Let $\cos A = \frac{1}{\sqrt{10}}$ With $A$ In Quadrant I. Find $\cot (2A) = \, \square$.

Let $\cos A = \frac{1}{\sqrt{10}}$ with $A$ in Quadrant I. Find $\cot (2A) = \, \square$

In this article, we will explore the problem of finding the cotangent of double angle $2A$, given that the cosine of angle $A$ is $\frac{1}{\sqrt{10}}$ and $A$ lies in Quadrant I. We will use trigonometric identities and double angle formulas to find the required value.

We are given that $\cos A = \frac{1}{\sqrt{10}}$ and $A$ lies in Quadrant I. This means that the angle $A$ is acute and lies between $0^\circ$ and $90^\circ$. We need to find the value of $\cot (2A)$.

Before we proceed, let's recall some basic trigonometric identities:

• $\sin^2 A + \cos^2 A = 1$
• $\tan A = \frac{\sin A}{\cos A}$
• $\cot A = \frac{\cos A}{\sin A}$

We will use these identities to find the required value.

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We are given that $\cos A = \frac{1}{\sqrt{10}}$. We can use the Pythagorean identity to find the value of $\sin A$:

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

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

$\sin^2 A + \left(\frac{1}{\sqrt{10}}\right)^2 = 1$

Simplifying, we get:

$\sin^2 A + \frac{1}{10} = 1$

Subtracting $\frac{1}{10}$ from both sides, we get:

$\sin^2 A = \frac{9}{10}$

Taking the square root of both sides, we get:

$\sin A = \pm \sqrt{\frac{9}{10}}$

Since $A$ lies in Quadrant I, $\sin A$ is positive. Therefore:

$\sin A = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}$

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We can now find the value of $\tan A$ using the formula:

$\tan A = \frac{\sin A}{\cos A}$

Substituting the values of $\sin A$ and $\cos A$, we get:

$\tan A = \frac{\frac{3}{\sqrt{10}}}{\frac{1}{\sqrt{10}}} = 3$

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We can now find the value of $\cot A$ using the formula:

$\cot A = \frac{\cos A}{\sin A}$

Substituting the values of $\cos A$ and $\sin A$, we get:

$\cot A = \frac{\frac{1}{\sqrt{10}}}{\frac{3}{\sqrt{10}}} = \frac{1}{3}$

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We can now use the double angle formula for cotangent to find the value of $\cot (2A)$:

$\cot (2A) = \frac{\cot^2 A - 1}{2 \cot A}$

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

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

Simplifying, we get:

$\cot (2A) = \frac{\frac{1}{9} - 1}{\frac{2}{3}} = \frac{-\frac{8}{9}}{\frac{2}{3}} = -\frac{8}{9} \times \frac{3}{2} = -\frac{4}{3}$

Therefore, the value of $\cot (2A)$ is $-\frac{4}{3}$.

In this article, we used trigonometric identities and double angle formulas to find the value of $\cot (2A)$, given that $\cos A = \frac{1}{\sqrt{10}}$ and $A$ lies in Quadrant I. We found that $\cot (2A) = -\frac{4}{3}$.
Q&A: Let $\cos A = \frac{1}{\sqrt{10}}$ with $A$ in Quadrant I. Find $\cot (2A) = \, \square$

In our previous article, we explored the problem of finding the cotangent of double angle $2A$, given that the cosine of angle $A$ is $\frac{1}{\sqrt{10}}$ and $A$ lies in Quadrant I. We used trigonometric identities and double angle formulas to find the required value. In this article, we will answer some frequently asked questions related to this problem.

Q: What is the value of $\sin A$?

A: We can find the value of $\sin A$ using the Pythagorean identity:

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

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

$\sin^2 A + \left(\frac{1}{\sqrt{10}}\right)^2 = 1$

Simplifying, we get:

$\sin^2 A + \frac{1}{10} = 1$

Subtracting $\frac{1}{10}$ from both sides, we get:

$\sin^2 A = \frac{9}{10}$

Taking the square root of both sides, we get:

$\sin A = \pm \sqrt{\frac{9}{10}}$

Since $A$ lies in Quadrant I, $\sin A$ is positive. Therefore:

$\sin A = \sqrt{\frac{9}{10}} = \frac{3}{\sqrt{10}}$

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

A: We can find the value of $\tan A$ using the formula:

$\tan A = \frac{\sin A}{\cos A}$

Substituting the values of $\sin A$ and $\cos A$, we get:

$\tan A = \frac{\frac{3}{\sqrt{10}}}{\frac{1}{\sqrt{10}}} = 3$

Q: What is the value of $\cot A$?

A: We can find the value of $\cot A$ using the formula:

$\cot A = \frac{\cos A}{\sin A}$

Substituting the values of $\cos A$ and $\sin A$, we get:

$\cot A = \frac{\frac{1}{\sqrt{10}}}{\frac{3}{\sqrt{10}}} = \frac{1}{3}$

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

A: We can use the double angle formula for cotangent to find the value of $\cot (2A)$:

$\cot (2A) = \frac{\cot^2 A - 1}{2 \cot A}$

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

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

Simplifying, we get:

$\cot (2A) = \frac{\frac{1}{9} - 1}{\frac{2}{3}} = \frac{-\frac{8}{9}}{\frac{2}{3}} = -\frac{8}{9} \times \frac{3}{2} = -\frac{4}{3}$

Q: What is the final answer?

A: The final answer is $\boxed{-\frac{4}{3}}$.

In this article, we answered some frequently asked questions related to the problem of finding the cotangent of double angle $2A$, given that the cosine of angle $A$ is $\frac{1}{\sqrt{10}}$ and $A$ lies in Quadrant I. We hope that this article has been helpful in clarifying any doubts that you may have had.