Problem 39
Question
\(23-44=\) Find the exact value of the expression, if it is defined. \(\tan \left(\sin ^{-1} \frac{1}{2}\right)\)
Step-by-Step Solution
Verified Answer
The value is \( \frac{\sqrt{3}}{3} \).
1Step 1: Understand the Expression
The expression we need to solve involves an inverse trigonometric function inside a trigonometric function: \( \tan \left( \sin^{-1} \frac{1}{2} \right) \). We need to first evaluate the inside part \( \sin^{-1} \frac{1}{2} \).
2Step 2: Evaluate the Inverse Sine
The expression \( \sin^{-1} \frac{1}{2} \) asks for the angle \( \theta \) whose sine is \( \frac{1}{2} \). From trigonometric knowledge, we know that the sine of \( \frac{\pi}{6} \) (or 30 degrees) is \( \frac{1}{2} \). Thus, \( \sin^{-1} \frac{1}{2} = \frac{\pi}{6} \).
3Step 3: Find the Tangent of the Angle
Substitute \( \frac{\pi}{6} \) into the tangent function: \( \tan \left( \frac{\pi}{6} \right) \). The tangent of \( \frac{\pi}{6} \) is known to be \( \frac{1}{\sqrt{3}} \), which can also be written as \( \frac{\sqrt{3}}{3} \) after rationalizing the denominator.
Key Concepts
Tangent of AnglesInverse Sine FunctionTrigonometric Identities
Tangent of Angles
Tangent, often abbreviated as "tan," is one of the primary trigonometric functions that relates the angles of a triangle to the lengths of its sides. In a right triangle, the tangent of an angle \( \theta \) is defined as the ratio of the length of the opposite side to the length of the adjacent side. Mathematically, this is expressed as:
\[\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\]This function is very useful in various real-world applications like physics, engineering, and even in calculating heights and distances.
When dealing with angles outside the context of a triangle, such as in the unit circle, the tangent function can be extended. For any angle \( \theta \), the tangent is the sine of \( \theta \) divided by the cosine of \( \theta \):
\[\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\]This function is very useful in various real-world applications like physics, engineering, and even in calculating heights and distances.
When dealing with angles outside the context of a triangle, such as in the unit circle, the tangent function can be extended. For any angle \( \theta \), the tangent is the sine of \( \theta \) divided by the cosine of \( \theta \):
- \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \)
- It is periodic, with a period of \( \pi \), meaning \( \tan(\theta + \pi) = \tan(\theta) \)
- The tangent function is undefined at angles where \( \cos(\theta) = 0 \)
Inverse Sine Function
The inverse sine function, denoted as \( \sin^{-1}(x) \) or "arcsin," is used to find the angle whose sine is a given number. This is particularly useful when you know the sine value but need to find the corresponding angle.
For example, if you have \( \sin^{-1}(\frac{1}{2}) \), you're looking for the angle \( \theta \) such that \( \sin(\theta) = \frac{1}{2} \). The angle that satisfies this in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\) radians is \( \frac{\pi}{6} \) (or 30 degrees).
Key properties of the inverse sine function include:
For example, if you have \( \sin^{-1}(\frac{1}{2}) \), you're looking for the angle \( \theta \) such that \( \sin(\theta) = \frac{1}{2} \). The angle that satisfies this in the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\) radians is \( \frac{\pi}{6} \) (or 30 degrees).
Key properties of the inverse sine function include:
- Its output range (also called the principal value) is restricted to \([-\frac{\pi}{2}, \frac{\pi}{2}]\)
- It is the inverse of the sine function, meaning \( \sin(\sin^{-1}(x)) = x \) for \(-1 \leq x \leq 1\)
- Helps determine angles from known functions, acting as a bridge in solving trigonometric problems.
Trigonometric Identities
Trigonometric identities are equations that are true for all values of the variables involved where both sides are defined. They are powerful tools in simplifying expressions and solving trigonometric equations.
One of the most commonly used identities is the Pythagorean identity:
In our original problem, knowing that the tangent of \( \tan(\frac{\pi}{6}) \) simplifies to \( \frac{\sqrt{3}}{3} \), we are leveraging a trigonometric identity that relates tangent to sine and cosine:
One of the most commonly used identities is the Pythagorean identity:
- \( \sin^2(\theta) + \cos^2(\theta) = 1 \)
In our original problem, knowing that the tangent of \( \tan(\frac{\pi}{6}) \) simplifies to \( \frac{\sqrt{3}}{3} \), we are leveraging a trigonometric identity that relates tangent to sine and cosine:
- \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \)
Other exercises in this chapter
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