Question

Не используя производной, найдите наибольшее значение функции y=sinx(12cosx-5sinx)

Answer 1

12cosx - 5sinx = 13( 12/13 cosx - 5/13 sinx) = {(12/13)^2 + (5/13)^2 = (144+25)/169 = 1 -> 12/13 = cos(a), 5/13 = sin(a)} = 13(cosa cosx - sina sinx) = 13cos(x + a)

sinx(12cosx - 5sinx) = 13 sinx cos(x+a) = 13/2[ sinx cos(x+a) + sin(x+a) cosx + sinx cos(x+a) - sin(x+a) cos(x) ] = 13/2[ sin(x + x + a) + sin(x - x - a) ] = 13/2 [ sin(2x + a) - sina ] = 13/2 sin(2x+a) - 13/2 * 5/13 = 13/2 sin(2x + a) - 5/2

sin(2x + a) <= 1; при x = (П/2 - a)/2, 2x + a = П/2, sin(2x + a) = 1

Макс. значение: 13/2 - 5/2 = 8/2 = 4

Answer 2

$y=sin(x)\cdot (12cos(x)-5sin(x))=13sin(x)\cdot (\frac{12}{13}cos(x)-\frac{5}{13}sin(x))\\\\(12/13)^2+(5/13)^2=1\\\\cos\varphi =12/13\\sin\varphi=5/13$

$y=13sin(x)(cos(\varphi)cos(x)-sin(x)sin(\varphi))=13sin(x)cos(x+\varphi)=\\\\=\frac{13}{2}(sin(x+x+\varphi)+sin(x-x-\varphi))=\frac{13}{2}(sin(2x+\varphi)-\frac{5}{13})=\\\\=\frac{13}{2}sin(2x+\varphi)-2.5$

$-1\leq sin(2x+\varphi) \leq 1\\\\-\frac{13}{2}\leq \frac{13}{2}sin(2x+\varphi) \leq \frac{13}{2}\\\\-6.5-2.5\leq \frac{13}{2}sin(2x+\varphi)-2.5 \leq 6.5-2.5\\\\-9\leq \frac{13}{2}sin(2x+\varphi)-2.5 \leq 4$

Значит наибольшее значение функции у равно 4.