Question

这几道数学题这么求啊

(1)若cosα=1/7,α∈(0,π/2),则cos(α+π/3)=
(2)化简cos(π/3+α)+sin(π/6+α)=
(3)求值(1+cos20)/(2sin20)-sin10[(1/tan5)-tan5]
(4)已知α∈(0,π/2),tanα=1/2,求tan2α和sin(2α+π/3)的值
(5)sin163sin223+sin253sin313=
(6)已知α∈(π/2,π),tan(α+π/4)=1/7，则cosα=
下面这个展开后为什么得到（tana+1）/（1-tana）)=1/7

（6）tan(α+π/4)=1/7展开（tana+1）/（1-tana）)=1/7
解tana=-3/4推出sina=3/5，cosa=-4/5

Answer 1

（1）先求sina取正的再代入，自己带吧

(2)化简cos(π/3+α)+sin(π/6+α)=cos(π/3+α)+cos[π/2-(π/6+α)]
=cos(π/3+α)+cos(π/3-α)
=cosπ/3casα-sinαsinπ/3+cosπ/3casα+sinαsinπ/3
=2cosπ/3casα=casα
（3）[(1+cos20°）/2sin20°]-sin10°(1/tan5°-tan5°)
=[(1+cos20°)/2sin20°]-sin10°(cot5°-tan5°)
=[(1+cos20°)/4sin10°cos10°]-sin10°(cot5°-tan5°)
=(2cos10°/4sin10°)-2sin5°cos5°(cot5°-tan5°)
=(cos10°/2sin10°)-2((cos5°)＾2-(sin5°)＾2)
=(cos10°/2sin10°)-2cos10°
=(cos10°-4sin10°cos10°)/2sin10°
=(sin80°-2sin20°)/2sin10°
=((sin80°-sin20°)-sin20°)/2sin10°
=(2cos50°sin30°-sin20°)/2sin10°
=(sin40°-sin20°)/2sin10°
={2cos30°sin10°)/2sin10°
=cos30°
=（根号3）/2

（4）tan2α直接用公式，sina+根3分之一，cosa=根下（2/3）带入求sin(2α+π/3)
（5)sin163sin223+sin253sin313
=sin(180-17)sin(180+43)+sin(180+73)sin(360-47)
=-sin17sin43+sin73sin47
=-cos73sin43+sin73cos43
=sin(73-43)
=sin30
=1/2
（6）tan(α+π/4)=1/7展开（tana+1）/（1-tana）)=1/7
解tana=-3/4推出sina=3/5，cosa=-4/5