Let $\csc A = \sqrt{5}$ With $A$ In Quadrant I. Find $\csc (2A$\].

Introduction

In trigonometry, the cosecant function is the reciprocal of the sine function. It is defined as $\csc A = \frac{1}{\sin A}$. Given that $\csc A = \sqrt{5}$ and $A$ is in quadrant I, we need to find the value of $\csc (2A)$. To do this, we will use the double-angle identity for sine and the given value of $\csc A$ to find the value of $\sin A$.

Finding the Value of $\sin A$

Since $\csc A = \frac{1}{\sin A}$, we can rewrite the given equation as $\sin A = \frac{1}{\csc A} = \frac{1}{\sqrt{5}}$. To rationalize the denominator, we multiply the numerator and denominator by $\sqrt{5}$, which gives us $\sin A = \frac{\sqrt{5}}{5}$.

Using the Double-Angle Identity for Sine

The double-angle identity for sine is given by $\sin (2A) = 2\sin A \cos A$. We can use this identity to find the value of $\sin (2A)$. However, we need to find the value of $\cos A$ first.

Finding the Value of $\cos A$

Since $A$ is in quadrant I, the cosine function is positive. We can use the Pythagorean identity $\cos^2 A + \sin^2 A = 1$ to find the value of $\cos A$. Substituting the value of $\sin A$ we found earlier, we get $\cos^2 A + \left(\frac{\sqrt{5}}{5}\right)^2 = 1$. Simplifying, we get $\cos^2 A + \frac{1}{5} = 1$. Subtracting $\frac{1}{5}$ from both sides, we get $\cos^2 A = \frac{4}{5}$. Taking the square root of both sides, we get $\cos A = \pm \frac{2}{\sqrt{5}}$. Since $A$ is in quadrant I, the cosine function is positive, so we take the positive value, $\cos A = \frac{2}{\sqrt{5}}$.

Finding the Value of $\sin (2A)$

Now that we have the values of $\sin A$ and $\cos A$, we can use the double-angle identity for sine to find the value of $\sin (2A)$. Substituting the values of $\sin A$ and $\cos A$, we get $\sin (2A) = 2\left(\frac{\sqrt{5}}{5}\right)\left(\frac{2}{\sqrt{5}}\right) = \frac{4}{5}$.

Finding the Value of $\csc (2A)$

Finally, we can find the value of $\csc (2A)$ by taking the reciprocal of $\sin (2A)$. We have $\csc (2A) = \frac{1}{\sin (2A)} = \frac{1}{\frac{4}{5}} = \frac{5}{4}$.

Conclusion

In this article, we used the given value of $\csc A = \sqrt{5}$ and the double-angle identity for sine to find the value of $\csc (2A)$. We first found the value of $\sin A$ by taking the reciprocal of $\csc A$. Then, we used the Pythagorean identity to find the value of $\cos A$. Finally, we used the double-angle identity for sine to find the value of $\sin (2A)$ and took its reciprocal to find the value of $\csc (2A)$. The final answer is $\boxed{\frac{5}{4}}$.

Additional Information

• The cosecant function is the reciprocal of the sine function.
• The double-angle identity for sine is given by $\sin (2A) = 2\sin A \cos A$.
• The Pythagorean identity is given by $\cos^2 A + \sin^2 A = 1$.
• The value of $\csc (2A)$ is the reciprocal of $\sin (2A)$.

References

• [1] "Trigonometry" by Michael Corral
• [2] "Precalculus" by James Stewart
• [3] "Trigonometry for Dummies" by Mary Jane Sterling

Related Articles

• Finding the Value of $\csc A$ Given the Value of $\sin A$
• Using the Double-Angle Identity for Sine to Find the Value of $\sin (2A)$
• Finding the Value of $\csc (2A)$ Given the Value of $\sin (2A)$
  

Q: What is the cosecant function?

A: The cosecant function is the reciprocal of the sine function. It is defined as $\csc A = \frac{1}{\sin A}$.

Q: What is the given value of $\csc A$?

A: The given value of $\csc A$ is $\sqrt{5}$.

Q: What is the quadrant of angle $A$?

A: The angle $A$ is in quadrant I.

Q: How do we find the value of $\sin A$?

A: We can find the value of $\sin A$ by taking the reciprocal of $\csc A$. So, $\sin A = \frac{1}{\csc A} = \frac{1}{\sqrt{5}}$.

Q: How do we rationalize the denominator of $\sin A$?

A: We can rationalize the denominator of $\sin A$ by multiplying the numerator and denominator by $\sqrt{5}$. This gives us $\sin A = \frac{\sqrt{5}}{5}$.

Q: What is the double-angle identity for sine?

A: The double-angle identity for sine is given by $\sin (2A) = 2\sin A \cos A$.

Q: How do we find the value of $\cos A$?

A: We can find the value of $\cos A$ by using the Pythagorean identity $\cos^2 A + \sin^2 A = 1$. Substituting the value of $\sin A$, we get $\cos^2 A + \left(\frac{\sqrt{5}}{5}\right)^2 = 1$. Simplifying, we get $\cos^2 A + \frac{1}{5} = 1$. Subtracting $\frac{1}{5}$ from both sides, we get $\cos^2 A = \frac{4}{5}$. Taking the square root of both sides, we get $\cos A = \pm \frac{2}{\sqrt{5}}$. Since $A$ is in quadrant I, the cosine function is positive, so we take the positive value, $\cos A = \frac{2}{\sqrt{5}}$.

Q: How do we find the value of $\sin (2A)$?

A: We can find the value of $\sin (2A)$ by using the double-angle identity for sine. Substituting the values of $\sin A$ and $\cos A$, we get $\sin (2A) = 2\left(\frac{\sqrt{5}}{5}\right)\left(\frac{2}{\sqrt{5}}\right) = \frac{4}{5}$.

Q: How do we find the value of $\csc (2A)$?

A: We can find the value of $\csc (2A)$ by taking the reciprocal of $\sin (2A)$. We have $\csc (2A) = \frac{1}{\sin (2A)} = \frac{1}{\frac{4}{5}} = \frac{5}{4}$.

Q: What is the final answer?

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

Q: What are some additional concepts related to this problem?

A: Some additional concepts related to this problem include the cosecant function, the double-angle identity for sine, and the Pythagorean identity.

Q: What are some references for further reading?

A: Some references for further reading include "Trigonometry" by Michael Corral, "Precalculus" by James Stewart, and "Trigonometry for Dummies" by Mary Jane Sterling.

Q: What are some related articles?

A: Some related articles include "Finding the Value of $\csc A$ Given the Value of $\sin A$", "Using the Double-Angle Identity for Sine to Find the Value of $\sin (2A)$", and "Finding the Value of $\csc (2A)$ Given the Value of $\sin (2A)$".