Where Does The Terminal Side Of $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right$\] Lie?

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Introduction

In trigonometry, the inverse sine function, denoted as $\sin^{-1}$, is used to find the angle whose sine is a given value. However, the range of the inverse sine function is restricted to the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. This means that the terminal side of the angle lies in the second or fourth quadrant. In this article, we will explore where the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$ lies.

Understanding the Inverse Sine Function

The inverse sine function is defined as the angle whose sine is a given value. In other words, if $\sin(\theta) = x$, then $\sin^{-1}(x) = \theta$. However, the range of the inverse sine function is restricted to the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. This means that the terminal side of the angle lies in the second or fourth quadrant.

Finding the Terminal Side of $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$

To find the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$, we need to find the angle whose sine is $-\frac{2}{\sqrt{11}}$. We can use the unit circle to find this angle.

The Unit Circle

The unit circle is a circle with a radius of 1 centered at the origin of the coordinate plane. The unit circle is used to define the trigonometric functions. The sine of an angle is defined as the y-coordinate of the point on the unit circle corresponding to that angle.

Finding the Angle on the Unit Circle

To find the angle on the unit circle corresponding to $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$, we need to find the point on the unit circle whose y-coordinate is $-\frac{2}{\sqrt{11}}$. We can use the Pythagorean theorem to find the x-coordinate of this point.

Using the Pythagorean Theorem

The Pythagorean theorem states that for a right triangle with legs of length $a$ and $b$, and a hypotenuse of length $c$, the following equation holds:

$$a^2 + b^2 = c^2$$

We can use this theorem to find the x-coordinate of the point on the unit circle corresponding to $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$.

Finding the x-Coordinate

Let $x$ be the x-coordinate of the point on the unit circle corresponding to $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$. Then, we have:

$$x^2 + \left(-\frac{2}{\sqrt{11}}\right)^2 = 1$$

Simplifying this equation, we get:

$$x^2 + \frac{4}{11} = 1$$

Subtracting $\frac{4}{11}$ from both sides, we get:

$$x^2 = \frac{7}{11}$$

Taking the square root of both sides, we get:

$$x = \pm \sqrt{\frac{7}{11}}$$

Since the x-coordinate is positive in the first quadrant and negative in the second quadrant, we take the negative value of $x$.

Finding the Terminal Side

Now that we have found the x-coordinate of the point on the unit circle corresponding to $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$, we can find the terminal side of this angle. The terminal side is the line segment connecting the origin to the point on the unit circle.

Conclusion

In this article, we explored where the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$ lies. We used the unit circle to find the angle whose sine is $-\frac{2}{\sqrt{11}}$. We found that the x-coordinate of the point on the unit circle corresponding to this angle is $-\sqrt{\frac{7}{11}}$. Therefore, the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$ lies in the second quadrant.

Final Answer

The terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$ lies in the second quadrant.

References

• [1] "Trigonometry" by Michael Corral
• [2] "Calculus" by Michael Spivak
• [3] "Mathematics for Computer Science" by Eric Lehman, F Thomson Leighton, and Albert R Meyer

Keywords

• Inverse sine function
• Unit circle
• Pythagorean theorem
• Trigonometry
• Calculus
• Mathematics for computer science
  

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Introduction

In our previous article, we explored where the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$ lies. We used the unit circle to find the angle whose sine is $-\frac{2}{\sqrt{11}}$. In this article, we will answer some frequently asked questions related to this topic.

Q1: What is the inverse sine function?

A1: The inverse sine function, denoted as $\sin^{-1}$, is used to find the angle whose sine is a given value. In other words, if $\sin(\theta) = x$, then $\sin^{-1}(x) = \theta$. However, the range of the inverse sine function is restricted to the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Q2: Why is the range of the inverse sine function restricted to the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$?

A2: The range of the inverse sine function is restricted to the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ because the sine function is periodic with period $2\pi$. This means that the sine function repeats itself every $2\pi$ radians. Therefore, we only need to consider the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ to find the angle whose sine is a given value.

Q3: How do we find the angle on the unit circle corresponding to $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$?

A3: To find the angle on the unit circle corresponding to $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$, we need to find the point on the unit circle whose y-coordinate is $-\frac{2}{\sqrt{11}}$. We can use the Pythagorean theorem to find the x-coordinate of this point.

Q4: What is the x-coordinate of the point on the unit circle corresponding to $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$?

A4: The x-coordinate of the point on the unit circle corresponding to $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$ is $-\sqrt{\frac{7}{11}}$.

Q5: Where does the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$ lie?

A5: The terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$ lies in the second quadrant.

Q6: Why does the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$ lie in the second quadrant?

A6: The terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$ lies in the second quadrant because the sine function is negative in the second quadrant.

Q7: What is the relationship between the inverse sine function and the unit circle?

A7: The inverse sine function is related to the unit circle because the unit circle is used to define the trigonometric functions. The sine of an angle is defined as the y-coordinate of the point on the unit circle corresponding to that angle.

Q8: How do we use the Pythagorean theorem to find the x-coordinate of the point on the unit circle corresponding to $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$?

A8: We use the Pythagorean theorem to find the x-coordinate of the point on the unit circle corresponding to $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$ by setting up the equation $x^2 + \left(-\frac{2}{\sqrt{11}}\right)^2 = 1$ and solving for $x$.

Q9: What is the significance of the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ in the context of the inverse sine function?

A9: The interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is significant in the context of the inverse sine function because it is the range of the inverse sine function. This means that the inverse sine function is only defined for angles in this interval.

Q10: How do we find the terminal side of an angle in the context of the inverse sine function?

A10: We find the terminal side of an angle in the context of the inverse sine function by using the unit circle and the Pythagorean theorem to find the x-coordinate and y-coordinate of the point on the unit circle corresponding to the angle.

Conclusion

In this article, we answered some frequently asked questions related to the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$. We explored the relationship between the inverse sine function and the unit circle, and we used the Pythagorean theorem to find the x-coordinate of the point on the unit circle corresponding to $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$. We also discussed the significance of the interval $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ in the context of the inverse sine function.

Final Answer

The terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{11}}\right)$ lies in the second quadrant.

References

• [1] "Trigonometry" by Michael Corral
• [2] "Calculus" by Michael Spivak
• [3] "Mathematics for Computer Science" by Eric Lehman, F Thomson Leighton, and Albert R Meyer

Keywords

• Inverse sine function
• Unit circle
• Pythagorean theorem
• Trigonometry
• Calculus
• Mathematics for computer science