Ответ:
Пошаговое объяснение:
√48cos^2 23pi/12-√48sin^223pi/12=
(√48)(cos^2 23pi/12- sin^223pi/12)=
'cos²a-sin²a=cos2a'
= (√48)(cos23pi/6)= (√48)(cos((3 5/6)pi)=(√48)(cos(2pi+pi+5/6pi)=
'cos(2п+а)=cosa'
=(√48)(cos(pi+5/6pi) =
'cos(п+а)=-cosa'
(√48)(cos(pi+5/6pi)=-(√48)(cos(5/6pi)=
= -(√48)(-√3)/2=4(√3)*(√3)/2=2*3=6
√75-√300sin^2 7pi/12=
'sin²a/2=(1-cos2a)/2'
=√75-√300(1-cos7pi/6)/2=
=√75-√300(1-cos(pi+(pi/6))/2=
=√75-√300(1+cos(pi/6))/2
=√75-√300((√3)/2)/2=√75-(√300)(√3)/4=
=√75-(√900)/4=5(√3)-30/4=5(√3)-7,5