高一数学题求教,要过程.sin^(π/11)+sin^(9π/22)=

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高一数学题求教,要过程.sin^(π/11)+sin^(9π/22)=
高一数学题求教,要过程.sin^(π/11)+sin^(9π/22)=

高一数学题求教,要过程.sin^(π/11)+sin^(9π/22)=
sin^(π/11)+sin^(9π/22)
=sin^(π/11)+sin^(11π/22-2π/22)
=sin^(π/11)+sin^(π/2-π/11)
=sin^(π/11)+cos^(π/11)
=1
完毕,请批评指正.

原式=sin(π/11)+sin(11π/22-2π/22)
=sin(π/11)+cos(π/11)=根号2sin(π/11+π/4)
最后一步,是叫做辅助角公式,根号下A^2+B^2乘上sin(α+任意角)
对于这题,都提出个根号2来

sin函数为周期奇函数,π/11=π/2-9π/22,即原式可化为sin(π/11)+sin(π/2-π/11),后面的用和差化积公式,求得sin(π/11)+sin(π/2)cos(π/11)-cos(π/2)sin(π/11)=sin(π/11)+cos(π/11)