Sin50+sin70=? sin27+cos63=? cos15-cos75=? cos(2п/3)-cos(3п/5)=? sin(п/12)-sin(5п/12)=? *п-число пи

1)

sin50+sin70 = 

= 2 sin (50+70)/2* cos(50-70)/2=

= 2 sin 60*cos10 = 2* √3/2 cos10= √3cos10

2)

sin27°+cos63°= sin(90°-63°)+cos63° = cos63°+ cos63°= 2 cos63°

3)

cos15° - cos75° = -2 sin(15°-75°)/2 sin(15°+75°)/2=

= -2 sin(-30°) sin45°= 2*0,5*√2/2 = √2/2

4)
cos(2П/3)-cos(3П/5) = -2 sin(2П/3-3П/5)/2 sin(2П/3+3П/5)/2=

= -2 sin(П/15)/2 sin(19П/15)/2=

= -2 sinП/30* sin19П/30

Возможно в условии одинаковые знаменатели, т.е. нужно cos(2П/5) вместо cos(2П/3).

Тогда решение такое:

cos(2П/5)-cos(3П/5)=

= -2 sin(2П/5-3П/5)/2 sin(2П/5+3П/5)/2=

= -2 sin(-П/10) sinП/2=

= 2*√2/2* sinП/10= √2/ sinП/10

5)

 sin(π/12)-sin5π/12 = 

= 2sin[(π/12-5π/12)/2]*cos[(π/12+5π/12)/2] =
=2sin(-π/6)*cosπ/4 = 

= - 2sinπ/6*cosπ/4 = -2*1/2*√2/2*= -√2/2