Problem 23 - Algebra and Trigonometry

Question

Find the exact value of the expression, if it is defined. \(\tan \left(\sin ^{-1} \frac{1}{2}\right)\)

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We need to find the exact value of \( \tan \left( \sin^{-1} \frac{1}{2} \right) \). This involves finding the tangent of an angle whose sine is \( \frac{1}{2} \).

Step 2: Determine the Related Angle

Since \( \sin(\theta) = \frac{1}{2} \), we recognize \( \theta \) as an angle for which the sine value is \( \frac{1}{2} \). In the unit circle, this corresponds to \( \theta = \frac{\pi}{6} \) because \( \sin \left( \frac{\pi}{6} \right) = \frac{1}{2} \).

Step 3: Calculate the Tangent of the Angle

We calculate \( \tan(\theta) \) for \( \theta = \frac{\pi}{6} \). We know that \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \). For \( \theta = \frac{\pi}{6} \), we have \( \cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2} \). Therefore, \( \tan \left( \frac{\pi}{6} \right) = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} \).

Step 4: Rationalize the Denominator

To express \( \frac{1}{\sqrt{3}} \) in a standard form, we multiply the numerator and the denominator by \( \sqrt{3} \). This gives \( \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3}}{3} \).

Key Concepts

Tangent Function

The tangent function, often expressed as \(\tan(\theta)\), is one of the basic trigonometric functions. It is defined as the ratio of the sine function to the cosine function:

• \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \)

Understanding the tangent function is crucial because it helps describe the relationship between different trigonometric functions. It is particularly useful in scenarios where you need to relate the height and base of a right triangle.

The tangent function has a range that spans all real numbers and shows a distinct pattern of repeating every \(\pi\) radians due to its periodic nature. This means if you start at one point on a graph of the tangent function and move \(\pi\) units, you will see the same shape repeat again.

Finally, the tangent function can have asymptotes, which are regions where the function approaches infinity. This occurs whenever the cosine of the angle is zero, leading to division by zero in the tangent definition.

Sine Function

The sine function, expressed as \(\sin(\theta)\), is another fundamental trigonometric function. It describes the y-coordinate of a point on the unit circle as a function of the angle \(\theta\) from the positive x-axis.

• For example, \(\sin(0) = 0\), \(\sin\left(\frac{\pi}{2}\right) = 1\), and \(\sin(\pi) = 0\).

The sine function oscillates smoothly between -1 and 1, creating a wave-like pattern which is periodic. This means it repeats every \(2\pi\) radians.

In the context of the problem, finding \(\sin^{-1}\left(\frac{1}{2}\right)\) implies finding the angle whose sine value is \(\frac{1}{2}\). Based on the unit circle, this angle is \(\theta = \frac{\pi}{6}\). This helps in calculating the tangent of this specific angle for the exercise we are trying to solve.

Unit Circle

The unit circle is a perfect tool to visualize trigonometric functions. It is a circle with a radius of 1 centered at the origin of a coordinate plane.

• Every angle \(\theta\) defined from the origin corresponds to a specific point on this circle.

The coordinates of any point on this circle are \((\cos(\theta), \sin(\theta))\), which maps directly to cosine and sine values.

The unit circle helps us understand the periodic nature and values of trigonometric functions across different quadrants. For example, any angle \(\theta\) in the first quadrant will have both positive sine and cosine values.

In our problem, the unit circle helps pinpoint \(\theta = \frac{\pi}{6}\), where \(\sin(\theta) = \frac{1}{2}\). This allows you to easily determine other trigonometric values, like \(\cos(\frac{\pi}{6})\) and subsequently \(\tan(\frac{\pi}{6})\), facilitating the calculation of angles and proving invaluable for understanding function behaviors in mathematics.