已知函数sin(70°+α)=1/3,则cos(2α-40°)=

问题描述:

已知函数sin(70°+α)=1/3,则cos(2α-40°)=

∵sin(70°+α)=1/3
∴sin[90°-(-α+20°)]=1/3
∴cos[(-α+20°)]=1/3
∴cos(α+20°)=1/3
∴原式=cos(2α-40°)=2cos²(α-20°)-1
=-7/9

sin(70°+α)=1/3
cos(90°-70°-α)=sin(70°+α)=1/3
即cos(20°-α)=1/3
所以cos(2α-40°)=2cos(α-20°)^2-1=2*1/9-1=-7/9

cos(2α-40)
=cos(40-2α)
=-cos[180-(40-2α)]
=-cos(140+2α)
=-cos[2(70+α)]
=-[1-2sin²(70+α)]
=-7/9