Преобразуем левую часть 2sin^2(45-1t)+sin(4t)=
1-cos(90-4t)+sin(4t)=1-sin(4t)+sin(4t)=1
Вот там вообщем кос=0 и кос=-1 это частные случаи
(15-2√15√33+33)/(8-√55)=(48-2√3√5√3√11)/(8-√55)=(48-6√55)/(8-√55)=6(6-√55)/(8-√55)=6
Найдём cosα с помощью основного тригонометрического тождества
![\cos^2\alpha + \sin^2\alpha = 1\\\\\cos^2\alpha = 1 - \sin^2\alpha\\\\\cos^2\alpha = 1 - \frac{144}{169}\\\\\cos^2\alpha = \frac{25}{169}\\\\\cos\alpha = \pm\;\frac{5}{13}](https://tex.z-dn.net/?f=%5Ccos%5E2%5Calpha+%2B+%5Csin%5E2%5Calpha+%3D+1%5C%5C%5C%5C%5Ccos%5E2%5Calpha+%3D+1+-+%5Csin%5E2%5Calpha%5C%5C%5C%5C%5Ccos%5E2%5Calpha+%3D+1+-+%5Cfrac%7B144%7D%7B169%7D%5C%5C%5C%5C%5Ccos%5E2%5Calpha+%3D+%5Cfrac%7B25%7D%7B169%7D%5C%5C%5C%5C%5Ccos%5Calpha+%3D+%5Cpm%5C%3B%5Cfrac%7B5%7D%7B13%7D)
Так как α ∈ (π, 3π/2) то cos(α) = -5/13
Найдём tgα
![tg\alpha = \dfrac{\sin\alpha}{cos\alpha}\\\\\\tg\alpha = -\frac{12}{13} : (-\frac{5}{13}) = \frac{12}{5}](https://tex.z-dn.net/?f=tg%5Calpha+%3D+%5Cdfrac%7B%5Csin%5Calpha%7D%7Bcos%5Calpha%7D%5C%5C%5C%5C%5C%5Ctg%5Calpha+%3D+-%5Cfrac%7B12%7D%7B13%7D+%3A+%28-%5Cfrac%7B5%7D%7B13%7D%29+%3D+%5Cfrac%7B12%7D%7B5%7D)