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)的單調(diào)遞增區(qū)間,即-π/2 ≤ 2x ≤ π/2,即-π/4 ≤ x ≤ π/4,寫成區(qū)間即為[-π/4,π/4]