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Chapter 5: Problem 35

Find the exact value of the given expression. If an exact value cannot be given, give the value to the nearest ten-thousandth. $$\tan \left(\sin ^{-1} \frac{1}{2}\right)$$

Short Answer

Expert verified
\(\frac{\sqrt{3}}{3}\)

Step by step solution

01

Identify the angle from its sine

The inverse sine function, \(\sin^{-1}\), takes a number and gives the angle whose sine is that number. So \(\sin^{-1}\frac{1}{2}\) is the angle whose sine is \(\frac{1}{2}\). From basic knowledge of trigonometric values of standard angles, we know that \(\sin{\frac{\pi}{6}} = \frac{1}{2}\). Hence, \(\sin^{-1}\frac{1}{2} = \frac{\pi}{6}\).
02

Find the tangent of the angle

Now, we need to find the tangent of the angle we just found, i.e., \(\tan{\frac{\pi}{6}}\). Again from our knowledge of trigonometric values, particularly the 30-60-90 degree triangle relationships, we know that \(\tan{\frac{\pi}{6}} = \frac{\sqrt{3}}{3}\).
03

Representation of the final solution

So, the given expression \(\tan (\sin ^{-1} \frac{1}{2})\) simplifies to \(\frac{\sqrt{3}}{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse Trigonometric Functions
Inverse trigonometric functions act as the gateways to angles from known trigonometric values. When you see \( \sin^{-1}(x) \), it is asking what angle has a sine of \( x \). Let's consider the given expression, \( \sin^{-1} \frac{1}{2} \). The function tells us to find an angle whose sine is precisely \( \frac{1}{2} \).
  • This particular inverse function restricts its range, meaning it only provides angles in the specific range \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \) for sine.
  • Within these limits, the angle where the sine gives \( \frac{1}{2} \) is \( \frac{\pi}{6} \), which is 30 degrees.
This step of identifying the angle is crucial since it allows us to then apply other trigonometric calculations, such as finding tangents or cosines of this angle. Thus, understanding how inverse trigonometric functions work helps us unlock precise angle values from simple ratios.
Exact Trigonometric Values
Exact trigonometric values are the known values of trigonometric functions at specific angles. These values are essential in solving many trigonometric problems without resorting to approximations. For example, once we have identified that \( \sin^{-1} \frac{1}{2} = \frac{\pi}{6} \), we immediately surround ourselves with a realm of known values.
  • For \( \frac{\pi}{6} \), we can recall:
    • \( \sin \frac{\pi}{6} = \frac{1}{2} \)
    • \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \)
    • \( \tan \frac{\pi}{6} = \frac{\sqrt{3}}{3} \)
Accessing these values enables us to determine the exact outputs for functions without approximating them on a calculator. These known angles typically include 30 degrees (\( \frac{\pi}{6} \)), 45 degrees (\( \frac{\pi}{4} \)), and 60 degrees (\( \frac{\pi}{3} \)), forming the backbone of many quick calculations.
Trigonometric Identities
Trigonometric identities are formulas involving trigonometric functions that are true for all angles they are defined on. They play a fundamental role in simplifying expressions and solving equations. For our problem, understanding trigonometric identities helps verify and derive exact values.
  • Let's revisit the original expression, \( \tan(\sin^{-1} \frac{1}{2}) = \tan\frac{\pi}{6} \).
  • The relationship for \( \tan \theta \) can always be defined as \( \frac{\sin \theta}{\cos \theta} \).
For the angle \( \frac{\pi}{6} \), the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) allows us to write: \[ \tan \frac{\pi}{6} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{\sqrt{3}}{3} \]This relationship is an example of how identities interlock to link various trigonometric functions through established equations, helping us derive specific known values from one another.

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Most popular questions from this chapter

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