已知tan(4/π-α)=3.求sin2α-sin²α+cos2α的值

Question

已知tan(4/π-α)=3.求sin2α-sin²α+cos2α的值

Answer 1

tan(4/π-α)=3
-tana=tan(-a)=tan[(4/π-α)-4/π]
=[tan(4/π-α)-tan(4/π)]/[1+tan(4/π-α) * tan(4/π)]
=(3-1)/(1+3)
=1/2;
tana=-1/2
sin2a=(2tana)/[1+(tana)^2]=-4/5
cos2a=[1-(tana)^2]/[1+(tana)^2]=3/5
sin2α-sin²α+cos2α=sin2α-(1-cos2)/2+cos2α=sin2α-1/2+3/2*cos2α
=-4/5-1/2+3/2*3/5=-2/5