试求函数y=1/2*cos(πx+π/3)-sin(πx+5π/6)的单调递减区间.

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试求函数y=1/2*cos(πx+π/3)-sin(πx+5π/6)的单调递减区间.

试求函数y=1/2*cos(πx+π/3)-sin(πx+5π/6)的单调递减区间.
试求函数y=1/2*cos(πx+π/3)-sin(πx+5π/6)的单调递减区间.

试求函数y=1/2*cos(πx+π/3)-sin(πx+5π/6)的单调递减区间.
y=1/2*cos(πx+π/3)-sin[π-(πx+5π/6)]
=1/2*cos(πx+π/3)-sin(-πx+π/6)
=1/2*cos(πx+π/3)-cos[π/2-(-πx+π/6)]
=1/2*cos(πx+π/3)-cos(πx+π/3)
=-1/2*cos(πx+π/3)
y递减则cos递增
cosx增区间是(2kπ-π,2kπ)
2kπ-π

a²+a-1=0
a²=-a+1
a^4=(-a+1)²
=a²-2a+1
=-a+1-2a+1
=-3a+2
a^5=a(-3a+2)=-3a²+2a
=-3(-a+1)+2a
=5a-3
b²=-b+1
b³=b(-b+1)
=-b²+b
=-(-b+1)+b
=2b-1
韦达定理
a+b=-1
原式=2(5a-3)+5(2b-1)
=10(a+b)-11
=-21