问题标题:
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+sin253sin31
问题描述:
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
任杰回答:
(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...
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