Как сделать уравнение x + 300 +​

1.) sin2α/ 2sinαcosα - cos2α = 2sinαcosα/2sinαcosα - cos2a = 1 - cos2a = cos^2a + sin^2a - cos^2a + sin^2a = 2sin^2a
2.) sin7α- sin5α / (sin7α+sin5α) = 2cos(12a/2)sin(2a/2)/2sin(12a/2)cos(2a/2) = 2cos(6a)sin(a)/2sin(6a)cos(a) = tg(a)*ctg(6a)
3.) cosα+ cos3α+ cos5α / sinα+ sin3α+ sin5α = cos(3a) + 2cos(6a/2)cos(4a/2)/sin(3a) + 2sin(6a/2)cos(4a/2) = cos(3a)(1+2cos(2a))/sin(3a)(1+2cos(2a)) = ctg(3a)
4.) 2sin40 cos10 cos8  - Между ними стоит знак умножения???
5.) sin2α, cos2α, tg2α если  cosα= - 5/13 и sinα> 0
cos(2a) = 2cos^2(a)-1 = 2*(25/169)-1 = -144/169
sin(a)=
sin(2a)=2sin(a)cos(a)=2*(-5/13)*(12/13) = -120/169
tg(2a) = sin(2a)/cos(2a) = (-120/169)*(-169/144) = 5/6

sin2α+cos2α =11+tg2α=1/cos2α1+ctg2α=1/sin2αtgα *ctgα=1sin(α+β)=sinαcosβ +cosαsinβsin(α-β)=sinαcosβ-cosαsinβcos(α+β)=cosαcosβ-sinαsinβcos(α-β)=cosαcosβ+sinαsinβtg(α+β)=(tgα+tgβ)/(1-tgαtgβ)tg(α-β)=(tgα-tgβ)/(1+tgαtgβ)sin2α=2sinαcosβcos2α=cos2α-sin2α= 2cos2α-1=1-2 sin2αtg2α=2tg/(1-tg2α)sin2α=(1-cos2α)/2cos2α=(1+cos2α)/2tg(α/2)=sinα/(1+cos α)=(1-cos α)/sin αsin3α=3sinα-4sin3αcos3α=4cos3α-3cosαtg3α=(3tgα-tg3α)/(1-3tg2α)ctg3α=(ctg3α-3ctgα)/(3ctg2α-1)sin α+sin β=2sin[(α+β)/2]cos[(α-β)/2]sin α-sin β=2sin[(α-β)/2]cos[(α+β)/2]cosα+cosβ=2cos[(α+β)/2]cos[(α-β)/2]cosα-cosβ=-2sin[(α+β)/2]sin[(α-β)/2]tgα±tgβ=sin(α±β)/cosαcosβctgα±ctgβ=sin(β±α)/sinαsinβsinαcosβ=1/2*(sin(α+β)+sin(α-β))cosαcosβ=1/2*(cos(α+β)+cos(α-β))sinαsinβ=1/2*(cos(α-β)+cos(α+β))