问题
解答题
设
(1)若
(2)若f(x)=(
|
答案
(1)∵
∥a
∴2cosx-sinx=0∴tanx=2b
(sinx+cosx)2=sin2x+2sinxcosx+cos2x=cos2x•(tan2x+2tanx+1)=
(tan2x+2tanx+1)=1 1+tan2x 9 5
(2)f(x)=cos2x-sinxcosx-1=-
sin(2x-2 2
)-π 4 1 2
∵2kπ-
≤2x-π 2
≤2kπ+π 4 π 2
∴kπ-
≤x≤kπ+π 8
k∈z3π 8
∵x∈[0,π]∴令k=0,1得f(x)在区间[0,π]上的递减区间是[0,
],[3π 8
,π]7π 8