Evaluate \sin \left(\cos ^{-1}\left(-\frac{15}{17}\right)\right ]. Enter Your Answer As A Fraction Using The Slash Bar (/).Answer Here:

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Introduction

Trigonometric expressions are a fundamental part of mathematics, and evaluating them requires a deep understanding of the underlying concepts. In this article, we will focus on evaluating the expression sin⁡(cos⁡−1(−1517))\sin \left(\cos ^{-1}\left(-\frac{15}{17}\right)\right). We will break down the problem into manageable steps, using trigonometric identities and properties to simplify the expression.

Understanding the Problem

The given expression involves the inverse cosine function, which is denoted by cos⁡−1x\cos^{-1}x. This function returns the angle whose cosine is equal to xx. In this case, we are given the value of xx as −1517-\frac{15}{17}.

To evaluate the expression, we need to find the angle whose cosine is equal to −1517-\frac{15}{17}. This angle is denoted by cos⁡−1(−1517)\cos^{-1}\left(-\frac{15}{17}\right).

Step 1: Finding the Angle

To find the angle, we can use the inverse cosine function. However, we need to be careful when using this function, as it returns an angle in the range [0,π][0, \pi].

Let's denote the angle as θ=cos⁡−1(−1517)\theta = \cos^{-1}\left(-\frac{15}{17}\right). We can then use the cosine function to find the value of cos⁡θ\cos \theta.

cos⁡θ=−1517\cos \theta = -\frac{15}{17}

Step 2: Using Trigonometric Identities

We can use the trigonometric identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 to find the value of sin⁡θ\sin \theta.

sin⁡2θ=1−cos⁡2θ\sin^2 \theta = 1 - \cos^2 \theta

sin⁡2θ=1−(−1517)2\sin^2 \theta = 1 - \left(-\frac{15}{17}\right)^2

sin⁡2θ=1−225289\sin^2 \theta = 1 - \frac{225}{289}

sin⁡2θ=64289\sin^2 \theta = \frac{64}{289}

Step 3: Finding the Value of sin⁡θ\sin \theta

We can now take the square root of both sides to find the value of sin⁡θ\sin \theta.

sin⁡θ=±64289\sin \theta = \pm \sqrt{\frac{64}{289}}

sin⁡θ=±817\sin \theta = \pm \frac{8}{17}

Since the angle θ\theta is in the range [0,π][0, \pi], the value of sin⁡θ\sin \theta is positive.

sin⁡θ=817\sin \theta = \frac{8}{17}

Conclusion

In this article, we evaluated the expression sin⁡(cos⁡−1(−1517))\sin \left(\cos ^{-1}\left(-\frac{15}{17}\right)\right). We broke down the problem into manageable steps, using trigonometric identities and properties to simplify the expression.

We found the angle whose cosine is equal to −1517-\frac{15}{17}, and then used the trigonometric identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 to find the value of sin⁡θ\sin \theta.

The final answer is 817\boxed{\frac{8}{17}}.

Additional Resources

For more information on trigonometric expressions and identities, please refer to the following resources:

Discussion

Please feel free to ask questions or provide feedback on this article. We would be happy to help you understand the concepts better.

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Introduction

In our previous article, we evaluated the expression sin⁡(cos⁡−1(−1517))\sin \left(\cos ^{-1}\left(-\frac{15}{17}\right)\right). We broke down the problem into manageable steps, using trigonometric identities and properties to simplify the expression.

In this article, we will provide a Q&A guide to help you understand the concepts better. We will cover common questions and topics related to evaluating trigonometric expressions.

Q: What is the inverse cosine function?

A: The inverse cosine function, denoted by cos⁡−1x\cos^{-1}x, returns the angle whose cosine is equal to xx. This function is also known as the arccosine function.

Q: How do I evaluate the expression sin⁡(cos⁡−1(−1517))\sin \left(\cos ^{-1}\left(-\frac{15}{17}\right)\right)?

A: To evaluate this expression, you need to follow these steps:

  1. Find the angle whose cosine is equal to −1517-\frac{15}{17}.
  2. Use the trigonometric identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 to find the value of sin⁡θ\sin \theta.
  3. Take the square root of both sides to find the value of sin⁡θ\sin \theta.

Q: What is the range of the inverse cosine function?

A: The range of the inverse cosine function is [0,π][0, \pi]. This means that the angle returned by the inverse cosine function is always between 0 and π\pi.

Q: How do I use the trigonometric identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1?

A: To use this identity, you need to substitute the value of cos⁡θ\cos \theta into the equation. For example, if cos⁡θ=−1517\cos \theta = -\frac{15}{17}, you can substitute this value into the equation to get:

sin⁡2θ=1−(−1517)2\sin^2 \theta = 1 - \left(-\frac{15}{17}\right)^2

sin⁡2θ=1−225289\sin^2 \theta = 1 - \frac{225}{289}

sin⁡2θ=64289\sin^2 \theta = \frac{64}{289}

Q: What is the final answer to the expression sin⁡(cos⁡−1(−1517))\sin \left(\cos ^{-1}\left(-\frac{15}{17}\right)\right)?

A: The final answer to this expression is 817\boxed{\frac{8}{17}}.

Q: Can I use the inverse sine function to evaluate the expression sin⁡(cos⁡−1(−1517))\sin \left(\cos ^{-1}\left(-\frac{15}{17}\right)\right)?

A: No, you cannot use the inverse sine function to evaluate this expression. The inverse sine function returns the angle whose sine is equal to xx, but the expression sin⁡(cos⁡−1(−1517))\sin \left(\cos ^{-1}\left(-\frac{15}{17}\right)\right) involves the inverse cosine function.

Conclusion

In this article, we provided a Q&A guide to help you understand the concepts better. We covered common questions and topics related to evaluating trigonometric expressions.

We hope this guide has been helpful in understanding the concepts of trigonometric expressions and identities. If you have any further questions or need additional clarification, please don't hesitate to ask.

Additional Resources

For more information on trigonometric expressions and identities, please refer to the following resources:

Discussion

Please feel free to ask questions or provide feedback on this article. We would be happy to help you understand the concepts better.

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