Cosα+cos2α+cos6α+cos7α=(cosα+cos7α)+(cos2α+cos6α)=
=2·cos<span>(α+7α)/2</span>·cos((α-7α)/2+2·cos(2α+6α)/2·cos(2α-6α)/2=
=2cos4α·cos(-3α)+2·cos4α·cos(-2α)=
=2·cos4α·(cos3α+cos2α)=2cos4α·2·cos5α/2·cosα/2=
=4cos4α·cos5α/2·cosα/2;
F(x)=(x²-6x+5)^7
f'(x)=14(x-3)(x²-6x+5)^6
Y = -x^3+3*x-33
[-3;3]
Находим первую производную функции:
y' = -3x2+3
Приравниваем ее к нулю:
-3x2+3 = 0
x1<span> = -1</span>
x2<span> = 1</span>
Вычисляем значения функции на концах отрезка
f(-1) = -35
f(1) = -31
f(-3) = -15
f(3) = -51
Ответ: <span>f</span>max<span> = -15</span>