/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 \(\sec 30^{\circ}\)... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\sec 30^{\circ}\)

Short Answer

Expert verified
\( \sec 30^{\circ} = \frac{2\sqrt{3}}{3} \)

Step by step solution

01

Recall the Definition of Secant

The secant function is the reciprocal of the cosine function. Therefore, the formula for secant is: \[ \sec \theta = \frac{1}{\cos \theta} \] In this exercise, we need to find \( \sec 30^{\circ} \).
02

Find the Cosine of the Angle

Next, we need to determine \( \cos 30^{\circ} \). From trigonometric tables or known values, \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \).
03

Apply the Secant Formula

Using the definition of secant, substitute \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \) into the secant formula:\[ \sec 30^{\circ} = \frac{1}{\cos 30^{\circ}} = \frac{1}{\frac{\sqrt{3}}{2}} \].
04

Simplify the Expression

To simplify \( \frac{1}{\frac{\sqrt{3}}{2}} \), multiply the numerator and the denominator by 2:\[ \sec 30^{\circ} = \frac{1 \times 2}{\frac{\sqrt{3}}{2} \times 2} = \frac{2}{\sqrt{3}} \].
05

Rationalize the Denominator

To write the expression without a radical in the denominator, multiply by \( \frac{\sqrt{3}}{\sqrt{3}} \):\[ \sec 30^{\circ} = \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \].

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Key Concepts

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

Secant Function
The secant function is one of the six fundamental trigonometric functions often used in mathematics. It is closely related to the cosine function. Specifically, the secant of an angle is defined as the reciprocal of the cosine of that angle.
This means that if you know the cosine of a particular angle, you can find the secant by just taking one divided by the cosine value.
More formally, the relationship is expressed as:
  • \( \sec \theta = \frac{1}{\cos \theta} \)
The secant function helps in understanding right-angled triangles, circles, and various geometrical shapes. It shows up in many physical applications such as engineering and physics, where periodicity and wave behaviors happen. Because these waves often repeat or cycle through certain values, using the reciprocal identity helps simplify calculations.
Cosine of 30 Degrees
The cosine function is vital in trigonometry, which helps link angles to side ratios in right triangles. For a widely used angle like 30 degrees, its cosine has a standard value:
  • \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \)
This value comes from either using special triangles or memorizing from trigonometric tables. Understanding this known cosine value makes it easier to determine the secant of the angle and proceed with calculations. The cosine of 30 degrees being equal to root 3 over 2 derives from the properties of an equilateral triangle, split into two 30-60-90 triangles.
30-60-90 triangles divide the angles in a triangle into specific ratios, making it easy to determine specific trigonometric values such as sine, cosine, and tangent for these angles. Remembering these ratios is a valuable tool in solving more complex trigonometric problems lengthily.
Rationalizing the Denominator
Rationalizing the denominator is a technique used to eliminate a radical (like a square root) from the denominator of a fraction. This process makes calculations less complex while providing a cleaner expression. In mathematics, fractions with no radicals in the denominator are often preferred.
To rationalize a denominator, multiply both the numerator and the denominator of the fraction by the radical found in the denominator.
  • For example, if you have \( \frac{2}{\sqrt{3}} \), you multiply by \( \frac{\sqrt{3}}{\sqrt{3}} \).
This effectively removes the square root in the denominator, leading to a new expression:
  • \( \frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \)
Using this method not only gives a simpler look to your formulas but also conforms to mathematical standards in solving fractional equations.

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Most popular questions from this chapter

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. \(\mathbf{V} \mid=425, \theta=36^{\circ} 10^{\prime}\)

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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.)

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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. \(V \mid=17.6, \theta=67.2^{\circ}\)

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