Keisha And David Each Found The Same Value For $\cos \theta$, Given $\sin \theta=-\frac{8}{17}$. Their Solutions Are Shown Below:$\[ \begin{array}{|c|c|} \hline \text{Keisha's Solution} & \text{David's Solution} \\ \hline \tan ^2

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Introduction

In trigonometry, the relationship between the sine, cosine, and tangent functions is crucial for solving problems involving right triangles. Given the value of $\sin \theta = -\frac{8}{17}$, Keisha and David were tasked with finding the value of $\cos \theta$. Their approaches, although different, led to the same solution. In this article, we will explore their methods and discuss the underlying principles of trigonometric functions.

Keisha's Solution

Keisha started by using the Pythagorean identity, $\sin^2 \theta + \cos^2 \theta = 1$, to find the value of $\cos \theta$. She first squared the given value of $\sin \theta$:

$$\sin^2 \theta = \left(-\frac{8}{17}\right)^2 = \frac{64}{289}$$

Next, she substituted this value into the Pythagorean identity:

$$\frac{64}{289} + \cos^2 \theta = 1$$

To isolate $\cos^2 \theta$, Keisha subtracted $\frac{64}{289}$ from both sides:

$$\cos^2 \theta = 1 - \frac{64}{289} = \frac{225}{289}$$

Taking the square root of both sides, Keisha obtained:

$$\cos \theta = \pm \sqrt{\frac{225}{289}} = \pm \frac{15}{17}$$

Keisha's solution is valid because the cosine function can take both positive and negative values.

David's Solution

David, on the other hand, used the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$ to find the value of $\cos \theta$. He first squared both sides of the equation:

$$\tan^2 \theta = \left(\frac{\sin \theta}{\cos \theta}\right)^2 = \frac{\sin^2 \theta}{\cos^2 \theta}$$

Substituting the given value of $\sin \theta$, David obtained:

$$\tan^2 \theta = \frac{\left(-\frac{8}{17}\right)^2}{\cos^2 \theta} = \frac{\frac{64}{289}}{\cos^2 \theta}$$

Rearranging the equation, David isolated $\cos^2 \theta$:

$$\cos^2 \theta = \frac{64}{289 \tan^2 \theta}$$

To find the value of $\cos \theta$, David needed to find the value of $\tan \theta$. He used the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$ again:

$$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{8}{17}}{\cos \theta}$$

Squaring both sides, David obtained:

$$\tan^2 \theta = \left(\frac{-\frac{8}{17}}{\cos \theta}\right)^2 = \frac{\frac{64}{289}}{\cos^2 \theta}$$

Substituting this expression into the equation for $\cos^2 \theta$, David obtained:

$$\cos^2 \theta = \frac{64}{289 \cdot \frac{64}{289}} = \frac{289}{289} = 1$$

Taking the square root of both sides, David obtained:

$$\cos \theta = \pm \sqrt{1} = \pm 1$$

However, David's solution is not valid because the value of $\cos \theta$ cannot be $\pm 1$ given the value of $\sin \theta = -\frac{8}{17}$.

Discussion

Keisha and David's approaches to finding $\cos \theta$ demonstrate the importance of understanding the underlying principles of trigonometric functions. Keisha's solution, although more straightforward, relies on the Pythagorean identity, which is a fundamental concept in trigonometry. David's solution, although more complex, demonstrates the use of the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$ to find the value of $\cos \theta$.

In conclusion, Keisha and David's solutions, although different, led to the same value of $\cos \theta = \pm \frac{15}{17}$. Their approaches highlight the importance of understanding the relationships between trigonometric functions and the use of identities to solve problems involving right triangles.

Conclusion

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Q: What is the relationship between the sine, cosine, and tangent functions?

A: The sine, cosine, and tangent functions are related through the Pythagorean identity, $\sin^2 \theta + \cos^2 \theta = 1$, and the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$.

Q: How did Keisha find the value of $\cos \theta$?

A: Keisha used the Pythagorean identity, $\sin^2 \theta + \cos^2 \theta = 1$, to find the value of $\cos \theta$. She first squared the given value of $\sin \theta$ and then substituted this value into the Pythagorean identity to isolate $\cos^2 \theta$. Finally, she took the square root of both sides to obtain the value of $\cos \theta$.

Q: How did David find the value of $\cos \theta$?

A: David used the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$ to find the value of $\cos \theta$. He first squared both sides of the equation and then rearranged the equation to isolate $\cos^2 \theta$. However, David's solution was not valid because the value of $\cos \theta$ cannot be $\pm 1$ given the value of $\sin \theta = -\frac{8}{17}$.

Q: What is the importance of understanding the relationships between trigonometric functions?

A: Understanding the relationships between trigonometric functions is crucial for solving problems involving right triangles. The Pythagorean identity and the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$ are fundamental concepts in trigonometry that can be used to find the values of sine, cosine, and tangent functions.

Q: Can you provide an example of how to use the Pythagorean identity to find the value of $\cos \theta$?

A: Yes, here is an example:

Suppose we are given the value of $\sin \theta = \frac{3}{5}$. We can use the Pythagorean identity to find the value of $\cos \theta$ as follows:

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

$$\left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1$$

$$\frac{9}{25} + \cos^2 \theta = 1$$

$$\cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}$$

Taking the square root of both sides, we obtain:

$$\cos \theta = \pm \sqrt{\frac{16}{25}} = \pm \frac{4}{5}$$

Q: Can you provide an example of how to use the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$ to find the value of $\cos \theta$?

A: Yes, here is an example:

Suppose we are given the value of $\sin \theta = \frac{3}{5}$ and the value of $\tan \theta = \frac{3}{4}$. We can use the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$ to find the value of $\cos \theta$ as follows:

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

$$\frac{3}{4} = \frac{\frac{3}{5}}{\cos \theta}$$

$$\cos \theta = \frac{5}{4} \cdot \frac{3}{5} = \frac{3}{4}$$

However, this solution is not valid because the value of $\cos \theta$ cannot be $\frac{3}{4}$ given the value of $\sin \theta = \frac{3}{5}$.

Conclusion

In this article, we explored Keisha and David's approaches to finding $\cos \theta$ given the value of $\sin \theta = -\frac{8}{17}$. We also provided examples of how to use the Pythagorean identity and the identity $\tan \theta = \frac{\sin \theta}{\cos \theta}$ to find the values of sine, cosine, and tangent functions. The discussion highlights the importance of understanding the relationships between trigonometric functions and the use of identities to solve problems involving right triangles.