Предмет: Алгебра, автор: Yaneznayuaaaa

найти множество значений функции y=2- sinxcosx
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Ответы

Автор ответа: dzhakypovaajrin
1

Ответ:

y = (sin x + cos x)^2 = sin^2 x + cos^2 x + 2sin x*cos x = 1 + sin 2x

Синус принимает значения от -1 до 1, поэтому

y(min) = 1 - 1 = 0; y(max) = 1 + 1 = 2

Ответ: y ∈ [0; 2]

Объяснение:

Автор ответа: bb573878
1

Ответ:

[1.5; 2.5]

Объяснение:

\displaystyle\bf\\y=2-sinxcosx=2-\frac{2sinxcosx}{2} =2-\frac{sin(2x)}{2}\\\\-1\leq  sin(2x)\leq 1\\\\-\frac{1}{2} \leq \frac{sin(2x)}{2} \leq \frac{1}{2} \\\\-\frac{1}{2} \leq -\frac{sin(2x)}{2} \leq \frac{1}{2} \\\\2-\frac{1}{2} \leq 2-\frac{sin(2x)}{2} \leq 2+\frac{1}{2} \\\\1.5 \leq 2-\frac{sin(2x)}{2} \leq 2.5\\\\1.5\leq 2-sinxcosx\leq 2.5

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