已知cos(π/2+α)=根号2cos(3π/2-β),根号3sin(3π/2+α)=-根号2sin(π/2-β),且0
问题描述：
已知cos(π/2+α)=根号2cos(3π/2-β),根号3sin(3π/2+α)=-根号2sin(π/2-β),且0
答
cos(π/2 + α) = √2cos(3π/2 - β) => -sinα = -√2sinβ => sinα = √2sinβ ①,√3sin(3π/2 + α) = -√2sin(π/2 - β) => -√3cosα = -√2cosβ => √3cosα = √2cosβ ②,把① + ②,可得sin2α + 3cos2...