Suppose That $\cot \alpha = -\frac{1}{3}$ And $90^{\circ} \ \textless \ \alpha \ \textless \ 180^{\circ}$. Find The Exact Values Of $\cos \frac{\alpha}{2}$ And $\tan \frac{\alpha}{2}$.$\cos \frac{\alpha}{2} =

Solving Trigonometric Identities: Finding Exact Values of Cosine and Tangent

Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles. It is a fundamental subject that has numerous applications in various fields, including physics, engineering, and navigation. In this article, we will focus on solving trigonometric identities, specifically finding the exact values of cosine and tangent of a half-angle.

Given Information

We are given that $\cot \alpha = -\frac{1}{3}$ and $90^{\circ} \ \textless \ \alpha \ \textless \ 180^{\circ}$. This information will be used to find the exact values of $\cos \frac{\alpha}{2}$ and $\tan \frac{\alpha}{2}$.

Recall of Half-Angle Formulas

To find the exact values of $\cos \frac{\alpha}{2}$ and $\tan \frac{\alpha}{2}$, we need to recall the half-angle formulas for cosine and tangent.

• $\cos \frac{\alpha}{2} = \pm \sqrt{\frac{1 + \cos \alpha}{2}}$
• $\tan \frac{\alpha}{2} = \frac{1 - \cos \alpha}{\sin \alpha}$

Finding the Value of Cosine

We are given that $\cot \alpha = -\frac{1}{3}$. We can rewrite this as $\tan \alpha = -3$. Since $90^{\circ} \ \textless \ \alpha \ \textless \ 180^{\circ}$, we know that $\tan \alpha$ is negative.

Using the half-angle formula for cosine, we can substitute the value of $\cos \alpha$ into the formula.

$\cos \alpha = \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha}$

Substituting the value of $\tan \alpha$, we get:

$\cos \alpha = \frac{1 - (-3)^2}{1 + (-3)^2} = \frac{1 - 9}{1 + 9} = \frac{-8}{10} = -\frac{4}{5}$

Now, we can substitute the value of $\cos \alpha$ into the half-angle formula for cosine.

$\cos \frac{\alpha}{2} = \pm \sqrt{\frac{1 + \cos \alpha}{2}} = \pm \sqrt{\frac{1 - \frac{4}{5}}{2}} = \pm \sqrt{\frac{\frac{1}{5}}{2}} = \pm \sqrt{\frac{1}{10}} = \pm \frac{1}{\sqrt{10}}$

Since $90^{\circ} \ \textless \ \alpha \ \textless \ 180^{\circ}$, we know that $\cos \frac{\alpha}{2}$ is negative.

Therefore, the exact value of $\cos \frac{\alpha}{2}$ is:

$\cos \frac{\alpha}{2} = -\frac{1}{\sqrt{10}}$

Finding the Value of Tangent

Using the half-angle formula for tangent, we can substitute the value of $\cos \alpha$ into the formula.

$\tan \frac{\alpha}{2} = \frac{1 - \cos \alpha}{\sin \alpha}$

We can rewrite this as:

$\tan \frac{\alpha}{2} = \frac{1 - \left(-\frac{4}{5}\right)}{\sin \alpha}$

Simplifying the expression, we get:

$\tan \frac{\alpha}{2} = \frac{1 + \frac{4}{5}}{\sin \alpha} = \frac{\frac{9}{5}}{\sin \alpha}$

We can rewrite this as:

$\tan \frac{\alpha}{2} = \frac{\frac{9}{5}}{\sqrt{1 - \cos^2 \alpha}}$

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

$\tan \frac{\alpha}{2} = \frac{\frac{9}{5}}{\sqrt{1 - \left(-\frac{4}{5}\right)^2}} = \frac{\frac{9}{5}}{\sqrt{1 - \frac{16}{25}}} = \frac{\frac{9}{5}}{\sqrt{\frac{9}{25}}} = \frac{\frac{9}{5}}{\frac{3}{5}} = 3$

Therefore, the exact value of $\tan \frac{\alpha}{2}$ is:

$\tan \frac{\alpha}{2} = 3$

In this article, we have found the exact values of $\cos \frac{\alpha}{2}$ and $\tan \frac{\alpha}{2}$ given that $\cot \alpha = -\frac{1}{3}$ and $90^{\circ} \ \textless \ \alpha \ \textless \ 180^{\circ}$. We have used the half-angle formulas for cosine and tangent to find the exact values.

The exact value of $\cos \frac{\alpha}{2}$ is:

$\cos \frac{\alpha}{2} = -\frac{1}{\sqrt{10}}$

The exact value of $\tan \frac{\alpha}{2}$ is:

$\tan \frac{\alpha}{2} = 3$

These values can be used to solve various trigonometric problems and applications.
Solving Trigonometric Identities: Q&A

In our previous article, we discussed how to find the exact values of cosine and tangent of a half-angle using the half-angle formulas. We also provided a step-by-step solution to a problem where $\cot \alpha = -\frac{1}{3}$ and $90^{\circ} \ \textless \ \alpha \ \textless \ 180^{\circ}$. In this article, we will provide a Q&A section to help you better understand the concepts and formulas discussed earlier.

Q: What is the half-angle formula for cosine?

A: The half-angle formula for cosine is:

$\cos \frac{\alpha}{2} = \pm \sqrt{\frac{1 + \cos \alpha}{2}}$

Q: What is the half-angle formula for tangent?

A: The half-angle formula for tangent is:

$\tan \frac{\alpha}{2} = \frac{1 - \cos \alpha}{\sin \alpha}$

Q: How do I find the value of cosine of a half-angle?

A: To find the value of cosine of a half-angle, you need to use the half-angle formula for cosine. You will need to substitute the value of $\cos \alpha$ into the formula and simplify the expression.

Q: How do I find the value of tangent of a half-angle?

A: To find the value of tangent of a half-angle, you need to use the half-angle formula for tangent. You will need to substitute the value of $\cos \alpha$ into the formula and simplify the expression.

Q: What is the relationship between the cotangent and tangent functions?

A: The cotangent function is the reciprocal of the tangent function. This means that $\cot \alpha = \frac{1}{\tan \alpha}$.

Q: How do I use the cotangent function to find the value of cosine of a half-angle?

A: To use the cotangent function to find the value of cosine of a half-angle, you need to first find the value of $\tan \alpha$. Then, you can use the half-angle formula for cosine to find the value of $\cos \frac{\alpha}{2}$.

Q: What is the significance of the half-angle formulas?

A: The half-angle formulas are important because they allow us to find the exact values of cosine and tangent of a half-angle. This is useful in solving various trigonometric problems and applications.

Q: Can I use the half-angle formulas to find the value of sine of a half-angle?

A: Yes, you can use the half-angle formulas to find the value of sine of a half-angle. However, you will need to use the Pythagorean identity $\sin^2 \alpha + \cos^2 \alpha = 1$ to find the value of $\sin \alpha$.

In this article, we have provided a Q&A section to help you better understand the concepts and formulas discussed earlier. We have covered topics such as the half-angle formulas, the relationship between the cotangent and tangent functions, and the significance of the half-angle formulas. We hope that this article has been helpful in clarifying any doubts you may have had about the subject.

If you are looking for additional resources to help you learn more about trigonometry, we recommend the following:

• Khan Academy: Trigonometry
• MIT OpenCourseWare: Trigonometry
• Wolfram Alpha: Trigonometry

These resources provide a wealth of information and examples to help you learn more about trigonometry.

To help you practice what you have learned, we have provided a set of practice problems below.

1. Find the exact value of $\cos \frac{\alpha}{2}$ given that $\cot \alpha = \frac{1}{2}$ and $90^{\circ} \ \textless \ \alpha \ \textless \ 180^{\circ}$.
2. Find the exact value of $\tan \frac{\alpha}{2}$ given that $\cot \alpha = -\frac{1}{2}$ and $90^{\circ} \ \textless \ \alpha \ \textless \ 180^{\circ}$.
3. Find the exact value of $\cos \frac{\alpha}{2}$ given that $\tan \alpha = 2$ and $90^{\circ} \ \textless \ \alpha \ \textless \ 180^{\circ}$.
4. Find the exact value of $\tan \frac{\alpha}{2}$ given that $\tan \alpha = -2$ and $90^{\circ} \ \textless \ \alpha \ \textless \ 180^{\circ}$.

We hope that these practice problems will help you to reinforce your understanding of the concepts and formulas discussed earlier.

1. $\cos \frac{\alpha}{2} = -\frac{1}{\sqrt{5}}$
2. $\tan \frac{\alpha}{2} = -2$
3. $\cos \frac{\alpha}{2} = -\frac{1}{\sqrt{5}}$
4. $\tan \frac{\alpha}{2} = 2$

We hope that this article has been helpful in clarifying any doubts you may have had about the subject. If you have any further questions or need additional help, please don't hesitate to ask.