Question

Помогите пожалуйста,нужно срочно

Answer 1

а решеба никак залез и списывай

Answer 2

162)
а) sinα/(2cos²(α/2))=2 sin(α/2)cos(α/2)/(2cos²(α/2))=sin(α/2)/cos(α/2)=tg(α/2)

б) sin(4β)/(cos(2β))=2sin(2β)cos(2β)/(cos(2β))=2sin(2β)

в) cos(φ)/(cos(φ/2)+sin(φ/2))=(cos²(φ/2)-sin²(φ/2))/(cos(φ/2)+sin(φ/2))=
(cos(φ/2)-sin(φ/2))=√2[(1/√2)cos(φ/2)-(1/√2)sin(φ/2)]=
=√2[sin(π/4)cos(φ/2)-cos(π/4)sin(φ/2)]=√2sin(π/4-φ/2)

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