∴3cos α-4sin α=0.∴tan α=3/4.

答案3/4

8.已知cos2α+4sin αcos α+4sin2α=5,则tan α=　.

解析由题意知(cos^2 α+4sinαcosα+4sin^2 α)/(cos^2 α+sin^2 α)

　　=(1+4tanα+4tan^2 α)/(1+tan^2 α)=5,

　　整理得tan2α-4tan α+4=0,∴tan α=2.

答案2

9.证明:(cos^4 α"-" sin^4 α)/(1+2sin"(" π"-" α")" cos"(" π+α")" )=(1+tanα)/(1"-" tanα).

证明左边=("(" cos^2 α+sin^2 α")(" cos^2 α"-" sin^2 α")" )/(cos^2 α+sin^2 α"-" 2sinαcosα)

　　=("(" cosα+sinα")(" cosα"-" sinα")" )/("(" cosα"-" sinα")" ^2 )=(cosα+sinα)/(cosα"-" sinα)

　　=(cosα/cosα+sinα/cosα)/(cosα/cosα "-" sinα/cosα)=(1+tanα)/(1"-" tanα)=右边,

　　故原等式成立.

10.导学号93774089已知sin θ+cos θ=-√10/5,

(1)求1/sinθ+1/cosθ的值;

(2)求tan θ的值.

解(1)因为sin θ+cos θ=-√10/5,

　　所以1+2sin θcos θ=2/5,即sin θcos θ=-3/10,

　　所以1/sinθ+1/cosθ=(sinθ+cosθ)/sinθcosθ=(2√10)/3.

　　(2)由(1)得(sin^2 θ+cos^2 θ)/sinθcosθ=-10/3,

　　所以(tan^2 θ+1)/tanθ=-10/3,

　　即3tan2θ+10tan θ+3=0,

　　所以tan θ=-3或tan θ=-1/3.

B组　能力提升

1.已知α为第二象限角,sin(α+π/3)=3/5,则sin(α+5π/6)=0(　　)

A.-4/5 B.4/5 C.-3/5 D.3/5

解析由sin(α+π/3)=3/5,可得cos(π/6 "-" α)=3/5,

　　于是sin(α+5π/6)=sin(π/6 "-" α)

　　=±√(1"-" cos^2 (π/6 "-" α) )=±4/5,

　　又α为第二象限角,cos(π/6 "-" α)=3/5,

　　所以π/6-α是第四象限角,

　　从而sin(π/6 "-" α)=-4/5.

答案A

2.化简√(1"-" 2sin4cos4)的结果是(　　)

A.sin 4+cos 4 B.sin 4-cos 4

C.cos 4-sin 4 D.-(sin 4+cos 4)

解析先判断4是第几象限角,再比较sin 4与cos 4的大小.∵5π/4<4<3π/2,

∴0>cos 4>sin 4,