Problem 11 - Thomas Calculus in SI Units

Question

Find the values. $$\tan \left(\sin ^{-1}\left(-\frac{1}{2}\right)\right)$$

Step-by-Step Solution

Answer

The value is \( -\frac{\sqrt{3}}{3} \).

Step 1: Identify the inverse trigonometric function

Recognize that you need to evaluate \( \sin^{-1}(-\frac{1}{2}) \). This means you are looking for an angle \( \theta \) such that \( \sin(\theta) = -\frac{1}{2} \).

Step 2: Determine the applicable angle

The angle \( \theta \) for which \( \sin(\theta) = -\frac{1}{2} \) is one of the standard angles found on the unit circle. It can be \( -\frac{\pi}{6} \). This is because the sine function has a value of \( -\frac{1}{2} \) for \( \theta = -30^\circ \).

Step 3: Find the tangent of the angle

Now that you have \( \theta = -\frac{\pi}{6} \), find \( \tan(\theta) \). The tangent of an angle \( \theta \) is the ratio \( \frac{\sin(\theta)}{\cos(\theta)} \). For \( \theta = -\frac{\pi}{6} \), \( \sin(-\frac{\pi}{6}) = -\frac{1}{2} \) and \( \cos(-\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \).

Step 4: Calculate the tangent

Use the values from the previous step to find \( \tan(-\frac{\pi}{6}) \), which is \( \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} \).

Step 5: Simplify the tangent value

To simplify \( \frac{-1}{\sqrt{3}} \), multiply the numerator and the denominator by \( \sqrt{3} \) to rationalize the denominator. This gives \( -\frac{\sqrt{3}}{3} \).

Key Concepts

Trigonometry

Trigonometry is a branch of mathematics that studies the relationships between the angles and sides of triangles. It's extensively used in various fields such as physics, engineering, and even computer graphics. Trigonometry involves several key functions: sine, cosine, tangent, and their respective inverses. The key functions have certain properties, often explored using right-angled triangles. For example, the sine of an angle is the ratio of the opposite side to the hypotenuse. Meanwhile, the cosine is the ratio of the adjacent side to the hypotenuse. The tangent is slightly different, as it represents the ratio of the sine to the cosine, making it the ratio of the opposite side to the adjacent side.Trigonometric identities also play a crucial role. These include helpful equations like the Pythagorean identity: \( \sin^2(\theta) + \cos^2(\theta) = 1 \)and angle addition formulas. Understanding and applying these relationships is fundamental when solving problems involving angles and lengths in triangles.

Unit Circle

The unit circle is a powerful tool in trigonometry that provides insights into the values of trigonometric functions for various angles. It is a circle with a radius equal to one, centered at the origin of a coordinate plane. On the unit circle:

• The x-coordinate of a point on the circle represents the cosine of the angle.
• The y-coordinate corresponds to the sine of the angle.
• The tangent can be found using the ratio of these y and x coordinates, since \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \).

Using the unit circle also helps to easily determine the values of trigonometric functions for angles commonly used, such as \( \frac{\pi}{6} \), \( \frac{\pi}{4} \), and \( \frac{\pi}{3} \). For instance, the angle \( \frac{\pi}{6} \) (or 30 degrees) has a cosine value of \( \frac{\sqrt{3}}{2} \) and sine value of \( \frac{1}{2} \), which are derived directly from the unit circle coordinates. Knowing these values simplifies many trigonometric calculations.

Tangent

The tangent function is one of the principal trigonometric functions in mathematics. It's often thought of as a measure of slope, representing how steep a line is. To understand the tangent, remember that it is defined as the ratio of the opposite side to the adjacent side in a right triangle:\( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \)This function appears frequently in problems involving angles and slopes. For example, if you have an angle \( \theta = -\frac{\pi}{6} \), the sine is \( -\frac{1}{2} \) and the cosine is \( \frac{\sqrt{3}}{2} \). By applying these values, you find \( \tan(-\frac{\pi}{6}) = -\frac{1}{\sqrt{3}} \), which can be further simplified to \( -\frac{\sqrt{3}}{3} \) by rationalizing the denominator.Understanding the tangent's geometric interpretation, as well as how to compute and simplify it, provides a comprehensive look at how this function relates to both trigonometry and real-world scenarios.