To find the coordinates of the point $ P(x, y) $ on the terminal ray of angle $ \theta $ given that $ \csc \theta = -\frac{5}{2} $, we can follow these steps:

1. **Understand the relationship between cosecant and sine**:
   \[
   \csc \theta = \frac{1}{\sin \theta}
   \]
   Therefore, if $ \csc \theta = -\frac{5}{2} $, then:
   \[
   \sin \theta = -\frac{2}{5}
   \]

2. **Determine the quadrant**:
   Since $ \theta $ is between $ \pi $ and $ \frac{3\pi}{2} $, it is in the third quadrant where both sine and cosine are negative.

3. **Use the Pythagorean identity**:
   We know that:
   \[
   \sin^2 \theta + \cos^2 \theta = 1
   \]
   Substituting $ \sin \theta $:
   \[
   \left(-\frac{2}{5}\right)^2 + \cos^2 \theta = 1
   \]
   This simplifies to:
   \[
   \frac{4}{25} + \cos^2 \theta = 1
   \]
   \[
   \cos^2 \theta = 1 - \frac{4}{25} = \frac{25}{25} - \frac{4}{25} = \frac{21}{25}
   \]
   Taking the square root gives:
   \[
   \cos \theta = -\sqrt{\frac{21}{25}} = -\frac{\sqrt{21}}{5}
   \]
   (negative because cosine is negative in the third quadrant).

4. **Find the coordinates**:
   The coordinates $ P(x, y) $ can be expressed in terms of the radius $ r $:
   \[
   x = r \cos \theta, \quad y = r \sin \theta
   \]
   We can choose $ r = 5 $ (the hypotenuse) for simplicity:
   \[
   x = 5 \left(-\frac{\sqrt{21}}{5}\right) = -\sqrt{21}
   \]
   \[
   y = 5 \left(-\frac{2}{5}\right) = -2
   \]

Thus, the coordinates of point $ P $ are $ P(-\sqrt{21}, -2) $.

The correct answer is:
$ P(-\sqrt{21}, -2) $

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Question

$The point $ P(x, y) $ is on the terminal ray of angle $ \theta $. If $ \theta $ is between $ \pi $ radians and $ \frac{3 \pi}{2} $ radians and \( \csc$

The point $ P(x, y) $ is on the terminal ray of angle $ \theta $. If $ \theta $ is between $ \pi $ radians and $ \frac{3 \pi}{2} $ radians and $ \csc \theta=-\frac{5}{2} $, what are the coordinates of $ P(x, y) $ ? $ P(-\sqrt{21},-2) $ $ P(\sqrt{21},-2) $ $ P(-2, \sqrt{21}) $ $ P(-2,-\sqrt{21}) $

Ask by Mcguire Bowen.

Mar 22,2025 00:23

UpStudy AI Solution

Tutor-Verified Answer

Answer

The coordinates of point $ P $ are $ (-\sqrt{21}, -2) $.

Solution

To find the coordinates of the point $ P(x, y) $ on the terminal ray of angle $ \theta $ given that $ \csc \theta = -\frac{5}{2} $, we can follow these steps: 1. **Understand the relationship between cosecant and sine**: \[ \csc \theta = \frac{1}{\sin \theta} \] Therefore, if $ \csc \theta = -\frac{5}{2} $, then: \[ \sin \theta = -\frac{2}{5} \] 2. **Determine the quadrant**: Since $ \theta $ is between $ \pi $ and $ \frac{3\pi}{2} $, it is in the third quadrant where both sine and cosine are negative. 3. **Use the Pythagorean identity**: We know that: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting $ \sin \theta $: \[ \left(-\frac{2}{5}\right)^2 + \cos^2 \theta = 1 \] This simplifies to: \[ \frac{4}{25} + \cos^2 \theta = 1 \] \[ \cos^2 \theta = 1 - \frac{4}{25} = \frac{25}{25} - \frac{4}{25} = \frac{21}{25} \] Taking the square root gives: \[ \cos \theta = -\sqrt{\frac{21}{25}} = -\frac{\sqrt{21}}{5} \] (negative because cosine is negative in the third quadrant). 4. **Find the coordinates**: The coordinates $ P(x, y) $ can be expressed in terms of the radius $ r $: \[ x = r \cos \theta, \quad y = r \sin \theta \] We can choose $ r = 5 $ (the hypotenuse) for simplicity: \[ x = 5 \left(-\frac{\sqrt{21}}{5}\right) = -\sqrt{21} \] \[ y = 5 \left(-\frac{2}{5}\right) = -2 \] Thus, the coordinates of point $ P $ are $ P(-\sqrt{21}, -2) $. The correct answer is: $ P(-\sqrt{21}, -2) $

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The Deep Dive

To find the coordinates of point $ P(x, y) $, we start with the given information about $ \csc \theta $. We know that $ \csc \theta = \frac{1}{\sin \theta} $, and since $ \csc \theta = -\frac{5}{2} $, we have: \[ \sin \theta = -\frac{2}{5} \] Since $ \theta $ is between $ \pi $ radians and $ \frac{3\pi}{2} $ radians, this places $ \theta $ in the third quadrant, where both sine and cosine are negative. We will use the identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting for $ \sin \theta $: \[ \left(-\frac{2}{5}\right)^2 + \cos^2 \theta = 1 \] \[ \frac{4}{25} + \cos^2 \theta = 1 \] \[ \cos^2 \theta = 1 - \frac{4}{25} = \frac{25}{25} - \frac{4}{25} = \frac{21}{25} \] Taking the square root, we find $ \cos \theta $: \[ \cos \theta = -\sqrt{\frac{21}{25}} = -\frac{\sqrt{21}}{5} \] Now that we have both sine and cosine values, we can express the coordinates of point $ P $ as follows, where the radius $ r $ can be taken as 5 (as it relates to $ c $ where $ r = \frac{1}{\csc \theta} $): \[ x = r \cos \theta = 5 \left(-\frac{\sqrt{21}}{5}\right) = -\sqrt{21} \] \[ y = r \sin \theta = 5 \left(-\frac{2}{5}\right) = -2 \] Thus, the coordinates of point $ P(x, y) $ are: \[ P(-\sqrt{21}, -2) \] So the correct answer is: $ P(-\sqrt{21},-2) $

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