1)а)2sinπ/8cosπ/8=sinπ/4=√2/2
б)(cosπ/12-sinπ/12)(cosπ/12+sinπ/12)=cos²π/12-sin²π/12=cosπ/6=1/2
2)cosα=-0,8
sin2α=2*0,6*(-0,8)=-0,96
cos2α=0,64-0,36=0,28
3)sin2α=(2sinα/cosα)/(1/cos²α)
sin2α=(2sinαcos²α)/cosα
2sinαcosα=2sinαcosα
2 задание. Пишу без условий.
1)а)2sin60°cos10°=√3*cos10°
б)2sin10°cos60°=sin10°
в)2cos60°cos10°=cos10°
г)-2sin60°sin10°=-√3*sin10°
2)2sin150°cos50°=sin40°
cos50°=sin40°
cos(90°-40°)=sin40°
sin40°=sin40°
3)sin7π/12+sinπ/12+cosπ/12-cos7π/12=2sinπ/3cosπ/4-2sinπ/3sinπ/4=0
4)sin13°+sin15°+sin17°2sin15°cos2°+sin15°=sin15°(cos2°+1)