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
答
tan（a+π/4）=2
(tana+1)/(1-tana*1)=2
tana+1=2-2tana
tana=1/3
1+sinacosa-2(cosa)^2
=1/2*2sinacosa-[2(cosa)^2-1]
=1/2sin2a-cos2a
=1/2*2tana/(1+(tana)^2)-(1-(tana)^2)/(1+(tana)^2)
=(1/3)/(1+1/9)-(1-1/9)/(1+1/9)
=-1/2
答
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...