Evaluate The Expression:$\[ \sin^{-1}\left(\sin\left(-\frac{7\pi}{2}\right)\right) \\]

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Introduction

In mathematics, trigonometric functions and their inverses play a crucial role in solving various problems. The sine function, in particular, is a fundamental component of trigonometry, and its inverse, the arcsine function, is used to find the angle whose sine is a given value. In this article, we will evaluate the expression sinβ‘βˆ’1(sin⁑(βˆ’7Ο€2))\sin^{-1}\left(\sin\left(-\frac{7\pi}{2}\right)\right) and explore its properties.

Understanding the Sine Function

The sine function is a periodic function that oscillates between -1 and 1. It is defined as the ratio of the length of the side opposite a given angle to the length of the hypotenuse in a right-angled triangle. The sine function has a period of 2Ο€2\pi, which means that the value of the sine function repeats every 2Ο€2\pi radians.

Understanding the Inverse Sine Function

The inverse sine function, denoted by sinβ‘βˆ’1\sin^{-1}, is used to find the angle whose sine is a given value. It is defined as the angle whose sine is equal to the given value. The range of the inverse sine function is [βˆ’Ο€2,Ο€2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], which means that the output of the inverse sine function is always between -Ο€2\frac{\pi}{2} and Ο€2\frac{\pi}{2}.

Evaluating the Expression

To evaluate the expression sinβ‘βˆ’1(sin⁑(βˆ’7Ο€2))\sin^{-1}\left(\sin\left(-\frac{7\pi}{2}\right)\right), we need to find the value of sin⁑(βˆ’7Ο€2)\sin\left(-\frac{7\pi}{2}\right). Since the sine function has a period of 2Ο€2\pi, we can rewrite βˆ’7Ο€2-\frac{7\pi}{2} as βˆ’(3Ο€+Ο€2)-\left(3\pi + \frac{\pi}{2}\right). Using the properties of the sine function, we can simplify this expression as follows:

sin⁑(βˆ’7Ο€2)=sin⁑(βˆ’(3Ο€+Ο€2))=sin⁑(βˆ’3Ο€βˆ’Ο€2)\sin\left(-\frac{7\pi}{2}\right) = \sin\left(-\left(3\pi + \frac{\pi}{2}\right)\right) = \sin\left(-3\pi - \frac{\pi}{2}\right)

Since the sine function is an odd function, we can rewrite this expression as:

sin⁑(βˆ’3Ο€βˆ’Ο€2)=βˆ’sin⁑(3Ο€+Ο€2)\sin\left(-3\pi - \frac{\pi}{2}\right) = -\sin\left(3\pi + \frac{\pi}{2}\right)

Using the properties of the sine function, we can simplify this expression as follows:

βˆ’sin⁑(3Ο€+Ο€2)=βˆ’sin⁑(Ο€2)=βˆ’1-\sin\left(3\pi + \frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right) = -1

Therefore, the value of sin⁑(βˆ’7Ο€2)\sin\left(-\frac{7\pi}{2}\right) is -1.

Finding the Value of the Inverse Sine Function

Now that we have found the value of sin⁑(βˆ’7Ο€2)\sin\left(-\frac{7\pi}{2}\right), we can find the value of the inverse sine function. Since the range of the inverse sine function is [βˆ’Ο€2,Ο€2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], we need to find the angle whose sine is equal to -1. The angle whose sine is equal to -1 is βˆ’Ο€2-\frac{\pi}{2}.

Conclusion

In conclusion, the value of the expression sinβ‘βˆ’1(sin⁑(βˆ’7Ο€2))\sin^{-1}\left(\sin\left(-\frac{7\pi}{2}\right)\right) is βˆ’Ο€2-\frac{\pi}{2}. This result is consistent with the properties of the sine function and the inverse sine function.

Properties of the Inverse Sine Function

The inverse sine function has several important properties that are useful in solving problems. Some of these properties include:

  • The range of the inverse sine function is [βˆ’Ο€2,Ο€2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].
  • The inverse sine function is an odd function, which means that sinβ‘βˆ’1(βˆ’x)=βˆ’sinβ‘βˆ’1(x)\sin^{-1}(-x) = -\sin^{-1}(x).
  • The inverse sine function is a decreasing function, which means that sinβ‘βˆ’1(x)\sin^{-1}(x) decreases as xx increases.

Applications of the Inverse Sine Function

The inverse sine function has several important applications in mathematics and other fields. Some of these applications include:

  • Finding the angle whose sine is a given value.
  • Solving trigonometric equations.
  • Finding the area of a triangle.
  • Finding the length of a side of a triangle.

Final Thoughts

In conclusion, the expression sinβ‘βˆ’1(sin⁑(βˆ’7Ο€2))\sin^{-1}\left(\sin\left(-\frac{7\pi}{2}\right)\right) is equal to βˆ’Ο€2-\frac{\pi}{2}. This result is consistent with the properties of the sine function and the inverse sine function. The inverse sine function has several important properties and applications that make it a useful tool in mathematics and other fields.

References

Further Reading

Introduction

In our previous article, we evaluated the expression sinβ‘βˆ’1(sin⁑(βˆ’7Ο€2))\sin^{-1}\left(\sin\left(-\frac{7\pi}{2}\right)\right) and found that it is equal to βˆ’Ο€2-\frac{\pi}{2}. In this article, we will answer some frequently asked questions related to this expression.

Q: What is the value of sin⁑(βˆ’7Ο€2)\sin\left(-\frac{7\pi}{2}\right)?

A: The value of sin⁑(βˆ’7Ο€2)\sin\left(-\frac{7\pi}{2}\right) is -1.

Q: Why is the value of sin⁑(βˆ’7Ο€2)\sin\left(-\frac{7\pi}{2}\right) equal to -1?

A: The value of sin⁑(βˆ’7Ο€2)\sin\left(-\frac{7\pi}{2}\right) is equal to -1 because the sine function has a period of 2Ο€2\pi, and βˆ’7Ο€2-\frac{7\pi}{2} is equivalent to βˆ’(3Ο€+Ο€2)-\left(3\pi + \frac{\pi}{2}\right), which is equal to βˆ’sin⁑(Ο€2)-\sin\left(\frac{\pi}{2}\right).

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

A: The range of the inverse sine function is [βˆ’Ο€2,Ο€2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Q: Why is the inverse sine function an odd function?

A: The inverse sine function is an odd function because it satisfies the property sinβ‘βˆ’1(βˆ’x)=βˆ’sinβ‘βˆ’1(x)\sin^{-1}(-x) = -\sin^{-1}(x).

Q: What are some applications of the inverse sine function?

A: Some applications of the inverse sine function include finding the angle whose sine is a given value, solving trigonometric equations, finding the area of a triangle, and finding the length of a side of a triangle.

Q: Can the inverse sine function be used to find the value of sinβ‘βˆ’1(sin⁑(x))\sin^{-1}\left(\sin\left(x\right)\right) for any value of xx?

A: No, the inverse sine function can only be used to find the value of sinβ‘βˆ’1(sin⁑(x))\sin^{-1}\left(\sin\left(x\right)\right) for values of xx in the range [βˆ’Ο€2,Ο€2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Q: What is the value of sinβ‘βˆ’1(sin⁑(x))\sin^{-1}\left(\sin\left(x\right)\right) for values of xx outside the range [βˆ’Ο€2,Ο€2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]?

A: For values of xx outside the range [βˆ’Ο€2,Ο€2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], the value of sinβ‘βˆ’1(sin⁑(x))\sin^{-1}\left(\sin\left(x\right)\right) is not defined.

Q: Can the inverse sine function be used to find the value of sinβ‘βˆ’1(sin⁑(βˆ’x))\sin^{-1}\left(\sin\left(-x\right)\right) for any value of xx?

A: Yes, the inverse sine function can be used to find the value of sinβ‘βˆ’1(sin⁑(βˆ’x))\sin^{-1}\left(\sin\left(-x\right)\right) for any value of xx.

Q: What is the value of sinβ‘βˆ’1(sin⁑(βˆ’x))\sin^{-1}\left(\sin\left(-x\right)\right) for any value of xx?

A: The value of sinβ‘βˆ’1(sin⁑(βˆ’x))\sin^{-1}\left(\sin\left(-x\right)\right) is equal to βˆ’sinβ‘βˆ’1(sin⁑(x))-\sin^{-1}\left(\sin\left(x\right)\right).

Conclusion

In conclusion, the expression sinβ‘βˆ’1(sin⁑(βˆ’7Ο€2))\sin^{-1}\left(\sin\left(-\frac{7\pi}{2}\right)\right) is equal to βˆ’Ο€2-\frac{\pi}{2}. We have also answered some frequently asked questions related to this expression, including the value of sin⁑(βˆ’7Ο€2)\sin\left(-\frac{7\pi}{2}\right), the range of the inverse sine function, and some applications of the inverse sine function.

References

Further Reading