已知tan(a+π/4)=2,则1+sinacosa-2(cosa)^2

问题描述:

已知tan(a+π/4)=2,则1+sinacosa-2(cosa)^2

tan(a+π/4)=2tan(a+π/4-π/4)= [tan(a+π/4)-tan(π/4)]/[1+tan(a+π/4)tan(π/4)]=(2-1)/3=1/3所以sina/cosa=1/3cosa=3sinacos²a+sin²a=110sin²a=1 sin²a=1/10,cos²a=9/10sina co...