已知函数f(x)=cosx/cos(π/6-x),则f(x)+f(π/3-x)的值为

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已知函数f(x)=cosx/cos(π/6-x),则f(x)+f(π/3-x)的值为已知函数f(x)=cosx/cos(π/6-x),则f(x)+f(π/3-x)的值为已知函数f(x)=cosx/co

已知函数f(x)=cosx/cos(π/6-x),则f(x)+f(π/3-x)的值为
已知函数f(x)=cosx/cos(π/6-x),则f(x)+f(π/3-x)的值为

已知函数f(x)=cosx/cos(π/6-x),则f(x)+f(π/3-x)的值为
f(π/3-x)=cos(π/3-x)/cos(π/6-(π/3-x))
=cos(π/3-x)/cos(x-π/6)
=cos(π/3-x)/cos(π/6-x) 注:cosx = cos(-x)
所以:
f(x)+f(π/3-x) = cosx/cos(π/6-x) + cos(π/3-x)/cos(π/6-x)
= [cosx + cos(π/3-x)]/cos(π/6-x)
= 2cos(π/6)cos(π/6-x)/cos(π/6-x)
= 2cos(π/6)
=√3

f(x)+f(π/3-x)=cosx/cos(π/6-x)+cos(π/3-x)/cos[π/6-(π/3-x)]=cosx/cos(π/6-x)+cos(π/3-x)/cos(-π/6+x)=[cosx+cos(π/3-x)]/cos(π/6-x)=2cos{[x+(π/3-x)]/2}cos{x-(π/3-x)]/2}/cos(π/6-x)=2cos(π/6)cos(π/6-x)/cos(π/6-x)=根号3