Решите пожалуйста 48cos^2 23pi/12-48sin^223pi/1275-300sin^2 7pi/12

Ответ:

Пошаговое объяснение:

48cos^2 23pi/12-48sin^223pi/12=

(48)(cos^2 23pi/12- sin^223pi/12)=

'cosa-sina=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=

'sina/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