已知tan(π-α)=2,(sin^2α-2sincosα-cos^2α)/(4cos^2α-3sin^2α+1)?

问题描述:

已知tan(π-α)=2,(sin^2α-2sincosα-cos^2α)/(4cos^2α-3sin^2α+1)?

一楼的对了,二楼错了

tan(π-α)=-tanα=2,解得tanα=-2
因为(sinα)^2+(cosα)^2=1,并且sinα=-2cosα,所以(sinα)^2=4/5,(cosα)^2=1/5
那么(sin^2α-2sincosα-cos^2α)/(4cos^2α-3sin^2α+1)
=(4/5-2*(-2/5)-1/5)/(4*1/5-3*4/5+1)
=-1
楼主看看答案对不对?

tan(π-α)=-tanα=2
tanα=-2
由万能公式得出:
sin2α=2tanα/(1+tan^α)=2*(-2)/(1+2^)=-4/5
cos2α=(1-tan^α)/(1+tan^α)=(1-2^)/(1+2^)=-3/5
原式分子=-2sinαcosα-(cos^α-sin^α)=-sin2α-cos2α
原式分母=4*(1+cos2α)/2 - 3*(1-cos2α)/2 +1
=2+2cos2α-(3/2)+(3/2)cos2α+1
=(3/2)+(7/2)cos2α
于是原式=(-sin2α-cos2α)/[(3+7cos2α)/2]
代入sin2α和cos2α的值:
最后求得原式=-7/3