tg(pai/4+a)=(1+tga)/(1-tga)=3 所以tga=1/2 2sina*cosa-1-cos2a =2sina*cosa-(cosa*cosa+sina*sina)-(cosa*cosa-sina*sina) =2sina*cosa-2cosa*cosa =(2sina*cosa-2cosa*cosa)/1 =(2sina*cosa-2cosa*cosa)/(cosa*cosa+sina*sina) =(2tga-1)/(1+tga) =(2*1/2-1)/(1+1/2) =0
答：tan(π/4+a)=[tan(π/4)+tanα]/[1-tan(π/4)*tanα] =(1+tanα)/(1-tanα) =(sinα+cosα)/(co...详情>>
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