已知tanα=2,且α为第三象限角,求cos(α+π/3),sin2α
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![已知tanα=2,且α为第三象限角,求cos(α+π/3),sin2α](/uploads/image/z/995059-19-9.jpg?t=%E5%B7%B2%E7%9F%A5tan%CE%B1%3D2%2C%E4%B8%94%CE%B1%E4%B8%BA%E7%AC%AC%E4%B8%89%E8%B1%A1%E9%99%90%E8%A7%92%2C%E6%B1%82cos%28%CE%B1%2B%CF%80%2F3%29%2Csin2%CE%B1)
已知tanα=2,且α为第三象限角,求cos(α+π/3),sin2α
已知tanα=2,且α为第三象限角,求cos(α+π/3),sin2α
已知tanα=2,且α为第三象限角,求cos(α+π/3),sin2α
第一个问题:
cos(α+π/3)
=cosαcos(π/3)-sinαsin(π/3)=(1/2)cosα-(√3/2)sinα
=(1/2)cosα/√[(sinα)^2+(cosα)^2]-(√3/2)sinα/√[(sinα)^2+(cosα)^2]
=-(1/2)/√[(tanα)^2+1]+(√3/2)tanα/√[(tanα)^2+1]
=-(1/2)/√(4+1)+(√3/2)×2/√(4+1)
=√15/5-√5/10.
第二个问题:
sin2α
=2sinαcosα=2sinαcosα/[(sinα)^2+(cosα)^2]
=2tanα/[(tanα)^2+1]=2×2/(4+1)=4/5.