Without Using A Calculator, Find The Matching Value For \[$ F \$\] In This Function. Give An Exact Answer. The Independent Variable Is In Radians.If $ F(x) = \sin \left(\sin^{-1} X\right) $, Find $ F\left(\frac{1}{2}\right) $.

Introduction

Trigonometric functions are a fundamental part of mathematics, and solving them without a calculator requires a deep understanding of the underlying concepts. In this article, we will explore how to find the matching value for $ f(x) $ in the given function $ f(x) = \sin \left(\sin^{-1} x\right) $, specifically when the independent variable is $ x = \frac{1}{2} $.

Understanding the Function

The given function is $ f(x) = \sin \left(\sin^{-1} x\right) $. This function involves the composition of the sine function and its inverse. The sine function takes an angle as input and returns a value between -1 and 1. The inverse sine function, denoted as $ \sin^{-1} x $, takes a value between -1 and 1 as input and returns an angle in radians.

Finding the Matching Value

To find the matching value for $ f(x) $ when $ x = \frac{1}{2} $, we need to substitute $ x = \frac{1}{2} $ into the function. This gives us:

$$f\left(\frac{1}{2}\right) = \sin \left(\sin^{-1} \frac{1}{2}\right)$$

Using the Definition of Inverse Sine

The inverse sine function $ \sin^{-1} x $ returns an angle in radians. When $ x = \frac{1}{2} $, the inverse sine function returns an angle $ \theta $ such that $ \sin \theta = \frac{1}{2} $. We know that $ \sin \frac{\pi}{6} = \frac{1}{2} $, so $ \sin^{-1} \frac{1}{2} = \frac{\pi}{6} $.

Substituting the Value

Now that we have found the value of $ \sin^{-1} \frac{1}{2} $, we can substitute it into the original function:

$$f\left(\frac{1}{2}\right) = \sin \left(\frac{\pi}{6}\right)$$

Evaluating the Sine Function

The sine function is periodic with a period of $ 2\pi $. The sine function is also symmetric about the origin, meaning that $ \sin (-\theta) = -\sin \theta $. We know that $ \sin \frac{\pi}{6} = \frac{1}{2} $, so:

$$f\left(\frac{1}{2}\right) = \frac{1}{2}$$

Conclusion

In this article, we have shown how to find the matching value for $ f(x) $ in the given function $ f(x) = \sin \left(\sin^{-1} x\right) $, specifically when the independent variable is $ x = \frac{1}{2} $. We used the definition of the inverse sine function and the properties of the sine function to evaluate the expression. The final answer is $ f\left(\frac{1}{2}\right) = \frac{1}{2} $.

Additional Examples

To further illustrate the concept, let's consider a few more examples.

Example 1

Find $ f(0) $ in the function $ f(x) = \sin \left(\sin^{-1} x\right) $.

Solution

We know that $ \sin^{-1} 0 = 0 $, so:

$$f(0) = \sin (0) = 0$$

Example 2

Find $ f(1) $ in the function $ f(x) = \sin \left(\sin^{-1} x\right) $.

Solution

We know that $ \sin^{-1} 1 = \frac{\pi}{2} $, so:

$$f(1) = \sin \left(\frac{\pi}{2}\right) = 1$$

Example 3

Find $ f(-1) $ in the function $ f(x) = \sin \left(\sin^{-1} x\right) $.

Solution

We know that $ \sin^{-1} (-1) = -\frac{\pi}{2} $, so:

$$f(-1) = \sin \left(-\frac{\pi}{2}\right) = -1$$

Conclusion

Introduction

In our previous article, we explored how to find the matching value for $ f(x) $ in the given function $ f(x) = \sin \left(\sin^{-1} x\right) $, specifically when the independent variable is $ x = \frac{1}{2} $. In this article, we will provide a Q&A section to further clarify any doubts and provide additional examples.

Q: What is the definition of the inverse sine function?

A: The inverse sine function, denoted as $ \sin^{-1} x $, takes a value between -1 and 1 as input and returns an angle in radians.

Q: How do I find the value of $ \sin^{-1} x $?

A: To find the value of $ \sin^{-1} x $, you need to find an angle $ \theta $ such that $ \sin \theta = x $. For example, if $ x = \frac{1}{2} $, then $ \sin^{-1} \frac{1}{2} = \frac{\pi}{6} $.

Q: What is the relationship between the sine function and its inverse?

A: The sine function and its inverse are related in that the inverse sine function returns an angle in radians, and the sine function takes an angle as input and returns a value between -1 and 1.

Q: Can I use a calculator to find the value of $ f(x) $?

A: While a calculator can be useful, it is not necessary to find the value of $ f(x) $. By understanding the definition of the inverse sine function and the properties of the sine function, you can evaluate the expression without a calculator.

Q: How do I evaluate the expression $ f(x) = \sin \left(\sin^{-1} x\right) $?

A: To evaluate the expression $ f(x) = \sin \left(\sin^{-1} x\right) $, you need to substitute the value of $ \sin^{-1} x $ into the sine function. For example, if $ x = \frac{1}{2} $, then $ f\left(\frac{1}{2}\right) = \sin \left(\frac{\pi}{6}\right) $.

Q: What is the final answer for $ f\left(\frac{1}{2}\right) $?

A: The final answer for $ f\left(\frac{1}{2}\right) $ is $ \frac{1}{2} $.

Q: Can I use this method to find the value of $ f(x) $ for any value of $ x $?

A: Yes, you can use this method to find the value of $ f(x) $ for any value of $ x $. However, you need to be careful when evaluating the expression, as the inverse sine function returns an angle in radians.

Q: What are some common mistakes to avoid when evaluating the expression $ f(x) = \sin \left(\sin^{-1} x\right) $?

A: Some common mistakes to avoid when evaluating the expression $ f(x) = \sin \left(\sin^{-1} x\right) $ include:

• Not understanding the definition of the inverse sine function
• Not substituting the value of $ \sin^{-1} x $ into the sine function
• Not being careful when evaluating the expression, as the inverse sine function returns an angle in radians

Conclusion

In this article, we have provided a Q&A section to further clarify any doubts and provide additional examples. We have also highlighted some common mistakes to avoid when evaluating the expression $ f(x) = \sin \left(\sin^{-1} x\right) $. By understanding the definition of the inverse sine function and the properties of the sine function, you can evaluate the expression without a calculator.

Additional Examples

To further illustrate the concept, let's consider a few more examples.

Example 1

Find $ f(0) $ in the function $ f(x) = \sin \left(\sin^{-1} x\right) $.

Solution

We know that $ \sin^{-1} 0 = 0 $, so:

$$f(0) = \sin (0) = 0$$

Example 2

Find $ f(1) $ in the function $ f(x) = \sin \left(\sin^{-1} x\right) $.

Solution

We know that $ \sin^{-1} 1 = \frac{\pi}{2} $, so:

$$f(1) = \sin \left(\frac{\pi}{2}\right) = 1$$

Example 3

Find $ f(-1) $ in the function $ f(x) = \sin \left(\sin^{-1} x\right) $.

Solution

We know that $ \sin^{-1} (-1) = -\frac{\pi}{2} $, so:

$$f(-1) = \sin \left(-\frac{\pi}{2}\right) = -1$$

Conclusion

In this article, we have provided a Q&A section to further clarify any doubts and provide additional examples. We have also highlighted some common mistakes to avoid when evaluating the expression $ f(x) = \sin \left(\sin^{-1} x\right) $. By understanding the definition of the inverse sine function and the properties of the sine function, you can evaluate the expression without a calculator.