Так как
![\frac{\sqrt{2}}{2}= sin( \frac{\pi}{4} )=cos( \frac{\pi}{4})](https://tex.z-dn.net/?f=+%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%3D+sin%28+%5Cfrac%7B%5Cpi%7D%7B4%7D+%29%3Dcos%28+%5Cfrac%7B%5Cpi%7D%7B4%7D%29)
то:
![sin( \frac{\pi}{4} )*sin\alpha-cos( \frac{\pi}{4} -\alpha)](https://tex.z-dn.net/?f=sin%28+%5Cfrac%7B%5Cpi%7D%7B4%7D+%29%2Asin%5Calpha-cos%28+%5Cfrac%7B%5Cpi%7D%7B4%7D+-%5Calpha%29)
применим формулу косинус разности двух углов
![sin( \frac{\pi}{4} )*sin\alpha-(cos( \frac{\pi}{4})*cos(\alpha)+sin( \frac{\pi}{4})*sin\alpha)=\\=sin( \frac{\pi}{4} )*sin\alpha-sin( \frac{\pi}{4})*sin\alpha-cos( \frac{\pi}{4})*cos(\alpha)=-cos( \frac{\pi}{4})*cos(\alpha)\\=- \frac{\sqrt{2}}{2} *cos(\alpha)=- \frac{\sqrt{2}*cos\alpha}{2}](https://tex.z-dn.net/?f=sin%28+%5Cfrac%7B%5Cpi%7D%7B4%7D+%29%2Asin%5Calpha-%28cos%28+%5Cfrac%7B%5Cpi%7D%7B4%7D%29%2Acos%28%5Calpha%29%2Bsin%28+%5Cfrac%7B%5Cpi%7D%7B4%7D%29%2Asin%5Calpha%29%3D%5C%5C%3Dsin%28+%5Cfrac%7B%5Cpi%7D%7B4%7D+%29%2Asin%5Calpha-sin%28+%5Cfrac%7B%5Cpi%7D%7B4%7D%29%2Asin%5Calpha-cos%28+%5Cfrac%7B%5Cpi%7D%7B4%7D%29%2Acos%28%5Calpha%29%3D-cos%28+%5Cfrac%7B%5Cpi%7D%7B4%7D%29%2Acos%28%5Calpha%29%5C%5C%3D-+%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D+%2Acos%28%5Calpha%29%3D-+%5Cfrac%7B%5Csqrt%7B2%7D%2Acos%5Calpha%7D%7B2%7D+)
Ответ:
![- \frac{\sqrt{2}*cos\alpha}{2}](https://tex.z-dn.net/?f=-+%5Cfrac%7B%5Csqrt%7B2%7D%2Acos%5Calpha%7D%7B2%7D)
<span>1)квадраты противоположных чисел равны.
числа -n и n
(-n)^2 = n^2
(-1)^2 * n^2 = n^2
n^2 = n^2
2)кубы противоположных чисел противоположны
числа -k и k
(-k) ^ 3 не равен k^3
-1^3 *k^3 не равен k^3
- k^3 не равен k^3
противоположны</span>
А^2 *x + 4ax = a + 5x + 5
x( a^2 + 4a - 5) = a + 5
x = (a + 5)/ a(a + 4)
12х²-7х+1=0
D=b²-4ac
D=49-4·12=1 ; √D=1
x¹,²=-b±√D/2a
x¹=7-1/24=¼ ; x²=7+1/24=⅓
Відповідь: ⅓ і ¼