[(sinx)^2+(cosx)^2]^2-2(sinx)^2(cosx)^2f(x)=1-(sin2x)/2-1


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sin4x=2 sin2x cos2x=2 sinx cosx(1-(sinx)^2)cos4x=[1-(sin2x)^2]/2=[1-4(sinx)^2 (cosx)^2]/2±¶½Ç¹«Ê½Ö»Òª¼Çס¶þ±¶½Ç¾ÍºÃ,¸ß±¶µÄÓöþ±¶½Ç¹«Ê½½µ½×...


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sin4x+cos4x

sin4x+cos4x =cos4x+sin4x =¡Ì2(cos4x¡Ì2/2+sin4x¡Ì2/2)=¡Ì2(cos4xcos¦Ð/4+sin4xsin¦Ð/4)=¡Ì2cos(4x-¦Ð/4)»¹¿ÉÒÔת»¯ ¡Ì2sin(4x+¦Ð/4)=¡Ì2sin(¦Ð...


ÒÑÖªº¯Êýf(x)=sin4xcos4x,Ôòf(x)µÄÖÜÆÚÊÇ

f(x)=sin4xcos4x =1/2*sin8x T=2¦Ð/8=¦Ð/4


ÇóÖ¤sin^4x+cos^4x=1 - 2sin^2xcos^2x

ÇóÖ¤sin^4x+cos^4x=1-2sin^2xcos^2xÖ¤:sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x=1-2sin^2xcos^2x


½«f(x)=sinËĴη½+cosËĴη½»¯³É±ê×¼Èý½Çº¯Êý

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