Question

已知向量a=(cos3x,sin3x),b=(cosx,sinx),x∈[-π/2,π/2]且f(x)=a乘b，g(x)=a+b的绝对值

1，求f（x）和g（x）的解析式
2.求函数f（x）的单调增区间
请写出过程 谢谢！！

Answer 1

1. f(x)=cos2x; g(x)=2cosx 2. [-π/4, π/4]
f(x)=a*b=(cos3x, sin3x)*(cosx, sinx)=cosxcos3x+sinxsin3x=cos(3x-x)=cos2x
g(x)=|a+b|=|(cos3x, sin3x) + (cosx, sinx)|
= |(cos3x+cosx, sin3x+sinx)|
=√[(cos3x+cosx)² + (sin3x+sinx)²]
=√(cos²3x+2cos3xcosx+cos²x + sin²3x+2sin3xsinx+sin²x)
=√(2+2cos2x)
=√[2(1+cos2x)]
=√(4cos²x)
x∈[-π/2, π/2]，所以，cosx≥0，所以，g(x)=2cosx

f(x)的单调递增区间，即-π/2 ≤ 2x ≤ π/2，即-π/4 ≤ x ≤ π/4，写成区间即为[-π/4, π/4]