Вопрос задан 02.12.2023 в 12:26. Предмет Алгебра. Спрашивает Бабийчук Валя.

Постройте график функции y=1-cos2x/sin2x​

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Отвечает Насырова Азалия.

Ответ:

Построил (см. рисунок)


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Отвечает нейросеть ChatGpt. Будьте внимательны, ответы нейросети могут содержать ошибочные или выдуманные данные.

Graph of the Function y = (1 - cos(2x)) / sin(2x)

To graph the function y = (1 - cos(2x)) / sin(2x), we can start by analyzing its behavior and identifying key points.

1. Behavior of the Function: - The function is undefined when sin(2x) = 0, which occurs when 2x = nπ, where n is an integer. - At these points, the function has vertical asymptotes. - The function is also undefined when cos(2x) = 1, which occurs when 2x = 2nπ, where n is an integer. - At these points, the function has vertical asymptotes.

2. Key Points: - The function has a period of π, as sin(2x) and cos(2x) have a period of π. - The function has a horizontal asymptote at y = 1, as x approaches positive or negative infinity.

Based on this information, we can now proceed to graphing the function.

Graph of y = (1 - cos(2x)) / sin(2x)

To graph the function, we will plot key points and sketch the curve based on its behavior.

1. Vertical Asymptotes: - The function has vertical asymptotes at x = nπ/2, where n is an integer. - These points occur when sin(2x) = 0, which happens when 2x = nπ. - Therefore, the vertical asymptotes occur at x = nπ/2, where n is an integer.

2. Horizontal Asymptote: - The function has a horizontal asymptote at y = 1, as x approaches positive or negative infinity.

3. Key Points: - To find key points, we can substitute some values of x into the function and calculate the corresponding y-values. - For example, when x = 0, y = (1 - cos(0)) / sin(0) = 0 / 0, which is undefined. - Similarly, when x = π/4, y = (1 - cos(π/2)) / sin(π/2) = (1 - 0) / 1 = 1. - We can find more key points by substituting different values of x into the function.

4. Sketching the Curve: - Using the information from the vertical asymptotes, horizontal asymptote, and key points, we can sketch the curve of the function. - The curve will approach the vertical asymptotes as x approaches the values of nπ/2. - The curve will approach the horizontal asymptote y = 1 as x approaches positive or negative infinity. - The curve will pass through the key points we calculated earlier.

Please note that without specific values for x, it is not possible to provide an accurate graph. However, the information provided above should give you a general idea of the behavior and shape of the graph.

I hope this helps! Let me know if you have any further questions.

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