А) 2ctg(π - 2α) + 2sin(π-α) = 2tg2α+ 2sinα =
2 sin(0.5π+α)+tgα*sin(-α) cosα-sin²α
cosα
По действиям:
1) ctg (π-2α) = cos(π/2 -2α) =-sin(-2α) = tg 2α
2 sin(π/2 - 2α) cos2α
2) sin(π-α)=-sin(-α)=sinα
3) sin(0.5π+α)=sin(π/2 + α)=cosα
4) tgα * sin(-α)=sinα * (-sinα) =-sin²α
cosα cosα
=2tg2α + 2sinα = 2tg2α + 2sinαcosα = 2tg2α + sin2α=2tg2α+tg2α=
cos²α-sin²α cos²α-sin²α cos2α
cosα
=3tg2α
При α=-π/12
3tg2α=3tg2*(-π/12)=3tg(-π/6)=-3tg(π/6)=-3 * √3 =-√3
3
Ответ: -√3.
б) cos(-2α) + 2sin(π-2α) = cos2α + 2sin2α =cos2α+2sin2α * tgα =
ctg(0.5π+α) +ctgα 1-tg²α 1-tg²α
tgα
По действиям:
1) cos(-2α)=cos2α
2) sin(π-2α)=sin2α
3) ctg(0.5π+α)=ctg(π/2 +α)=cos(π/2+α) =-sinα = -tgα
sin(π/2+α) cosα
4) -tgα+ctgα= 1 - tgα =1-tg²α
tgα tgα
=cos2α+2sin2α * 1 *tg2α =cos2α+sin2α tg2α=cos2α+sin2αsin2α =
2 cos2α
=cos2α + sin²2α =cos²2α+sin²2α = 1
cos2α cos2α cos2α
При α=π/8
1 = 1 = 1 = 1 = 2 = 2√2 =2√2 =√2
cos2α cos2*(π/8) cos(π/4) √2 √2 √2*√2 2
2
Ответ: √2.