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Chapter 2: Problem 51

\(\csc 60^{\circ}\)

Short Answer

Expert verified
\(\csc 60^{\circ} = \frac{2\sqrt{3}}{3}.\)

Step by step solution

01

Understand the Cosecant Function

The cosecant function, denoted as \(\csc\), is the reciprocal of the sine function. This means that to find \(\csc \theta\), you take the reciprocal of \(\sin \theta\). So the relationship is \(\csc \theta = \frac{1}{\sin \theta}\).
02

Determine the Sine of 60 Degrees

Recall from trigonometric tables or the unit circle that \(\sin 60^{\circ} = \frac{\sqrt{3}}{2}\). Knowing this value is essential to find the cosecant of the same angle.
03

Calculate the Cosecant of 60 Degrees

Using the relationship \(\csc \theta = \frac{1}{\sin \theta}\), substitute \(\sin 60^{\circ}\) into the equation: \[ \csc 60^{\circ} = \frac{1}{\sin 60^{\circ}} = \frac{1}{\frac{\sqrt{3}}{2}}. \]
04

Simplify the Expression

To simplify \(\frac{1}{\frac{\sqrt{3}}{2}}\), multiply the numerator and denominator by 2 to get rid of the fraction in the denominator: \[ \csc 60^{\circ} = \frac{2}{\sqrt{3}}. \]
05

Rationalize the Denominator

To rationalize the denominator, multiply the numerator and the denominator by \(\sqrt{3}\): \[ \csc 60^{\circ} = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}. \]

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Key Concepts

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

Rationalizing the Denominator
Rationalizing the denominator is a process used to eliminate square roots or irrational numbers from the bottom of a fraction. When dealing with trigonometric functions, especially involving the reciprocal identity, it often means simplifying expressions like \[ \frac{2}{\sqrt{3}}. \]Here's how we handle such expressions:
  • First, identify the irrational denominator. In this case, it's \(\sqrt{3}\).
  • Multiply both the numerator and the denominator by the same square root present in the denominator, which is \(\sqrt{3}\). This process involves creating an equivalent fraction that looks different but has the same value.
  • After multiplication, the expression becomes \(\frac{2\sqrt{3}}{3}\). This results because multiplying \(\sqrt{3} \times \sqrt{3} = 3\). Here, the denominator is rational because \(3\) is a perfect square, and hence rational.
This way, the denominator becomes a rational number, making calculations simpler and cleaner for further mathematical operations.
Trigonometric Tables
Trigonometric tables are essential tools in mathematics used to find the values of trigonometric functions for standard angles. These tables list values like sine, cosine, tangent, and their reciprocals, thus enabling quick calculations.For instance, when you need to find \(\sin 60^\circ\) or \(\csc 60^\circ\), you can refer to these tables:
  • They typically have pre-calculated exact values for common angles such as \(0^\circ, 30^\circ, 45^\circ, 60^\circ,\) and \(90^\circ\).
  • For \(60^\circ\), the trigonometric table gives \(\sin 60^\circ = \frac{\sqrt{3}}{2}\).
  • This exact value is crucial because to find \(\csc 60^\circ\), you take the reciprocal of \(\sin\) which is \(\frac{1}{\sin 60^\circ}\).
These tables help students and educators avoid errors and speed up the process of solving trigonometric equations, so they are a valuable reference in both academic and practical applications.
Unit Circle
The unit circle is a foundational concept in trigonometry that helps to understand angles and their corresponding trigonometric function values. Essentially, it is a circle with a radius of one unit, centered at the origin of an x-y coordinate plane.Here鈥檚 why the unit circle is so helpful:
  • Each angle in the unit circle corresponds to a point, where the x-coordinate is the cosine of the angle, and the y-coordinate is the sine.
  • For instance, at \(60^\circ\), the coordinates are \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\), explaining why \(\sin 60^\circ = \frac{\sqrt{3}}{2}\).
  • The unit circle covers all angles from \(0^\circ\) to \(360^\circ\) (or \(0\) to \(2\pi\) radians), which assists in visualizing the periodicity of trigonometric functions.
Using the unit circle, we can derive many trigonometric identities and solve equations more intuitively by connecting algebraic functions to geometric placements. This makes it an indispensable tool for anyone studying mathematics.

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Most popular questions from this chapter

Force Danny and Stacey have gone from the swing (Example 5) to the slide at the park. The slide is inclined at an angle of \(52.0^{\circ}\). Danny weighs \(42.0\) pounds. He is sitting in a cardboard box with a piece of wax paper on the bottom. Stacey is at the top of the slide holding on to the cardboard box (Figure 23). Find the magnitude of the force Stacey must pull with, in order to keep Danny from sliding down the slide. (We are assuming that the wax paper makes the slide into a frictionless surface, so that the only force keeping Danny from sliding is the force with which Stacey pulls.)

Refer to right triangle \(A B C\) with \(C=90^{\circ}\). Begin each problem by drawing a picture of the triangle with both the given and asked for information labeled appropriately. Also, write your answers for angles in decimal degrees.If \(A=42^{\circ}\) and \(c=15 \mathrm{ft}\), find \(a\).

Each problem below refers to a vector \(\mathbf{V}\) with magnitude \(|\mathbf{V}|\) that forms an angle \(\theta\) with the positive \(x\)-axis. In each case, give the magnitudes of the horizontal and vertical vector components of \(\mathbf{V}\), namely \(\mathbf{V}_{x}\) and \(\mathbf{V}_{y}\), respectively. \(|\mathbf{V}|=383, \theta=16^{\circ} 40^{\prime}\)

Refer to right triangle \(A B C\) with \(C=90^{\circ}\). Begin each problem by drawing a picture of the triangle with both the given and asked for information labeled appropriately. Also, write your answers for angles in decimal degrees.If \(b=6.7 \mathrm{~m}\) and \(c=7.7 \mathrm{~m}\), find \(A\).

For each problem below, the magnitudes of the horizontal and vertical vector components, \(\mathbf{V}_{x}\) and \(\mathbf{V}_{y}\), of vector \(\mathbf{V}\) are given. In each case find the magnitude of \(\mathbf{V}\). $$ \left|\mathbf{V}_{x}\right|=2.2,\left|\mathbf{V}_{y}\right|=5.8 $$
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