7) Find the exact value of the trig function. Don't forget. to rationalize any denominators. sec 8 (-√15,7)
7) Find the exact value of the trig function. Don't forget. to rationalize any denominators. sec 8 (-√15,7)
Holt Mcdougal Larson Pre-algebra: Student Edition 2012
1st Edition
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
ChapterCSR: Contents Of Student Resources
Section: Chapter Questions
Problem 1.38EP
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![### Trigonometry - Finding Exact Values
#### Problem
"Find the exact value of the trig function. Don’t forget to rationalize any denominators."
#### Diagram Explanation
The diagram presents a coordinate system with the \(x\)-axis and \(y\)-axis. A point \((- \sqrt{15}, 7)\) on the coordinate system is indicated, located in the second quadrant. From the origin, a vector points towards this coordinate, showing the direction of the angle \(\theta\). The vector forms an angle \(\theta\) with the positive \(x\)-axis, sweeping counter-clockwise, depicted by a red arc.
Please proceed to solve for the secant (\(\sec\)) of \(\theta\) using the provided coordinates. The solution involves the following steps:
1. **Identify \(x\) and \(y\):**
\[
x = -\sqrt{15}, \quad y = 7
\]
2. **Find the hypotenuse \(r\):**
Using the Pythagorean theorem for a right-angled triangle:
\[
r = \sqrt{x^2 + y^2} = \sqrt{(\sqrt{15})^2 + 7^2} = \sqrt{15 + 49} = \sqrt{64} = 8
\]
3. **Calculate \(\sec \theta\):**
\[
\sec \theta = \frac{r}{x} = \frac{8}{-\sqrt{15}} = -\frac{8}{\sqrt{15}}
\]
4. **Rationalize the denominator:**
\[
-\frac{8}{\sqrt{15}} \times \frac{\sqrt{15}}{\sqrt{15}} = -\frac{8\sqrt{15}}{15}
\]
Thus, the exact value of \(\sec \theta\) is:
\[
\sec \theta = -\frac{8\sqrt{15}}{15}
\]
Remember, rationalizing the denominator is a crucial step in expressing trigonometric functions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F24a6a4c4-f26c-45ad-82dd-c449c9817221%2F46b46f67-d02b-43dc-b246-182be64bce97%2F57e5djv_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Trigonometry - Finding Exact Values
#### Problem
"Find the exact value of the trig function. Don’t forget to rationalize any denominators."
#### Diagram Explanation
The diagram presents a coordinate system with the \(x\)-axis and \(y\)-axis. A point \((- \sqrt{15}, 7)\) on the coordinate system is indicated, located in the second quadrant. From the origin, a vector points towards this coordinate, showing the direction of the angle \(\theta\). The vector forms an angle \(\theta\) with the positive \(x\)-axis, sweeping counter-clockwise, depicted by a red arc.
Please proceed to solve for the secant (\(\sec\)) of \(\theta\) using the provided coordinates. The solution involves the following steps:
1. **Identify \(x\) and \(y\):**
\[
x = -\sqrt{15}, \quad y = 7
\]
2. **Find the hypotenuse \(r\):**
Using the Pythagorean theorem for a right-angled triangle:
\[
r = \sqrt{x^2 + y^2} = \sqrt{(\sqrt{15})^2 + 7^2} = \sqrt{15 + 49} = \sqrt{64} = 8
\]
3. **Calculate \(\sec \theta\):**
\[
\sec \theta = \frac{r}{x} = \frac{8}{-\sqrt{15}} = -\frac{8}{\sqrt{15}}
\]
4. **Rationalize the denominator:**
\[
-\frac{8}{\sqrt{15}} \times \frac{\sqrt{15}}{\sqrt{15}} = -\frac{8\sqrt{15}}{15}
\]
Thus, the exact value of \(\sec \theta\) is:
\[
\sec \theta = -\frac{8\sqrt{15}}{15}
\]
Remember, rationalizing the denominator is a crucial step in expressing trigonometric functions.
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