cos 5П/12 - cos П/12 + sin(arccos 1/2) Помогите пожалуйста

Question

cos 5П/12 - cos П/12 + sin(arccos 1/2)

Помогите пожалуйста

Answer 1

cos 5π/12 - cos π/12 + sin(arccos 1/2) =

= cos (π/12 + 4π/12) - cos π/12 + sin π/3 =

= cos π/12· cos π/3 - sin π/12· sin π/3 - cos π/12 + sin π/3 =

= 0.5cos π/12 - 0.5√3 sin π/12 - cos π/12 + sin π/3 =

= -0.5cos π/12 - 0.5√3 sin π/12 + sin π/3 =

= - (cos π/3 · cos π/12 + sin π/3 · sin π/12) + sin π/3 =

= - cos (π/3 - π/12) + sin π/6 = - cos π/4 + sin π/3 =

= -0.5·√2 + sin π/3 =  -0.5·√2 + 0.5·√3 = 0.5 (√3 - √2)

 

Answer 2

cos 5П/12 - cos П/12 + sin(arccos 1/2)=2sinП/4sin(-П/3)+sin(+-П/3)=

1)=-2*sqrt(2)/2*sqrt(3)/2+sqrt(3)/2=sqrt(3)/2(1-sqrt(2))

2)=-2*sqrt(2)/2*sqrt(3)/2-sqrt(3)/2=-sqrt(3)/2(1+sqrt(2))