1) cos (a+pi/3) cos a= -15/17 2) sin (a-pi/4) - вопрос №911919613151724063 от JHopeЧонХосочек 09.02.2023 13:00

Question

Алгебра: 1) cos (a+pi/3) cos a= -15/17 2) sin (a-pi/4) sin a= 0.6 3) sin (a-b) + sin (pi/2-a)*sin b

JHopeЧонХосочек · Answer

1. По формуле косинуса суммы:cos(a+π/3)=сos(a)*cos(π/3)-sin(π/3)*sin(a)=cos(a)/2-(√3*sin(a))/2=(cos(a)-(√3*sin(a)))/2=(cos(a)-(√3*√(1-cos(a/2==(-15/17-(√3*√(1-15/17))/2=-(15+√102/17)/2=7.5+√102/322.Формула разности синусовsin(a-π/4)=sin(a)*cos(π/4)-sin(π/4)*cos(a)=√2/2*(sin(a)-cos(a))Применив основное тригонометрическое тождество=√2/2*(1-2cos(a))=√2/2-√2*соs(a)=√2/2-√2*√(1-sin(a))==√2/2-√2*√(1-0.6)=√2/2-(2√5)/53. sin(a-b) + sin(π/2-a)*sin(b)=Формула суммы синусовsin(a)*cos(b)-sin(b)*cos(a)+(sin(π/2)*cos(a)-sin(a)*cos(π/2))*sinb=sin(a)*(cos(b)-sin(b)*(cos(a)-(sin(π/2)*cos(a)-sin(a)*cos(π/2))=sin(a)*cos(b)-sin(b)*(cos(a)-(sin(π/2)*cos(a)-sin(a)*cos(π/2)))=sin(a)*cos(b)-sin(b)*(cos(a)-sin(π/2)*cos(a)+sin(a)*cos(π/2))=sin(a)*cos(b)-sin(b)*(cos(a)(1-2*sin(π/2))+cos(π/2)*(sin(a)+cos(a))==sin(a)*cos(b)-sin(b)*(cos(a)*(1-√2)+√2/2*(sin(a)+cos(a))=sin(a)*cos(b)-sin(b)*(cos(a)-√2/2*cos(a)+√2/2*sin(a)=sin(a)*cos(b)-sin(b)*(cos(a)*(1-√2)+√2/2*(sin(a)+cos(a))=sin(a)*(cos(b)-√2/2*sin(b))-sin(b)*(cos(a)*(1-√2)+√2/2*(cos(a)))=sin(a)*(cos(b)-√2/2*sin(b))-sin(b)*((2-√2)*cos(a))=-sin(b)*((4+√2)/2)+sin(a)*cos(b)-sin(b)*cos(a)=sin(a-b)-sin(b)*((4+√2)/2)...