cos(2x-π)/[根号2cos(x+π/4)]=-1/5,0
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cos(2x-π)/[根号2cos(x+π/4)]=-1/5,0cos(2x-π)/[根号2cos(x+π/4)]=-1/5,0cos(2x-π)/[根号2cos(x+π/4)]=-1/5,0-cos
cos(2x-π)/[根号2cos(x+π/4)]=-1/5,0
cos(2x-π)/[根号2cos(x+π/4)]=-1/5,0
cos(2x-π)/[根号2cos(x+π/4)]=-1/5,0
-cos2x/根号2cos(x+π/4)=-1/5
cosx+sinx=1/5
结合cos^2(x)+sin^2(x)=1
解得：cosx=-3/5,sinx=4/5
根号2cos(x-π/4)化简
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cos(2x-π)/[根号2cos(x+π/4)]=-1/5,0
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