Where Does The Terminal Side Of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right$\] 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 $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$ 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 $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$

To find the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$, we need to find the angle whose sine is $-\frac{2}{\sqrt{15}}$. We can use the unit circle to find this angle.

Using 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 the angle.

Finding the Angle

To find the angle whose sine is $-\frac{2}{\sqrt{15}}$, we need to find the point on the unit circle corresponding to this value. We can do this by using the Pythagorean theorem.

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:

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

Applying the Pythagorean Theorem

We can apply the Pythagorean theorem to the right triangle formed by the point on the unit circle corresponding to the angle whose sine is $-\frac{2}{\sqrt{15}}$. Let the legs of the triangle be $a$ and $b$, and the hypotenuse be $c$. Then, we have:

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

Solving for $a$ and $b$

We know that the sine of the angle is $-\frac{2}{\sqrt{15}}$. This means that the y-coordinate of the point on the unit circle corresponding to the angle is $-\frac{2}{\sqrt{15}}$. We can use this information to find the values of $a$ and $b$.

Finding the Values of $a$ and $b$

We have:

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

$$\left(-\frac{2}{\sqrt{15}}\right)^2 + b^2 = 1$$

$$\frac{4}{15} + b^2 = 1$$

$$b^2 = \frac{11}{15}$$

$$b = \pm \sqrt{\frac{11}{15}}$$

Finding the Angle

We can now find the angle whose sine is $-\frac{2}{\sqrt{15}}$. We have:

$$\sin(\theta) = -\frac{2}{\sqrt{15}}$$

$$\cos(\theta) = \pm \sqrt{\frac{11}{15}}$$

Using the Pythagorean Identity

We can use the Pythagorean identity to find the value of $\theta$.

Pythagorean Identity

The Pythagorean identity states that for any angle $\theta$, the following equation holds:

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

Applying the Pythagorean Identity

We can apply the Pythagorean identity to the angle whose sine is $-\frac{2}{\sqrt{15}}$. We have:

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

$$\left(-\frac{2}{\sqrt{15}}\right)^2 + \left(\pm \sqrt{\frac{11}{15}}\right)^2 = 1$$

$$\frac{4}{15} + \frac{11}{15} = 1$$

$$\frac{15}{15} = 1$$

Finding the Value of $\theta$

We can now find the value of $\theta$.

Finding the Terminal Side

We have found the value of $\theta$. We can now find the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$.

Conclusion

In this article, we explored where the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$ lies. We used the unit circle and the Pythagorean theorem to find the angle whose sine is $-\frac{2}{\sqrt{15}}$. We then used the Pythagorean identity to find the value of $\theta$. Finally, we found the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$.

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Introduction

In our previous article, we explored where the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$ lies. We used the unit circle and the Pythagorean theorem to find the angle whose sine is $-\frac{2}{\sqrt{15}}$. In this article, we will answer some frequently asked questions about the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$.

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

A: 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.

Q: How do I find the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$?

A: To find the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$, you need to find the angle whose sine is $-\frac{2}{\sqrt{15}}$. You can use the unit circle and the Pythagorean theorem to find this angle.

Q: What is the value of $\theta$?

A: The value of $\theta$ is the angle whose sine is $-\frac{2}{\sqrt{15}}$. We found that $\theta = \sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$.

Q: What is the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$?

A: The terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$ is the line segment that extends from the point on the unit circle corresponding to the angle $\theta$ to the point on the unit circle corresponding to the angle $\theta + 2\pi$.

Q: How do I find the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$ using the unit circle?

A: To find the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$ using the unit circle, you need to find the point on the unit circle corresponding to the angle $\theta$. You can do this by using the coordinates of the point on the unit circle.

Q: What are the coordinates of the point on the unit circle corresponding to the angle $\theta$?

A: The coordinates of the point on the unit circle corresponding to the angle $\theta$ are $\left(\cos(\theta), \sin(\theta)\right)$.

Q: How do I find the coordinates of the point on the unit circle corresponding to the angle $\theta$?

A: To find the coordinates of the point on the unit circle corresponding to the angle $\theta$, you need to use the values of $\cos(\theta)$ and $\sin(\theta)$. We found that $\cos(\theta) = \pm \sqrt{\frac{11}{15}}$ and $\sin(\theta) = -\frac{2}{\sqrt{15}}$.

Q: What is the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$ in terms of the unit circle?

A: The terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$ is the line segment that extends from the point on the unit circle corresponding to the angle $\theta$ to the point on the unit circle corresponding to the angle $\theta + 2\pi$.

Q: How do I find the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$ in terms of the unit circle?

A: To find the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$ in terms of the unit circle, you need to find the point on the unit circle corresponding to the angle $\theta$. You can do this by using the coordinates of the point on the unit circle.

Q: What are the coordinates of the point on the unit circle corresponding to the angle $\theta$ in terms of the unit circle?

A: The coordinates of the point on the unit circle corresponding to the angle $\theta$ are $\left(\cos(\theta), \sin(\theta)\right)$.

Q: How do I find the coordinates of the point on the unit circle corresponding to the angle $\theta$ in terms of the unit circle?

A: To find the coordinates of the point on the unit circle corresponding to the angle $\theta$ in terms of the unit circle, you need to use the values of $\cos(\theta)$ and $\sin(\theta)$. We found that $\cos(\theta) = \pm \sqrt{\frac{11}{15}}$ and $\sin(\theta) = -\frac{2}{\sqrt{15}}$.

Conclusion

In this article, we answered some frequently asked questions about the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$. We used the unit circle and the Pythagorean theorem to find the angle whose sine is $-\frac{2}{\sqrt{15}}$. We then used the coordinates of the point on the unit circle corresponding to the angle to find the terminal side of $\sin^{-1}\left(-\frac{2}{\sqrt{15}}\right)$.