Предмет: Алгебра
Автор: Пазенко
Вопрос задан 2 года назад

Помогите решить:
1) (sin3a cos^3a + cos3a sin^3a)/sin4a=
2) cos2a/(1-sin2a) - (1+tga)/(1-tga)=

Ответы

Ответ дал: DimaPuchkov

$\frac{\sin{3\alpha}\cdot \cos^3{\alpha}+\cos{3\alpha}\cdot \sin^3{\alpha}}{\sin{4\alpha}} =\frac{(3\sin{\alpha}-4\sin^3{\alpha})\cdot \cos^3{\alpha}+(4\cos^3{\alpha}-3\cos{\alpha})\cdot \sin^3{\alpha}}{\sin{4\alpha}} =\\ \\ = \frac{3\sin{\alpha}\cdot \cos^3{\alpha}-4\sin^3{\alpha}\cdot \cos^3{\alpha}+4\cos^3{\alpha}\cdot \sin^3{\alpha}-3\cos{\alpha}\cdot \sin^3{\alpha}}{\sin{4\alpha}} =\frac{3\sin{\alpha}\cdot\cos{\alpha}\cdot (\cos^2{\alpha}-\sin^2{\alpha})}{\sin{4\alpha}}=$

$=\frac{3\sin{\alpha}\cdot\cos{\alpha}\cdot \cos{2a}}{2\cdot \sin{2\alpha}\cdot\cos{2\alpha}}=\frac{3\sin{\alpha}\cdot \cos{\alpha}}{2\sin{2\alpha}}=\frac{3\sin{\alpha}\cdot \cos{\alpha}}{2\cdot 2 \cdot \sin{\alpha}\cdot \cos{\alpha}}=\frac{3}{4}$

$\frac{\cos{2\alpha}}{1-\sin{2\alpha}} -\frac{1+ tg \, \alpha}{1- tg \, \alpha}=\frac{\cos^2{\alpha}-\sin^2{\alpha}}{1-2\sin{\alpha}\cdot\cos{\alpha}}-\frac{1+\frac{\sin{\alpha}}{\cos{\alpha}}}{1-\frac{\sin{\alpha}}{\cos{\alpha}}}=\frac{\cos^2{\alpha}-\sin^2{\alpha}}{1-2\sin{\alpha}\cdot\cos{\alpha}}-\frac{\frac{\cos{\alpha}-\sin{\alpha}}{\cos{\alpha}}}{\frac{\cos{\alpha}-\sin{\alpha}}{\cos{\alpha}}}=$

$=\frac{(\cos{\alpha}-\sin{\alpha})\cdot (\cos{\alpha}+\sin{\alpha}) }{\cos^2{\alpha}+\sin^2{\alpha}-2\sin{\alpha}\cdot\cos{\alpha}}-\frac{\cos{\alpha}+\sin{\alpha}}{\cos{\alpha}-\sin{\alpha}}= \frac{(\cos{\alpha}-\sin{\alpha})\cdot (\cos{\alpha}+\sin{\alpha}) }{(\cos{\alpha}-\sin{\alpha})^2}-\frac{\cos{\alpha}+\sin{\alpha}}{\cos{\alpha}-\sin{\alpha}}= \\ \\ =\frac{\cos{\alpha}+\sin{\alpha}}{\cos{\alpha}-\sin{\alpha}}-\frac{\cos{\alpha}+\sin{\alpha}}{\cos{\alpha}-\sin{\alpha}}=0$

Klayzi: Помогите пожалуйста решить: https://znanija.com/task/43511043

Ответ дал: hote

$\displaystyle \frac{(sin3a*cos^3a+cos3a*sin^3a)}{sin4a}=\frac{(3sina-4sin^3a)*cos^3a+(4cos^3a-3cosa)*sin^3a}{2sin2a*cos2a}=\\\\=\frac{3sina*cos^3a-4sin^3a*cos^3a+4cos^3a*sin^3a-3cosa*sin^3a}{4sina*cosa*cos2a}=\\\\=\frac{3sina*cosa(cos^2a-sin^2a)}{4sina*cosa*cos2a}=\frac{3}{4}$

$\displaystyle \frac{cos2a}{1-sin2a}-\frac{1+tga}{1-tga}=\frac{cos2a}{1-sin2a}-\frac{1+\frac{sina}{cosa}}{1-\frac{sina}{cosa}}=\frac{cos2a}{1-sin2a}-\frac{cosa+sina}{cosa-sina}=\\\\=\frac{cos^2a-sin^2a}{1-sin2a}-\frac{cosa+sina}{cosa-sina}=\\\\=(cosa+sina)*(\frac{cosa-sina}{1-sin2a}-\frac{1}{cosa-sina})=\\\\=(cosa+sina)*(\frac{cos^2a-2cosa*sina+sin^2a-1+sin2a}{(1-sin2a)(cosa-sina)})=\\\\=(cosa+sina)*(\frac{0}{(1-sin2a)(cosa-sina)})=0$