Question

ПОМОГИТЕ, с алгеброй!!!
Задание прикрепил!

Answer 1

1.= Sinα Cosβ+CosαSinβ+SinαCosβ - CosαSinβ = 2SinαCosβ
3. 2SinαCosα / (Cos²α+Sin²α+Cos²α - Sin²α) = 2SinαCosα /2Cos²α= Sinα/Cosα =tgα

Answer 2

1)  sin(α+β)+sin(α-β)=
=sin(α)cos(β)+sin(β)cos(α)+sin(α)cos(β)-sin(β)cos(α)=
=2sin(α)cos(β)

2) [sin(3π/2+α)+sin(2π+α)]/[2sin(-α)cos(-α)+1]=
=[-cos(α)+sin(α)]/[-2sin(α)cos(α)+sin²(α)+cos²(α)]=
=[-cos(α)+sin(α)]/[sin(α)-cos(α)]²=1/[sin(α)-cos(α)]=
=1/[√2sin(α-π/4)]
или -1/[√2cos(α+π/4)]

3) sin(2α)/[1+cos(2α)]= sin(2α)/[2cos²(α)]=
=2sin(α)cos(α)/[2cos²(α)]=1/cos(α)