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Chapter 7: Problem 30

Simplify and check using a graphing calculator. $$\frac{9 \cos ^{2} \alpha-25}{2 \cos \alpha-2} \cdot \frac{\cos ^{2} \alpha-1}{6 \cos \alpha-10}$$

Short Answer

Expert verified
\frac{(3 \cos \alpha + 5)(\cos \alpha + 1)}{4}

Step by step solution

01

Identify the Original Expression

Consider the given expression: $$\frac{9 \cos^{2} \alpha - 25}{2 \cos \alpha - 2} \cdot \frac{\cos^{2} \alpha -1}{6 \cos \alpha - 10}$$
02

Factor the Numerators and Denominators

Rewrite the expression using factorization: $$9 \cos^{2} \alpha - 25 = (3 \cos \alpha - 5)(3 \cos \alpha + 5)$$ $$2 \cos \alpha - 2 = 2(\cos \alpha - 1)$$ $$\cos^{2} \alpha - 1 = (\cos \alpha - 1)(\cos \alpha + 1)$$ $$6 \cos \alpha - 10 = 2(3 \cos \alpha - 5)$$
03

Substitute Factored Terms in the Expression

Substitute these factored forms into the original expression: $$\frac{(3 \cos \alpha - 5)(3 \cos \alpha + 5)}{2(\cos \alpha - 1)} \cdot \frac{(\cos \alpha - 1)(\cos \alpha + 1)}{2(3 \cos \alpha - 5)}$$
04

Cancel Out Common Factors

Cancel the common factors from the numerator and denominator: $$ = \frac{\cancel{(3 \cos \alpha - 5)}(3 \cos \alpha + 5)}{2\cancel{(\cos \alpha - 1)}} \cdot \frac{\cancel{(\cos \alpha - 1)}(\cos \alpha + 1)}{2\cancel{(3 \cos \alpha - 5)}}$$ $$ = \frac{(3 \cos \alpha + 5)(\cos \alpha + 1)}{2 \cdot 2}$$
05

Simplify the Expression

Simplify the remaining terms: $$ = \frac{(3 \cos \alpha + 5)(\cos \alpha + 1)}{4}$$
06

Verify Using a Graphing Calculator

To verify, graph the original and simplified expressions on a graphing calculator. Ensure that both graphs overlap completely, confirming the simplification.

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Key Concepts

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

trigonometric identities
Understanding trigonometric identities is key to simplifying trigonometric expressions efficiently. These identities are formulas involving trigonometric functions that are true for all values of the variable. For instance, the Pythagorean identities like \( \cos^2 \alpha + \sin^2 \alpha = 1 \) and \( \cos^2 \alpha - 1 = -\sin^2 \alpha \) are foundational. Recognizing and using these identities can make factorization and simplification much easier. In our exercise, identifying that \( \cos^2 \alpha - 1 = (\cos \alpha - 1)(\cos \alpha + 1) \) helped simplify the expression.
factorization
Factorization involves breaking down expressions into products of simpler expressions. In our given problem, we factorized several terms to simplify the expression. Here's a breakdown:
\(9 \cos^2 \alpha - 25 = (3 \cos \alpha - 5)(3 \cos \alpha + 5)\)
\(2 \cos \alpha - 2 = 2(\cos \alpha - 1)\)
\(\cos^2 \alpha - 1 = (\cos \alpha - 1)(\cos \alpha + 1) \)
\(6 \cos \alpha - 10 = 2(3 \cos \alpha - 5)\)
Breaking down these terms was essential for the next steps of simplification. Learning factorization techniques, such as recognizing patterns like the difference of squares, can significantly enhance your solving skills.
graphing calculators
Graphing calculators are powerful tools for verifying the accuracy of your algebraic manipulations. After simplifying an expression, you can use a graphing calculator to plot both the original and simplified equations. If the graphs overlap completely, it validates that your simplification is correct. Here's a simple way to use it:
  • Enter the original expression and plot the graph.
  • Enter the simplified expression and plot it over the original.
The two graphs should coincide if the simplification is accurate. This provides a visual and computational confirmation of your work.
simplified fractions
Simplified fractions are easier to handle and interpret. We reduced our complex trigonometric expression to a simpler fraction. The final expression after factorization and simplification is:
\( \frac{(3 \cos \alpha + 5)(\cos \alpha + 1)}{4} \)
Note that cancelling common factors is crucial. It reduced the original complicated fraction to a more straightforward form. Simplifying fractions often involves:
  • Identifying and canceling common factors.
  • Rewriting the fraction in the lowest terms.
This makes subsequent calculations and verifications more manageable.

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Most popular questions from this chapter

Nautical Mile. (See Exercise 60 in Exercise Set 7.2 ) In Great Britain, the nautical mile is defined as the length of a minute of arc of the earth's radius. since the earth is flattened at the poles, a British nautical mile varies with latitude. In fact, it is given, in feet, by the function $$N(\phi)=6066-31 \cos 2 \phi$$ where \(\phi\) is the latitude in degrees. At what latitude north is the length of a British nautical mile found to be \(6040 \mathrm{ft} ?\)

Find \(\sin 15^{\circ}\) first using a difference identity and then using a half-angle identity. Then compare the results.

Solve, finding all solutions in \([0,2 \pi)\). $$\cos 2 x \sin x+\sin x=0$$

The acceleration due to gravity is often denoted by \(g\) in a formula such as \(S=\frac{1}{2} g t^{2},\) where \(S\) is the distance that an object falls in time \(t .\) The number \(g\) relates to motion near the earth's surface and is generally considered constant. In fact, however, \(g\) is not constant, but varies slightly with latitude. Latitude is used to measure north-south location on the earth between the equator and the poles. If \(\phi\) stands for latitude, in degrees, \(g\) is given with good approximation by the formula \(g=9.78049\left(1+0.005288 \sin ^{2} \phi-0.000006 \sin ^{2} 2 \phi\right)\), where \(g\) is measured in meters per second per second at sea level. a) Chicago has latitude \(42^{\circ} \mathrm{N}\). Find \(g .\) b) Philadelphia has latitude \(40^{\circ} \mathrm{N}\). Find \(g\). c) Express \(g\) in terms of \(\sin \phi\) only. That is, eliminate the double angle.

Use the sum and difference identities to evaluate exactly. Then check using a graphing calculator. $$\sin \frac{7 \pi}{12}$$
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