Question

数学问题!!!!!!!???????非常急!!!!!!!在线等!!!!!!!要过程???

1.csc40°+cot80°=?
2.已知：tanα=7,tanβ=3，α、β为锐角.求证：α+2β=5π/4
注：5π/4即四分之五派
3.(cos2π/7)*(cos4π/7)*(cos6π/7)=?
注：题目中所有的“/”都代表分数线
4.若2sinα=1+cosα,则tanα/2的值为多少？
注：题目中所有的“/”都代表分数线

Answer 1

1.=csc40°+cot80°=1/sin40°+cos80°/sin80° =(2cos40°+cos80°)/sin80 °=(2cos(
30°+10°)+cos80°)/sin80°= (√3cos10°-sin10°+cos80°)/sin80°
=√3sin80°/sin80°=√3

2.因为tanβ=3>1,所以π/2>β>π/4,π>2β>π/2,因为tanα=7>1,所以π/2>α>π/4,3π/2>α+2β>3π/4
tan2β=2tanβ/(1-(tanβ)^2=-3/4,tan(α+2β)=(tanα+tan2β)/(1-tanαtan2β)=1,因为3π/2>α+2β>3π/4,所以α+2β=5π/4

3.(cos2π/7)(cos4π/7)(cos6π/7)=-cosπ/7cos2π/7cos4π/7=(-1/2sinπ/7)2sinπ/7cos2π/7cos4π/7=(-1/2sinπ/7)sin4π/7cos4π/7=(-1/4sinπ/7)sin8π/7=1/4

4.2sinα=1+cosα→ sinα/(1+cosα)=1/2
tanα/2=(sinα/2)/(cosα/2)=(2sinα/2cosα/2)/(2cosα/2)^2=sinα/(1+cosα)=1/2

Answer 2

1.csc40°+cot80°=1/sin40+cos40/sin40=(1+cos40)/sin40=2cos20^2/2sin20*cos20=cot20
2 用二倍角公式求出tan2β=-3/4 于是知2β是钝角
用两角和公式求出tan(α+2β)=1 于是得结论
3 自己用积化和差公式去做
4 tanα/2=sin/(1+cos)=1/2