Question

(sin3x+sinx)/(cos3x+cosx)=

Answer 1

sinα+sinβ=2sin[(α+β)/2]cos[(α-β)/2]

cosα+cosβ=2cos[(α+β)/2]cos[(α-β)/2]

所以：

(sin3x+sinx)/(cos3x+cosx)

=2sin2xcosx/(2cos2xcosx)

=sin2x/cos2x

=tam2x