Находим первую производную функции:
y' = 2x - 25/x²
или
y' = (2x³ - 25)/x²
Приравниваем ее к нулю:
2x - 25/x²<span> = 0</span>
x1<span> = 2, 32</span>
Вычисляем значения функции на концах отрезка
f(2, 32) = 16, 16
f(-10) = 97, 5
f(-1) = - 24
Ответ:fmin<span> = - 24, f</span>max<span> = 97, 5</span>
sin3π = 0
cos3π = -1
tg3π = 0
ctg3π не существует
<span>sin(-4π) = 0</span>
<span>cos(<span>-4π) = 1</span></span>
<span><span>tg(<span>-4π) = 0</span></span></span>
<span><span><span>ctg(<span>-4π) не существует</span></span></span></span>
<span><span><span><span>sin(-π/2) = -1</span></span></span></span>
<span><span><span><span>cos(-π/2) = 0</span></span></span></span>
<span><span><span><span>tg(-π/2) не существyет</span></span></span></span>
ctg(-π/2) = 0
sin(5π/2) = 1
cos(5π/2) = 0
tg(5π/2) не существует
cos(5π/2) = 0
sin(-5π/6) = -1/2
cos(-5π/6) = -√3/2
tg(-5π/6) = 1/√3
ctg(-5π/6) = √3
sin(3π/4) = √2/2
cos(3π/4) = - √2/2
tg(3π/4) = -1
ctg(3π/4) = -1
Ищи косинус по основному тригонометрическому тождеству.