已知0<α<π,tanα=-2. (1)求sin(α+π/6 )的值; (2)求2cos(π/2 +α)-cos(π-α) sin(π/2-α已知0<α<π,tanα=-2.(1)求sin(α+π/6)的值;(2)求2cos(π/2+α)-cos(π-α)sin(π/2-α)-3sin(π+α)的值;(3)2sin²α-sinαcosα+cos²α
问题描述:
已知0<α<π,tanα=-2. (1)求sin(α+π/6 )的值; (2)求2cos(π/2 +α)-cos(π-α) sin(π/2-α
已知0<α<π,tanα=-2.
(1)求sin(α+π/6)的值;
(2)求2cos(π/2+α)-cos(π-α)sin(π/2-α)-3sin(π+α)的值;
(3)2sin²α-sinαcosα+cos²α
答
tanα=-2 -> sinα/cosα=-2 -> sinα=-2cosα
sinα^2+cosα^2=1 -> (-2cosα)^2+cosα^2=1 -> cosα^2=1/5
又因为0所以α为第二象限角
故sinα>0,cosα cosα=-√5/5,sinα=-2cosα=2√5/5
(1)sin(α+π/6)=sinαcosπ/6+cosαsinπ/6=(2√15-√5)/10
(2)2cos(π/2+α)-cos(π-α)sin(π/2-α)-3sin(π+α)
=-2sinα+cosαcosα+3sinα
=sinα+cosα^2=(1+2√5)/5
(3)2sinα^2-sinαcosα+cos^α
=1+sinα^2-sinαcosα
=7/5