化简sin(x+7π/4)+cos(x-3π/4)
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化简sin(x+7π/4)+cos(x-3π/4)
步骤我已经找到撒
sin(x+7π/4)+cos(x-3π/4)=sin(x+2π-π/4)+cos(x-π+π/4)我问下这一步是在干嘛?=sin(x-π/4)+cos[-(π-x-π/4)]=sin(x-π/4)+cos(π-x-π/4)=sin(x-π/4)-cos(x+π/4)=sin(x-π/4)-cos(x+π/2-π/4)=sin(x-π/4)-cos(π/2+x-π/4)=sin(x-π/4)+sin(x-π/4)=2sin(x-π/4)
sin(x+2π-π/4)+cos(x-π+π/4)我问下这一步是在干嘛?sin(x+2π-π/4为什么给以化成sin(x-π/4)
步骤我已经找到撒
sin(x+7π/4)+cos(x-3π/4)=sin(x+2π-π/4)+cos(x-π+π/4)我问下这一步是在干嘛?=sin(x-π/4)+cos[-(π-x-π/4)]=sin(x-π/4)+cos(π-x-π/4)=sin(x-π/4)-cos(x+π/4)=sin(x-π/4)-cos(x+π/2-π/4)=sin(x-π/4)-cos(π/2+x-π/4)=sin(x-π/4)+sin(x-π/4)=2sin(x-π/4)
sin(x+2π-π/4)+cos(x-π+π/4)我问下这一步是在干嘛?sin(x+2π-π/4为什么给以化成sin(x-π/4)
你好
Sin(x)的周期为2丌
即sin(x)=sin(x+2丌)
可以从正弦函数曲线上清楚的看见这种关系
再问: 我问的是..sin(x+2π-π/4为什么给以化成sin(x-π/4),为什么2π没了。。
再答: 假设y=x-π/4 .sin(x+2π-π/4)=sin(y+2π)=sin(y)=sin(x-π/4)
Sin(x)的周期为2丌
即sin(x)=sin(x+2丌)
可以从正弦函数曲线上清楚的看见这种关系
再问: 我问的是..sin(x+2π-π/4为什么给以化成sin(x-π/4),为什么2π没了。。
再答: 假设y=x-π/4 .sin(x+2π-π/4)=sin(y+2π)=sin(y)=sin(x-π/4)
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