赞 mflsuccess 幼苗
共回答了12个问题采纳率：75% 举报
(1-cos^6x-sin^6x)/(1-cos^4x-sin^4x)
=[1-(cos^6x+sin^6x)]/[1-(cos^4x-sin^4x)]
=[1-(cos^2x+sin^2x)(cos^4x-cos^2xsin^2x+sin^4x)]/[1-(cos^2x-sin^2x)(cos^2x+sin^2x)]
=[1-(cos^4x-cos^2xsin^2x+sin^4x)]/[1-(cos^2x-sin^2x)]
=[1-(cos^4x+2cos^2xsin^2x+sin^4x-3cos^2xsin^2x)]/[1-cos^2x+sin^2x]
={1-[(cos^2x+sin^2x)^2-3cos^2xsin^2x]}/[sin^2x+sin^2x]
={1-[1-3cos^2xsin^2x]}/(2sin^2x)
=(1-1+3cos^2xsin^2x)/(2sin^2x)
=3cos^2xsin^2x/(2sin^2x)
=3cos^2x/2
=[1-(cos^6x+sin^6x)]/[1-(cos^4x-sin^4x)]
=[1-(cos^2x+sin^2x)(cos^4x-cos^2xsin^2x+sin^4x)]/[1-(cos^2x-sin^2x)(cos^2x+sin^2x)]
=[1-(cos^4x-cos^2xsin^2x+sin^4x)]/[1-(cos^2x-sin^2x)]
=[1-(cos^4x+2cos^2xsin^2x+sin^4x-3cos^2xsin^2x)]/[1-cos^2x+sin^2x]
={1-[(cos^2x+sin^2x)^2-3cos^2xsin^2x]}/[sin^2x+sin^2x]
={1-[1-3cos^2xsin^2x]}/(2sin^2x)
=(1-1+3cos^2xsin^2x)/(2sin^2x)
=3cos^2xsin^2x/(2sin^2x)
=3cos^2x/2
1年前 追问
5
用平方差和立方差
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你能帮帮他们吗