If the function y = sin (2x + φ) (0 is less than or equal to φ, less than or equal to π) is an even function on R, then= The function y = sin (2x + φ) is an even function on R, Then sin (- 2x + φ) = sin (2x + φ) The expansion is: sin φ cos2x cos φ sin2x = sin φ cos2x + cos φ sin2x So cos φ sin2x = 0. Cos φ = 0, φ = π / 2 "Ask," how is it expanded: sin φ cos2x cos φ sin2x = sin φ cos2x + cos φ sin2x "

If the function y = sin (2x + φ) (0 is less than or equal to φ, less than or equal to π) is an even function on R, then= The function y = sin (2x + φ) is an even function on R, Then sin (- 2x + φ) = sin (2x + φ) The expansion is: sin φ cos2x cos φ sin2x = sin φ cos2x + cos φ sin2x So cos φ sin2x = 0. Cos φ = 0, φ = π / 2 "Ask," how is it expanded: sin φ cos2x cos φ sin2x = sin φ cos2x + cos φ sin2x "

∵sin（2x+ φ ）＝ sinφcos2x＋cosφsin2x
sin（-2x+ φ ）＝sinφcos2x－cosφsin2x
∵sin（-2x+ φ ）＝sin（2x+ φ ）
∴sinφcos2x＋cosφsin2x＝sinφcos2x－cosφsin2x