已知函数f(x)=根号3sinx/2cosx/2-cos2x/2+1/2,x属于[0,π/2]f(x)跟号3/3求cosx值
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f(x)=√3sinx/2cosx/2-cos2x/2+1/2
＝√3/2sinx-1/2(1+cosx)+1/2
=√3/2sinx-1/2cosx
=sin(x- π/6)
f(x)=sin(x- π/6)
x属于[0,π/2]
x- π/6属于[- π/6,π/3]
sin(x- π/6)=√3/3
cos(x- π/6)=2√2/3
cosx=cos[(x- π/6)+π/6]=cos(x- π/6)cosπ/6+sin(x- π/6)sinπ/6
=2√2/3*√3/2-√3/3*1/2
=(2√6-√3)/6
＝√3/2sinx-1/2(1+cosx)+1/2
=√3/2sinx-1/2cosx
=sin(x- π/6)
f(x)=sin(x- π/6)
x属于[0,π/2]
x- π/6属于[- π/6,π/3]
sin(x- π/6)=√3/3
cos(x- π/6)=2√2/3
cosx=cos[(x- π/6)+π/6]=cos(x- π/6)cosπ/6+sin(x- π/6)sinπ/6
=2√2/3*√3/2-√3/3*1/2
=(2√6-√3)/6
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