函数y=a-bcosx(b

题目

函数y=a-bcosx(b

答案

y=a-bcosx最大值为a-b=3/2,最小值为a+b=1/2
∴a=1,b=-1/2
∴y=-4asin(bx)
=-4sin[(-1/2)x]
=4sin(x/2)
最大值为4,最小值为-4
周期为T=2π/(1/2)=4π
…………………………………………………………
y=sin²a+2sinxcosx+3cos²x
=1+sin2x+2cos²x
=1+sin2x+1+cos2x
=2+√2sin(2x+π/4)
周期T=2π/2=π
递增区间：2x+π/4∈[-π/2+2kπ,π/2+2kπ],即x∈[-3π/8+kπ,π/8+kπ],k∈Z
递减区间：2x+π/4∈[π/2+2kπ,3π/2+2kπ],即x∈[π/8+kπ,5π/8+kπ],k∈Z
最小值为2-√2
此时2x+π/4=-π/2+2kπ,x=-3π/8+kπ,k∈Z
即x∈{x|x=-3π/8+kπ,k∈Z}