已知函数fx=cos²+sinxcosx,求fx单调递增区间
问题描述:
已知函数fx=cos²+sinxcosx,求fx单调递增区间
答
f(x)=cos²x+sinxcosx
=(cos2x+1)/2+1/2sin2x
=(1/2cos2x+1/2sin2x)+1/2
=√2/2*(√2/2cos2x+√2/2sin2x)+1/2
=√2/2*sin(2x+π/4)+1/2
令-π/2+2kπ≤2x+π/4≤π/2+2kπ (k∈Z)
-3π/4+2kπ≤2x≤π/4+2kπ
-3π/8+kπ≤x≤π/8+kπ
所以fx单调递增区间为[-3π/8+kπ,π/8+kπ] (k∈Z)