tanα=3,求sin2α-cos2α的值

tanα=3,求sin2α-cos2α的值
tanα=3,求sin2α-cos2α的值

tanα=3,求sin2α-cos2α的值
方法一：
sin2α-cos2α=2sinαcosα-(cos^2α-sin^2α)=2cos^2αtanα-(2cos^2α-1)
=6/(3^2+1)-2/(3^2+1)+1
=6/10-2/10+1
=7/5
方法二：
万能公式sin2α=2tanα/1+(tanα)^2=3/5
cos2α=1-(tanα)^2/1+(tanα)^2=-4/5
所以原式=7/5

sin2a-cos2a=2sinacosa+1-2cosa^2;
2sinacosa-2cosa^2 =(2sinacosa-2cosa^2 )/(cosa^2+sina^2)
=(2tana-2)/(1+tana^2)=2/5
sin2a-coa2a=2sinacosa-2cosa^2 +1=7/5