cos2x-sin2x化简为一角一函数怎么化?
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/08/26 09:47:27
cos2x-sin2x化简为一角一函数怎么化?
cos2x-sin2x
=√2*(cos2x*√2/2 - sin2x*√2/2)
=√2*[cos2x*cos(π/4) - sin2x*sin(π/4)]
=√2*cos(2x+ π/4)
再问: 我做出来答案是-√2*sin(2x- π/4)对吗?
再答: 正确! cos2x-sin2x =√2*(cos2x*√2/2 - sin2x*√2/2) =√2*[cos2x*sin(π/4) - sin2x*cos(π/4)] =-√2*[sin2x*cos(π/4) - cos2x*sin(π/4)] =-√2*sin(2x- π/4)
再问: 如果tanA=-1,A=( )
再答: A=-π/4 + kπ,k属于Z
=√2*(cos2x*√2/2 - sin2x*√2/2)
=√2*[cos2x*cos(π/4) - sin2x*sin(π/4)]
=√2*cos(2x+ π/4)
再问: 我做出来答案是-√2*sin(2x- π/4)对吗?
再答: 正确! cos2x-sin2x =√2*(cos2x*√2/2 - sin2x*√2/2) =√2*[cos2x*sin(π/4) - sin2x*cos(π/4)] =-√2*[sin2x*cos(π/4) - cos2x*sin(π/4)] =-√2*sin(2x- π/4)
再问: 如果tanA=-1,A=( )
再答: A=-π/4 + kπ,k属于Z
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