Question

已知向量a=(sinX,cosX),b=(1,1).当a//b时,求tanX的值:若f（x）=aXb,求函数f（x）的单

Answer 1

当向量a//b时,可设向量a = kb,（k是非零实数）,所以(sinx,cosx) = k(1,1) = (k,k) =>sinx = cosx = k,因为sin 2 x+ cos 2 x = 1 => k 2 + k 2 = 1 => 2k 2 = 1 => k 2 = 1/2 => k = ±√2/2 => sinx = cosx = √2/2或者-√2/2,所以tanx = sinx/cosx = 1 ； 函数f(x) = a·b = (sinx,cosx)·(1,1) = sinx +cosx = √2sin(x + π/4),所以函数在(x + π/4)∈[-π/2 + 2kπ,π/2 + 2kπ],k∈Z上单调递增,即x∈[-3π/4 + 2kπ,π/4 + 2kπ],k∈Z,所以函数f(x)的单调递增区间是 [-3π/4 + 2kπ ,π/4 + 2kπ] ,k ∈ Z .