Problem 5

Question

Write the sums in Exercises \(1-6\) without sigma notation. Then evaluate them. $$\sum_{k=1}^{3}(-1)^{k+1} \sin \frac{\pi}{k}$$

Step-by-Step Solution

Verified
Answer
The answer is \(-1 + \frac{\sqrt{3}}{2}\).
1Step 1: Interpret the Sigma Notation
The given expression is \(\sum_{k=1}^{3}(-1)^{k+1} \sin \frac{\pi}{k}\). This requires summing from \(k=1\) to \(k=3\). Each term in the series can be written as \((-1)^{k+1} \sin \frac{\pi}{k}\).
2Step 2: Expand the Summation
We need to replace the sigma notation with the individual terms:- For \(k=1\), the term is \((-1)^{1+1} \sin \frac{\pi}{1}\) = \(\sin \pi\) - For \(k=2\), the term is \((-1)^{2+1} \sin \frac{\pi}{2}\) = \(-\sin \frac{\pi}{2}\)- For \(k=3\), the term is \((-1)^{3+1} \sin \frac{\pi}{3}\) = \(\sin \frac{\pi}{3}\)
3Step 3: Evaluate Each Term
Now, evaluate each trigonometric function:- \(\sin \pi = 0\)- \(-\sin \frac{\pi}{2} = -1\)- \(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\)
4Step 4: Sum the Evaluated Terms
Add the evaluated terms together: \(0 - 1 + \frac{\sqrt{3}}{2} = -1 + \frac{\sqrt{3}}{2}\)
5Step 5: Final Result
The sum of the terms is \(-1 + \frac{\sqrt{3}}{2}\), which is the final answer to the problem.

Key Concepts

Trigonometric FunctionsSeries EvaluationSin Function
Trigonometric Functions
Trigonometric functions are a fundamental concept in mathematics, particularly in the context of angles and periodic phenomena. There are six primary trigonometric functions, but the most common ones are sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)). These functions are based on the relationships between the angles and sides of right triangles. The sine function is particularly important because it describes the ratio of the length of the side opposite the angle to the hypotenuse of the triangle.
  • Sine (\( \sin \)): Represents the vertical coordinate of a point on the unit circle.
  • Cosine (\( \cos \)): Represents the horizontal coordinate on the unit circle.
  • Tangent (\( \tan \)): The ratio of sine to cosine.
Understanding these functions is essential for solving problems involving waves, oscillations, and circular motion. In the context of the given problem, we focus on evaluating the sine function at specific angles derived from dividing \( \pi \) by integer values, which are related to a fraction of a circle.
Series Evaluation
In mathematics, a series is the sum of the terms of a sequence, where each term is defined based on an index, often represented with sigma notation. Sigma notation (\( \Sigma \)) is a convenient way of writing long sums in a compact form. It tells us three crucial components:
  • Where the series starts (the lower index).
  • Where it ends (the upper index).
  • The general form of each term in the series.
In the original exercise, the series \( \sum_{k=1}^{3}(-1)^{k+1} \sin \frac{\pi}{k} \) involves a combination of trigonometric functions and alternating signs. By first interpreting the sigma notation, it can be expanded into individual components. This leads to an easier evaluation since each term can be tackled independently. Breaking it down:- Substitute the relevant values of \( k \) from 1 to 3 into the general term.- Evaluate these terms step by step.This approach simplifies the series and helps to understand the underlying pattern of alternating signs with each term.
Sin Function
The sine function (\( \sin \)) is one of the cornerstones of trigonometry. It provides the ability to define and analyze the properties related to periodic phenomena like waves and harmonic motion. The sine function oscillates between -1 and 1, reaching these maximum values at specific multiples of \( \frac{\pi}{2} \):
  • \( \sin 0 = 0 \), \( \sin \frac{\pi}{2} = 1 \), \( \sin \pi = 0 \)
  • After \( \pi \), the cycle repeats with equivalent values.
In this exercise, the evaluation of \( \sin \frac{\pi}{k} \) is a crucial step. Each value:- For \( k=1 \), \( \sin \frac{\pi}{1} = \sin \pi = 0 \).- For \( k=2 \), \( \sin \frac{\pi}{2} = 1 \) with a negative sign due to the term structure.- For \( k=3 \), \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \).These results are combined using the series' alternating signs to ultimately compute the sum. Understanding the specific values of the sine function at these key angles is essential for exact calculations and interpretations of trigonometric expressions.