Question

Math

Posted 4 months ago

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Below is the graph of a trigonometric function. It intersects its midline at $\left(\frac{1}{2} \pi,-1\right)$ and again at $\left(\frac{5}{4} \pi,-1\right)$.
What is the period of the function? Give an exact value.
$\square$ units
```

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Answer from Sia

Posted 4 months ago

Solution by Steps

step 1

Identify the two points where the function intersects its midline

step 2

The points given are $\left(\frac{1}{2} \pi, -1\right)$ and $\left(\frac{5}{4} \pi, -1\right)$

step 3

Calculate the distance along the x-axis between these two points to find the period

step 4

The distance is $\frac{5}{4} \pi - \frac{1}{2} \pi = \frac{5}{4} \pi - \frac{2}{4} \pi = \frac{3}{4} \pi$

step 5

Since the function intersects its midline twice within one period, the total period is twice the distance calculated

step 6

The period of the function is $2 \times \frac{3}{4} \pi = \frac{3}{2} \pi$

[1] Answer

$\frac{3}{2} \pi$ units

Key Concept

Period of a Trigonometric Function

Explanation

The period of a trigonometric function is the horizontal length of one complete cycle of the wave. It can be determined by finding the distance between two consecutive points where the function intersects its midline and then multiplying by the number of times the function intersects the midline in one period.

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