Let $\csc A=\frac{\sqrt{10}}{3}$ With $A$ In Quadrant I. Find $\sec (2A)=$

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Introduction

In trigonometry, the secant function is the reciprocal of the cosine function. It is an essential function in trigonometric identities and formulas. In this article, we will explore how to find the value of $\sec (2A)$ given that $\csc A=\frac{\sqrt{10}}{3}$ with $A$ in Quadrant I.

Understanding the Given Information

We are given that $\csc A=\frac{\sqrt{10}}{3}$, where $A$ is in Quadrant I. This means that $\sin A = \frac{3}{\sqrt{10}}$. Since $A$ is in Quadrant I, both $\sin A$ and $\cos A$ are positive.

Finding $\cos A$

To find $\cos A$, we can use the Pythagorean identity: $\sin^2 A + \cos^2 A = 1$. We can rearrange this identity to solve for $\cos A$:

$$\cos^2 A = 1 - \sin^2 A$$

Substituting the value of $\sin A$, we get:

$$\cos^2 A = 1 - \left(\frac{3}{\sqrt{10}}\right)^2$$

Simplifying, we get:

$$\cos^2 A = 1 - \frac{9}{10} = \frac{1}{10}$$

Taking the square root of both sides, we get:

$$\cos A = \pm \sqrt{\frac{1}{10}}$$

Since $A$ is in Quadrant I, $\cos A$ is positive. Therefore, we have:

$$\cos A = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}}$$

Finding $\sec A$

The secant function is the reciprocal of the cosine function. Therefore, we can find $\sec A$ by taking the reciprocal of $\cos A$:

$$\sec A = \frac{1}{\cos A} = \frac{1}{\frac{1}{\sqrt{10}}} = \sqrt{10}$$

Finding $\sec (2A)$

To find $\sec (2A)$, we can use the double-angle formula for cosine:

$$\cos (2A) = 2\cos^2 A - 1$$

Substituting the value of $\cos A$, we get:

$$\cos (2A) = 2\left(\frac{1}{\sqrt{10}}\right)^2 - 1 = 2\left(\frac{1}{10}\right) - 1 = \frac{1}{5} - 1 = -\frac{4}{5}$$

Taking the reciprocal of both sides, we get:

$$\sec (2A) = \frac{1}{\cos (2A)} = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}$$

Conclusion

In this article, we found the value of $\sec (2A)$ given that $\csc A=\frac{\sqrt{10}}{3}$ with $A$ in Quadrant I. We used the Pythagorean identity to find $\cos A$, and then used the double-angle formula for cosine to find $\cos (2A)$. Finally, we took the reciprocal of $\cos (2A)$ to find $\sec (2A)$. The value of $\sec (2A)$ is $-\frac{5}{4}$.

Additional Information

The secant function is an essential function in trigonometry, and it is used in many formulas and identities. In this article, we used the double-angle formula for cosine to find $\sec (2A)$. This formula is a fundamental concept in trigonometry, and it is used to find the value of cosine for any angle.

Final Answer

The final answer is: $\boxed{-\frac{5}{4}}$

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Introduction

In our previous article, we found the value of $\sec (2A)$ given that $\csc A=\frac{\sqrt{10}}{3}$ with $A$ in Quadrant I. In this article, we will answer some common questions related to this problem.

Q: What is the relationship between $\csc A$ and $\sec A$?

A: The secant function is the reciprocal of the cosine function. Therefore, we can find $\sec A$ by taking the reciprocal of $\cos A$. In this case, we have $\csc A = \frac{\sqrt{10}}{3}$, and we can use the Pythagorean identity to find $\cos A$. Once we have $\cos A$, we can take its reciprocal to find $\sec A$.

Q: How do we find $\cos A$ given that $\csc A=\frac{\sqrt{10}}{3}$?

A: We can use the Pythagorean identity: $\sin^2 A + \cos^2 A = 1$. We can rearrange this identity to solve for $\cos A$:

$$\cos^2 A = 1 - \sin^2 A$$

Substituting the value of $\sin A$, we get:

$$\cos^2 A = 1 - \left(\frac{3}{\sqrt{10}}\right)^2$$

Simplifying, we get:

$$\cos^2 A = 1 - \frac{9}{10} = \frac{1}{10}$$

Taking the square root of both sides, we get:

$$\cos A = \pm \sqrt{\frac{1}{10}}$$

Since $A$ is in Quadrant I, $\cos A$ is positive. Therefore, we have:

$$\cos A = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}}$$

Q: How do we find $\sec (2A)$ given that $\csc A=\frac{\sqrt{10}}{3}$?

A: We can use the double-angle formula for cosine:

$$\cos (2A) = 2\cos^2 A - 1$$

Substituting the value of $\cos A$, we get:

$$\cos (2A) = 2\left(\frac{1}{\sqrt{10}}\right)^2 - 1 = 2\left(\frac{1}{10}\right) - 1 = \frac{1}{5} - 1 = -\frac{4}{5}$$

Taking the reciprocal of both sides, we get:

$$\sec (2A) = \frac{1}{\cos (2A)} = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}$$

Q: What is the final answer to the problem?

A: The final answer to the problem is $\boxed{-\frac{5}{4}}$.

Q: What is the significance of the double-angle formula for cosine?

A: The double-angle formula for cosine is a fundamental concept in trigonometry. It is used to find the value of cosine for any angle. In this case, we used the double-angle formula to find $\cos (2A)$, and then took its reciprocal to find $\sec (2A)$.

Q: How do we know that $A$ is in Quadrant I?

A: We are given that $A$ is in Quadrant I. This means that both $\sin A$ and $\cos A$ are positive.

Q: What is the relationship between $\sin A$ and $\cos A$?

A: The sine and cosine functions are related by the Pythagorean identity: $\sin^2 A + \cos^2 A = 1$. In this case, we have $\sin A = \frac{3}{\sqrt{10}}$, and we can use the Pythagorean identity to find $\cos A$.

Q: How do we find $\sin A$ given that $\csc A=\frac{\sqrt{10}}{3}$?

A: We can use the definition of the cosecant function: $\csc A = \frac{1}{\sin A}$. Therefore, we can find $\sin A$ by taking the reciprocal of $\csc A$:

$$\sin A = \frac{1}{\csc A} = \frac{1}{\frac{\sqrt{10}}{3}} = \frac{3}{\sqrt{10}}$$

Q: What is the final answer to the problem in terms of $\sin A$ and $\cos A$?

A: The final answer to the problem is $\boxed{-\frac{5}{4}}$. This can be expressed in terms of $\sin A$ and $\cos A$ as:

$$\sec (2A) = \frac{1}{\cos (2A)} = \frac{1}{2\cos^2 A - 1} = \frac{1}{2\left(\frac{1}{\sqrt{10}}\right)^2 - 1} = \frac{1}{2\left(\frac{1}{10}\right) - 1} = \frac{1}{\frac{1}{5} - 1} = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}$$