Решите √48cos^2 23pi/12-√48sin^223pi/12 √75-√300sin^2 7pi/12

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