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已知tan(x/2)=2,求[cos^4x-sin^4x]/[√2cos(π/4+x)*sinx]的值
人气:259 ℃ 时间:2020-06-03 07:10:58
解答
tanx=2tan(x/2)/(1-tan^2x)=-4/3cos^4x-sin^4x=(cos^2x+sin^2x)(cos^2x-sin^2x)=cos^2x-sin^2xcos(π/4+x)=cosπ/4cosx-sinπ/4sinx=√2/2(cosx-sinx)原式=(cos^2x-sin^2x)/(cosx-sinx)sinx=(1-tan^2x)/(tanx-tan^2)...=(1-tan^2x)/(tanx-tan^2)这一步是不是有点问题??分子分母同除以cos^2x
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