1
π(2x-1)/3=-π/3+2πn U π(2x-1)/3=π/3+2πn
2x-1=-1+6n U 2x-1=1+6n
2x=6n U 2x=2+6n
x=3n U x=1+3n,n∈z
3
4sin76/sin(360-76)=4sin76/(-sin76)=-4
4
-24tg(90-20)*tg(180-20)=-24*ctg20*(-tg20)=-24*(-1)=24
При решение 1-4 используем в основном формулы приведения
1.
a) cos(-210°)=cos210°=cos(180°+30°)=-cos30°=-√3/2
б) tg(4π/3)=tg(3π/2-π/6)=ctgπ/6 =√3
2.
a) sin(3π/2-α)-cos(π+α)=-cosα+cosα=0
б) tg(π+α)-ctg(π/2-α) =tgα-tgα=0
в) sin2α+(sinα-cosα)^2= 2sinαcosα+sin^2α-2sinαcosα+cos^2α= 1
3.
a) sin(π-2α)/(1+cos2α)=tgα
sin2α/(sin^2α+cos^2α+cos^2α-sin^2α)=tgα
2sinαcosα/2cos^2α=tgα
sinα/cosα=tgα
tgα=tgα
б) 4sinαcosα/(cos^2α-sin^2α) = 2tg2α
2*2sinαcosα/cos2α=2tg2α
2sin2α/cos2α=2tg2α
2tg2α=2tg2α
4.
a) (ctgα-tgα)tg2α (1)
Учитывая, ctgα=1/tgα и tg2α=2tgα/(1-tg^2α)
Подставим в (1)
(ctgα-tgα)tg2α = (1/tgα-tgα)*2tgα/(1-tg^2α)=2tgα(1-tg^2α)/tgα(1-tg^2α)=2
б)(1+cos2α)/sin(π/2-α)=(sin^2α+cos^2α+cos^2α-sin^2α)/cosα = =2cos^2α/cosα=2cosα
5.
sin(290°+α)-cos(340°-α)/sin(110°+α)=-2
sin(270°+(20°+α))-cos(360°-(20°+α))/sin(90°+(20°+α))=-2
-cos((20°+α)-cos((20°+α)/cos((20°+α) =-2
-2cos((20°+α)/cos((20°+α)=-2
-2 = -2
А) = -2+8=6
б) -1 - ∛-2³ = -1 - (-2) = -1+2=1
7a-10b-7a= -10b; 30c+15d-6c-12d=24c+3d; 25k-12+7k+7n+12n=32k+19n-12.