Решение смотри во вложенном файле
Найдите значения выражений sin2α , cos3α , если
а) α =π/12 ; б) α =π/6 ; в) α =π /2 ; г) α = 2π/3 .
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2α = a) π/6 ; б) α =π/3 ; в) α =π ; <span> г) </span>α = 4<span>π/3 </span>
3α = a) π/4 ; б) α =π/2 ; в) α =3π /2 ; г) α = 2π
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a) sin2α =sinπ/6 =1/2 ; cos3α =cosπ/4 =√2 /2 .
б) sin2α =sinπ/3 =√3 /2 ; cos3α =cos<span>π./2 =0 .
</span>в) sin2α =sinπ =0 ; cos3α =cos3π/2 =0
г) sin2α =sin4π/3 =sin(π+π/3) = -sinπ/3 = -√3 /2 ; <span>cos3α =cos2</span>π = 1.
1 действие- возведение в квадрат:
-0,4^2= 0,16
2 действие:
1,4: 1 целая 3/7= 0,98
3 действие:
0,16/0,98=8/49
Составим матрицу и приведем её в ступенчатый вид:
![\displaystyle \left[\begin{array}{cccc}1&2&1&3\\1&2&-2&-3\\2&-3&-1&0\end{array}\right] \rightarrow R_2-R_1\rightarrow \left[\begin{array}{cccc}1&2&1&3\\0&0&-3&-6\\2&-3&-1&0\end{array}\right]\rightarrow \\\\\rightarrow R_3-2R_1 \rightarrow \left[\begin{array}{cccc}1&2&1&3\\0&0&-3&-6\\0&-7&-3&-6\end{array}\right] \rightarrow R_3\leftrightarrow R_2\rightarrow \\\\\left[\begin{array}{cccc}1&2&1&3\\0&-7&-3&-6\\0&0&-3&-6\end{array}\right]\rightarrow R_2-R_3](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%261%263%5C%5C1%262%26-2%26-3%5C%5C2%26-3%26-1%260%5Cend%7Barray%7D%5Cright%5D%20%5Crightarrow%20R_2-R_1%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%261%263%5C%5C0%260%26-3%26-6%5C%5C2%26-3%26-1%260%5Cend%7Barray%7D%5Cright%5D%5Crightarrow%20%5C%5C%5C%5C%5Crightarrow%20R_3-2R_1%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%261%263%5C%5C0%260%26-3%26-6%5C%5C0%26-7%26-3%26-6%5Cend%7Barray%7D%5Cright%5D%20%5Crightarrow%20R_3%5Cleftrightarrow%20R_2%5Crightarrow%20%5C%5C%5C%5C%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%261%263%5C%5C0%26-7%26-3%26-6%5C%5C0%260%26-3%26-6%5Cend%7Barray%7D%5Cright%5D%5Crightarrow%20R_2-R_3)
![\displaystyle \rightarrow \left[\begin{array}{cccc}1&2&1&3\\0&-7&0&0\\0&0&-3&-6\end{array}\right] \rightarrow R_3\rightarrow -\frac{1}{3}R_3\rightarrow \left[\begin{array}{cccc}1&2&1&3\\0&-7&0&0\\0&0&1&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%261%263%5C%5C0%26-7%260%260%5C%5C0%260%26-3%26-6%5Cend%7Barray%7D%5Cright%5D%20%5Crightarrow%20R_3%5Crightarrow%20-%5Cfrac%7B1%7D%7B3%7DR_3%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%262%261%263%5C%5C0%26-7%260%260%5C%5C0%260%261%262%5Cend%7Barray%7D%5Cright%5D)
Отсюда:
![\displaystyle \begin{cases}x+2y+z=3\\-7y=0\\z=2\end{cases} \Rightarrow \begin{cases}x=1\\y=0\\z=2\end{cases}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Bcases%7Dx%2B2y%2Bz%3D3%5C%5C-7y%3D0%5C%5Cz%3D2%5Cend%7Bcases%7D%20%5CRightarrow%20%5Cbegin%7Bcases%7Dx%3D1%5C%5Cy%3D0%5C%5Cz%3D2%5Cend%7Bcases%7D)
Сначала сократим неравенство, выполнив возможные действия:
![4x - 28 \ \textgreater \ 3x + 5 + m](https://tex.z-dn.net/?f=4x+-+28+%5C+%5Ctextgreater+%5C++3x+%2B+5+%2B+m)
![4x - 3x \ \textgreater \ 28 + 5 + m](https://tex.z-dn.net/?f=4x+-+3x++%5C+%5Ctextgreater+%5C+++28+%2B+5+%2B+m)
![x \ \textgreater \ 33 + m](https://tex.z-dn.net/?f=x+%5C+%5Ctextgreater+%5C++33+%2B+m)
1) (5; +∞)
Получаем 33+m=5 следовательно m = -28
2) (-2; +∞)
Получаем 33+m=-2, следовательно m = -35
3) (0; +<span>∞</span>)
Получаем 33+m=0, следовательно m=-33