1、已知cos4α=2/3,则(sin^4 α-cos^4 α)^2= 2、(1-cosx+sinx)/(1+cosx+sinx)=-2,则tanx=
问题描述：
1、已知cos4α=2/3,则(sin^4 α-cos^4 α)^2= 2、(1-cosx+sinx)/(1+cosx+sinx)=-2,则tanx=
答
1.cos4α=cos^22a-sin^22a=2/3
(sin^4 α-cos^4 α)^2
=(cos^22a+sin^22a)(cos^22a-sin^22a)
=cos^22a-sin^22a=2/3
2.
(1-cosx+sinx)/(1+cosx+sinx)
=(2sin^2x/2+2sinx/2cosx/2)/(2cos^2x/2+2sinx/2cosx/2)
=2sinx/2(sinx/2+cosx/2)/2cosx/2(sinx/2+cosx/2)
=tanx/2
=-2
tanx=2tanx/2/(1-ta/^2x/2)=-4/(1-4)=-4/3第一题答案好像不对。。。1.cos4α=cos^22a-sin^22a=2/3(sin^4 α-cos^4 α)^2=-(cos^22a+sin^22a)(cos^22a-sin^22a)=-(cos^22a-sin^22a)=-2/3