У=2+2х
2х-4*(2+2х)=10
2х-8-8х=10
-6х=18
х=-3
у=2+2*-3=2-6=-4
4х+8=0
4х=-8
х=-2
у=4*0+8
у=8
Ответ: координаты точки пересечения графика функции с осью Ох (-2;0) с осью Оу (0;8)
X-6<√10(x-6)
(x -6)²<10x - 60
x²-12x+36-10x+60<0
x²-22x+96<0
D1= 121-96=25
x1;2=11+-5
x1= 6
x2=16
Ответ: (6;16)
1) ![\frac{1-tga}{1+tga}= tg(45-a)\\ \frac{1-tga}{1+tga}= \frac{tg45-tga}{1+tg45tga}\\ \frac{1-tga}{1+tga}= \frac{1-tga}{1+tga}\\](https://tex.z-dn.net/?f=%5Cfrac%7B1-tga%7D%7B1%2Btga%7D%3D+tg%2845-a%29%5C%5C+%5Cfrac%7B1-tga%7D%7B1%2Btga%7D%3D+%5Cfrac%7Btg45-tga%7D%7B1%2Btg45tga%7D%5C%5C+%5Cfrac%7B1-tga%7D%7B1%2Btga%7D%3D+%5Cfrac%7B1-tga%7D%7B1%2Btga%7D%5C%5C)
верно
2) ![tg(a-B) = \frac{tga-tgB}{1+tga\cdot tgB}](https://tex.z-dn.net/?f=tg%28a-B%29+%3D+%5Cfrac%7Btga-tgB%7D%7B1%2Btga%5Ccdot+tgB%7D)
подставляем tga =3
![tg(a-B) = \frac{tga-tgB}{1+tga\cdot tgB}=1\\ \frac{tga-tgB}{1+tga\cdot tgB}=1\\ \frac{3-tgB}{1+3\cdot tgB}=1\\ 3-tgB=1+3tgB\\ 4tgB=2\\ tgB = \frac{1}{2}](https://tex.z-dn.net/?f=tg%28a-B%29+%3D+%5Cfrac%7Btga-tgB%7D%7B1%2Btga%5Ccdot+tgB%7D%3D1%5C%5C+%5Cfrac%7Btga-tgB%7D%7B1%2Btga%5Ccdot+tgB%7D%3D1%5C%5C+%5Cfrac%7B3-tgB%7D%7B1%2B3%5Ccdot+tgB%7D%3D1%5C%5C+3-tgB%3D1%2B3tgB%5C%5C+4tgB%3D2%5C%5C+tgB+%3D+%5Cfrac%7B1%7D%7B2%7D)
аналогично следующая
![tg(a-B) = \frac{tga-tgB}{1+tga\cdot tgB}=2\\ \frac{tga-tgB}{1+tga\cdot tgB}=2\\ \frac{\frac{1}{4}-tgB}{1+\frac{1}{4}\cdot tgB}=2\\ \frac{1}{4}-tgB=2+\frac{1}{2}tgB\\ \frac{3}{2}tgB=-\frac{7}{4}\\ tgB = -\frac{7}{6}](https://tex.z-dn.net/?f=tg%28a-B%29+%3D+%5Cfrac%7Btga-tgB%7D%7B1%2Btga%5Ccdot+tgB%7D%3D2%5C%5C+%5Cfrac%7Btga-tgB%7D%7B1%2Btga%5Ccdot+tgB%7D%3D2%5C%5C+%5Cfrac%7B%5Cfrac%7B1%7D%7B4%7D-tgB%7D%7B1%2B%5Cfrac%7B1%7D%7B4%7D%5Ccdot+tgB%7D%3D2%5C%5C+%5Cfrac%7B1%7D%7B4%7D-tgB%3D2%2B%5Cfrac%7B1%7D%7B2%7DtgB%5C%5C+%5Cfrac%7B3%7D%7B2%7DtgB%3D-%5Cfrac%7B7%7D%7B4%7D%5C%5C+tgB+%3D+-%5Cfrac%7B7%7D%7B6%7D)
3) Поскольку косинус положительный, то угол принадлежит первой четверти. Значит все остальные функции тоже положительные.
Найдем синус:
![sina=\sqrt{1-cos^{2}a}= \sqrt{1-\frac{9}{25}}= \frac{4}{5}](https://tex.z-dn.net/?f=sina%3D%5Csqrt%7B1-cos%5E%7B2%7Da%7D%3D+%5Csqrt%7B1-%5Cfrac%7B9%7D%7B25%7D%7D%3D+%5Cfrac%7B4%7D%7B5%7D)
Разложим формулы тангенса и подставим значения синуса и косинуса:
![tg(a+\frac{\pi}{3}) = \frac{tga+tg\frac{\pi}{3}}{1-tgatg\frac{\pi}{3}}= \frac{tga+\sqrt{3}}{1-\sqrt{3}tga} = \frac{\frac{sina}{cosa}+\sqrt{3}}{1-\sqrt{3}\frac{sina}{cosa}}=\frac{\frac{4}{5}\cdot \frac{5}{3}+\sqrt{3}}{1-\sqrt{3} \cdot \frac{4}{5} \cdot \frac{5}{3}} = \frac{\frac{4}{3}+\sqrt{3}}{1-\frac{4}{3}\sqrt{3}}](https://tex.z-dn.net/?f=tg%28a%2B%5Cfrac%7B%5Cpi%7D%7B3%7D%29+%3D+%5Cfrac%7Btga%2Btg%5Cfrac%7B%5Cpi%7D%7B3%7D%7D%7B1-tgatg%5Cfrac%7B%5Cpi%7D%7B3%7D%7D%3D+%5Cfrac%7Btga%2B%5Csqrt%7B3%7D%7D%7B1-%5Csqrt%7B3%7Dtga%7D+%3D+%5Cfrac%7B%5Cfrac%7Bsina%7D%7Bcosa%7D%2B%5Csqrt%7B3%7D%7D%7B1-%5Csqrt%7B3%7D%5Cfrac%7Bsina%7D%7Bcosa%7D%7D%3D%5Cfrac%7B%5Cfrac%7B4%7D%7B5%7D%5Ccdot+%5Cfrac%7B5%7D%7B3%7D%2B%5Csqrt%7B3%7D%7D%7B1-%5Csqrt%7B3%7D+%5Ccdot+%5Cfrac%7B4%7D%7B5%7D+%5Ccdot+%5Cfrac%7B5%7D%7B3%7D%7D+%3D+%5Cfrac%7B%5Cfrac%7B4%7D%7B3%7D%2B%5Csqrt%7B3%7D%7D%7B1-%5Cfrac%7B4%7D%7B3%7D%5Csqrt%7B3%7D%7D)
![tg(a-\frac{5\pi}{4}) = \frac{tga-tg\frac{5\pi}{4}}{1+tgatg\frac{5\pi}{4}}= \frac{tga-1}{1-tga} = - \frac{1-tga}{1-tga} = -1](https://tex.z-dn.net/?f=tg%28a-%5Cfrac%7B5%5Cpi%7D%7B4%7D%29+%3D+%5Cfrac%7Btga-tg%5Cfrac%7B5%5Cpi%7D%7B4%7D%7D%7B1%2Btgatg%5Cfrac%7B5%5Cpi%7D%7B4%7D%7D%3D+%5Cfrac%7Btga-1%7D%7B1-tga%7D+%3D+-+%5Cfrac%7B1-tga%7D%7B1-tga%7D+%3D+-1)