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myway
@myway
July 2022
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Найдите наибольшее значение ф-ции у=8(1/2 sin x/3+1/2 cos x/3)^2 +8
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nKrynka
Найдите наибольшее значение функции:
у=8(1/2 sin x/3+1/2 cos x/3)^2 +8
y = 8*1/2(( sin x/3)^2 + 2 sin x/3*cos x/3 + (cos x/3)^2) + 8 = 4*(1 + sin 2x/3) + 8 =
= 4 + 4sin 2x/3 + 8 = 12 + 4sin 2x/3
y = 12 + 4sin 2x/3
-1<=12 + 4sin 2x/3<=1
-1 - 12 <= 4sin 2x/3 <= 1 - 12
-13<= 4sin 2x/3 <= -11
-13/4 <= sin 2x/3 <= -11/4
Наибольшее значение функции y = 12 + 4sin 2x/3 равно -11/4
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Answers & Comments
у=8(1/2 sin x/3+1/2 cos x/3)^2 +8
y = 8*1/2(( sin x/3)^2 + 2 sin x/3*cos x/3 + (cos x/3)^2) + 8 = 4*(1 + sin 2x/3) + 8 =
= 4 + 4sin 2x/3 + 8 = 12 + 4sin 2x/3
y = 12 + 4sin 2x/3
-1<=12 + 4sin 2x/3<=1
-1 - 12 <= 4sin 2x/3 <= 1 - 12
-13<= 4sin 2x/3 <= -11
-13/4 <= sin 2x/3 <= -11/4
Наибольшее значение функции y = 12 + 4sin 2x/3 равно -11/4