cos2x/(√2cos(x+π/4))=1/5,0
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cos2x/(√2cos(x+π/4))=1/5,0
![cos2x/(√2cos(x+π/4))=1/5,0](/uploads/image/z/15225940-28-0.jpg?t=cos2x%2F%28%E2%88%9A2cos%28x%2B%CF%80%2F4%29%29%3D1%2F5%2C0)
(cos²x-sin²x)/[√2(cosxcosπ/4-sinxsinπ/4)]=1/5
(cosx+sinx)(cosx-sinx)/(cosx-sinx)=1/5
cosx+sinx=1/5
cosx=1/5-sinx
平方
cos²x=1-sin²x=sin²x-2/5*sinx+1/25
由x范围
sinx>0
所以sinx=4/5,cosx=-3/5
tanx=sinx/cosx=-4/3
(cosx+sinx)(cosx-sinx)/(cosx-sinx)=1/5
cosx+sinx=1/5
cosx=1/5-sinx
平方
cos²x=1-sin²x=sin²x-2/5*sinx+1/25
由x范围
sinx>0
所以sinx=4/5,cosx=-3/5
tanx=sinx/cosx=-4/3
cos2x/根号2cos(x+π/4)=1/5,0
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