Question

asin2x+bcos2x 如何化简?

Answer 1

asin2x+bcos2x

=(a^2+b^2)^0.5[(a/(a^2+b^2)^0.5)sin2x+(b/(a^2+b^2)^0.5)cos2x]

=(a^2+b^2)^0.5[cosθsin2x+sinθcos2x]

=(a^2+b^2)^0.5sin(2x+θ)

其中：cosθ=(a/(a^2+b^2)^0.5)

sinθ=(b/(a^2+b^2)^0.5)