2.2: Algebra with Trigonometric Functions

Note to the Instructor (click to expand) This section focuses on performing our old Algebra (factoring, distribution, powers, etc.) on expressions involving trigonometric functions. The results will not require any trigonometric identities. The point here is for the student to realize that Algebra still works, even though we are not using \( x\). To succeed in this section, you'll need to use some skills from previous courses. While you should already know them, this is the first time they've been required. You can review these skills in CRC's Corequisite Codex. If you have a support class, it might cover some, but not all, of these topics. Prerequisite Skills and Support Topics (click to expand) Simplifying Expressions Laws of Exponents: Once we start working with powers, a mastery of the Laws of Exponents must be assumed. Multiplying and Distributing Polynomial Expressions: This skill is foundational for multiplying expressions containing trigonometric functions, as shown in Examples 3 and 4. Factoring Techniques Greatest Common Factor: This skill is reviewed and applied to trigonometric expressions in Example 5. Factoring by Grouping: This skill is reviewed and applied to trigonometric expressions in Example 6. Factoring Trinomials: This skill is reviewed and applied to trigonometric expressions in Example 7. Factoring Difference of Squares: This skill is reviewed and applied to trigonometric expressions in Example 8. Factoring Sums and Differences of Cubes: This skill is reviewed and applied to trigonometric expressions in Example 9. The following is a list of learning objectives for this section. Learning Objectives (click to expand) Evaluate expressions involving trigonometric functions. Compute powers of trigonometric functions. Evaluate products of trigonometric functions. Perform factorization on expressions involving trigonometric functions.

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As we move forward, we will need to apply some techniques from Algebra to analyze more complicated trigonometric expressions. Therefore, let's review some algebraic terminology.

Definition: Algebraic Expression

An algebraic expression is a well-formed mathematical phrase that combines numbers, variables, and mathematical operations. It does not include an equals sign.

For example, the height of a golf ball is given in feet by the expression \(−16t^2 + 64t\), where \(t\) is the number of seconds after the ball is hit.

Definition: Evaluate (an algebraic expression)

To evaluate an algebraic expression means to substitute a specific value for the variable or variables involved.

Thus, after 1 second, the height of the golf ball is\[ -16(1)^2+64(1)=-16+64=48 \text { feet, }\nonumber \]and after 2 seconds, the height is\[-16(2)^2+64(2)=-64+128=64 \text { feet }\nonumber \]and so on.

Evaluating Trigonometric Expressions

The value of a trigonometric function represents a number, and it may appear as part of an algebraic expression. Expressions containing trigonometric functions can be simplified or evaluated like other algebraic expressions.

Example \( \PageIndex{ 1 } \)

Evaluate each expression for \(X=30^{\circ}\) and \(Y=240^{\circ}\).

1. \(\frac{1}{\sqrt{3}} \tan \left(Y\right)+3 \sin \left(X\right)\)
2. \(6 \tan \left(X\right) \cos \left(Y\right)\)

Solutions

1. Substituting the values for \(X\) and \(Y\), we get the following:\[\begin{array}{rrrclcl}
   \scriptscriptstyle\xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \frac{1}{\sqrt{3}} \tan \left(Y\right)+3 \sin \left(X\right) & = & \dfrac{1}{\sqrt{3}} \tan \left(240^{\circ}\right)+3 \left(\sin 30^{\circ}\right) & \quad & \left( \text{substitute} \right) \\[6pt] \scriptscriptstyle\xcancel{\mathrm{Arithmetic}} \to \xcancel{\mathrm{Algebra}} \to \mathrm{Trigonometry} & \implies & & = & \dfrac{1}{\sqrt{3}}(\sqrt{3})+3\left(\dfrac{1}{2}\right) & \quad & \left( \text{evaluating the trigonometric functions}\right) \\[6pt] \scriptscriptstyle\mathrm{Arithmetic} & \implies & & = & 1+\dfrac{3}{2} & \quad & \left( \text{canceling like factors} \right) \\[6pt] \scriptscriptstyle\mathrm{Arithmetic} & \implies & & = & \dfrac{5}{2} & \quad & \left( \text{addition} \right) \\[6pt] \end{array}\nonumber \]
2. This expression includes the product of two trigonometric functions.\[ \begin{array}{rrrclcl}
   \scriptscriptstyle\xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra}& \implies &6 \tan \left(X\right) \cos \left(Y\right) & = & 6\left(\tan \left(30^{\circ}\right)\right)\left(\cos \left(240^{\circ}\right)\right) & \quad & \left( \text{substitute} \right)\\[6pt] \scriptscriptstyle\xcancel{\mathrm{Arithmetic}} \to \xcancel{\mathrm{Algebra}} \to \mathrm{Trigonometry} & \implies & & = & 6\left(\dfrac{1}{\sqrt{3}}\right)\left(-\dfrac{1}{2}\right) & \quad & \left( \text{evaluating the trigonometric functions}\right) \\[6pt] \scriptscriptstyle\mathrm{Arithmetic} & \implies & & = & -\dfrac{3}{\sqrt{3}} & \quad & \left( \text{simplify} \right) \\[6pt] \scriptscriptstyle\xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & & = & -\sqrt{3} & \quad & \left( \text{rationalizing the denominator} \right)\\[6pt] \end{array}\nonumber \]

Checkpoint \( \PageIndex{ 1 } \)

Evaluate each expression for \(X=30^{\circ}, Y=60^{\circ}\).

1. \(4 \sin \left(3 X+45^{\circ}\right)\)
2. \(1-\cos (4 Y)\)

Answers

1. \(2 \sqrt{2}\)
2. \(\dfrac{3}{2}\)

Caution: Do NOT "Split" Arguments of Functions

In Checkpoint \( \PageIndex{ 1 } \), \(\sin \left(3 X+45^{\circ}\right)\) is not equal to \(\sin \left(3 X\right)+\sin \left(45^{\circ}\right)\). That is,\[\sin \left(90^{\circ}+45^{\circ}\right) \neq \sin \left(90^{\circ}\right)+\sin \left(45^{\circ}\right)\nonumber \](You can check this for yourself.) Before applying the sine function, we must follow the Order of Operations and evaluate the expression \(3 X+45^{\circ}\) inside parentheses.

Powers of Trigonometric Functions

Compare the two expressions\[(\cos \left(\theta\right))^2 \quad \text{and} \quad\cos \left(\theta^2\right)\nonumber \]They are not the same.

• The first expression, \((\cos \left(\theta\right))^2\), says to compute \(\cos \left(\theta\right)\) and then square the result.
• \(\cos \left(\theta^2\right)\), on the other hand, says to square the angle first and then compute the cosine.

For example, if \(\theta=30^{\circ}\), then\[ \left(\cos \left(30^{\circ}\right)\right)^2 = \left(\dfrac{\sqrt{3}}{2}\right)^2 = \dfrac{3}{4} \nonumber \]but\[ \cos \left(\left(30^{\circ}\right)^2\right) = \cos \left(900^{\circ}\right) \nonumber \]We have not delved into the details of how to compute the cosine of such a large angle (yet). Still, it should make sense that if the point \( P\left( x,y \right) \) is on the terminal side of \( \theta = 900^{ \circ } \), then \( P \) must be on the terminal side of any angle coterminal with \( \theta = 900^{ \circ } \). Specifically, \( P \) must be on the terminal side of \( 900^{ \circ } - 360^{ \circ } = 540^{ \circ }\). Still, this angle is too unwieldy to deal with, so we find another, smaller angle coterminal to \( 540^{ \circ } \). Luckily, \( 540^{ \circ } - 360^{ \circ } = 180^{ \circ } \) fits that bill! Hence, if \( P(x,y) \) is on the terminal side of \( 900^{ \circ } \), it must also be on the terminal side of \( 180^{ \circ } \). Since the point \( \left( -1,0 \right) \) is on the terminal side of \( 180^{ \circ } \), we can now compute the cosine of \( 900^{ \circ } \).\[ \cos\left( \left( 30^{ \circ } \right)^2 \right) = \cos\left( 900^{ \circ } \right) = \cos\left( 180^{ \circ } \right) = \dfrac{x}{r} = \dfrac{-1}{1} = -1. \nonumber \]Notice this is not the same value as \( \cos\left( 30^{ \circ } \right) \).

This brings us to a new notation reserved specifically for trigonometric functions.

Notation: Powers of Trigonometric Functions

\[ \begin{array}{rclc|crcl}
\sin^2\left( \theta \right) & \text{ means } & \left( \sin\left( \theta \right) \right)^2 & \quad & \quad & \csc^2\left( \theta \right) & \text{ means } & \left( \csc\left( \theta \right) \right)^2 \\[6pt]
\cos^2\left( \theta \right) & \text{ means } & \left( \cos\left( \theta \right) \right)^2 & \quad & \quad & \sec^2\left( \theta \right) & \text{ means } & \left( \sec\left( \theta \right) \right)^2 \\[6pt]
\tan^2\left( \theta \right) & \text{ means } & \left( \tan\left( \theta \right) \right)^2 & \quad & \quad & \cot^2\left( \theta \right) & \text{ means } & \left( \cot\left( \theta \right) \right)^2 \\[6pt]
\end{array} \nonumber \]

Example \( \PageIndex{ 2 } \)

Evaluate \(\sin ^2 \left(45^{\circ}\right)\).

Solution

\[\begin{array}{rrrclcl}
& & \sin^2 \left(45^{\circ}\right) & = & \left(\sin \left(45^{\circ}\right)\right)^2 & \quad & \left( \text{notational definition} \right) \\[6pt] \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \xcancel{\mathrm{Algebra}} \to \mathrm{Trigonometry} & \implies & & = & \left(\dfrac{1}{\sqrt{2}}\right)^2 & \quad & \left( \text{evaluating the trigonometric function} \right) \\[6pt] \scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & & = & \dfrac{1}{2} & \quad & \left( \text{Laws of Exponents} \right) \\[6pt] \end{array} \nonumber \]

Other powers are written in the same fashion. Thus, for example, \(\sin ^3 \left(\theta\right)=\left(\sin \left(\theta\right)\right)^3\).

Checkpoint \( \PageIndex{ 2 } \)

Evaluate \(\tan ^4 \left(60^{\circ}\right)\)

Answer

\(9\)

Caution: Negative Exponents

We have already stated that\[ \left[ \cos\left( \theta \right) \right]^2 = \cos^2\left( \theta \right) \nonumber \]and\[ \left[ \sin\left( \theta \right) \right]^5 = \sin^5\left( \theta \right). \nonumber \]However, we will not adopt such notation for negative exponents. That is,\[ \left[ \cos\left( \theta \right) \right]^{-1} \neq \cos^{-1}\left( \theta \right). \nonumber \]The reason for excluding the negative exponents from this "shorthand" notation will become apparent when we discuss inverse trigonometric functions.

Products of Trigonometric Functions

We can multiply together trigonometric expressions, just as we multiply algebraic expressions. Recall that we use the Distributive Law in computing products such as\[x(3 x-2)=3 x^2-2 x\nonumber \]and\[(x-3)(x+5)=x^2+2 x-15.\nonumber \]

Example \( \PageIndex{ 3 } \)

Using the Distributive Law, multiply \[\cos \left(t\right)\left(3 \cos \left(t\right)-2\right).\nonumber \]

Solution

Think of \(\cos \left(t\right)\) as a single variable, and multiply by each term inside parentheses. (The algebraic form of the calculation is shown on the right in bold).\[ \begin{array}{rrrclcl}
\scriptscriptstyle \xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & \cos \left(t\right)\left(3\cos \left(t\right) - 2\right) & = & \left(\cos \left(t\right)\right)\left(3\cos \left(t\right)\right) - \left(\cos \left(t\right) \cdot 2\right) & \quad & \left( \text{distribution: }\boldsymbol{x(3x-2) =x \cdot 3x - x \cdot 2}\right) \\[6pt] \scriptscriptstyle\xcancel{\mathrm{Arithmetic}} \to \mathrm{Algebra} & \implies & & = & 3\cos^2 \left(t\right) - 2\cos \left(t\right) & \quad & \left(\text{Commutative Property of Multiplication}\right) \\[6pt] \end{array}\nonumber \]Notice that we write \(\left(\cos \left(t\right)\right)^2\) as \(\cos ^2 \left(t\right)\).

Checkpoint \( \PageIndex{ 3 } \)

Multiply \(2 \tan \left(\beta\right)\left(4 \tan ^2 \left(\beta\right)+\tan \left(\alpha\right)\right)\)

Answer

\(8 \tan ^3 \left(\beta\right)+2 \tan \left(\beta\right) \tan \left(\alpha\right)\)

We can also use the Distributive Law to multiply binomials that include trigonometric functions.

Example \( \PageIndex{ 4 } \)

Multiply \(\left(4 \sin \left(C\right)-1\right)\left(3 \sin \left(C\right)+2\right)\).

Solution

This calculation is similar to the product \((4 x-1)(3 x+2)\), except that the variable \(x\) has been replaced by \(\sin \left(C\right)\). Compare the calculations for the two products; first the familiar algebraic product:\[ \begin{array}{rcl}
(4 x-1)(3 x+2) & = & 4 x \cdot 3 x+4 x \cdot 2-1 \cdot 3 x-1 \cdot 2 \\[6pt] & = & 12 x^2+8 x-3 x-2=12 x^2+5 x-2 \\[6pt] \end{array} \nonumber \]We compute the product in this example in the same way, but replacing \(x\) by \(\sin \left(C\right)\).\[\begin{array}{rcl}
\left(4 \sin \left(C\right)-1\right)\left(\sin \left(C\right)+2\right) & = & \left(4 \sin \left(C\right)\right)\left(3 \sin \left(C\right)\right)+\left(4 \sin \left(C\right) \cdot 2\right)-1\left(3 \sin \left(C\right)\right)-1 \cdot 2 \\[6pt] & = & 12 \sin ^2 \left(C\right)+5 \sin\left(C\right)-2 \\[6pt] \end{array}\nonumber \]

Checkpoint \( \PageIndex{ 4 } \)

Expand \(\left(4 \cos \left(\alpha\right)+3\right)^2\)

Answer

\(16 \cos ^2 \left(\alpha\right)+24 \cos \left(\alpha\right)+9\)

Factoring Expressions Involving Trigonometric Functions

We can factor trigonometric expressions with the same techniques we use for algebraic expressions. The next few subsections of this topic quickly review all our previous factorization techniques. It is important to note that each of these factoring techniques will be needed at some point during Trigonometry.

Factoring Out a Common Expression

Likely, the first technique you encountered when learning to factor was factoring out the greatest common factor (GCF). I call this factoring out a common expression.

In the next two examples, compare the familiar algebraic factoring with a similar trigonometric expression.

Example \( \PageIndex{ 5 } \)

Factor.

1. \(6 w^2-9 w\)
2. \(6 \sin ^2 \left(\theta\right)-9 \sin \left(\theta\right)\)

Solutions

1. We factor out the common factor, \(3 w\).\[6 w^2-9 w=3 w(2 w-3)\nonumber \]
2. We factor out the common factor, \(3 \sin \left(\theta\right)\).\[6 \sin ^2 \left(\theta\right)-9 \sin \left(\theta\right)=3 \sin \left(\theta\right)\left(2 \sin\left( \theta\right)-3\right)\nonumber \]

Checkpoint \( \PageIndex{ 5 } \)

Factor.

1. \(2 a^2-a b\)
2. \(2 \cos ^2 \left(\phi\right)-\cos \left(\phi\right) \sin \left(\phi\right)\)

Answers

1. \(a(2 a-b)\)
2. \(\cos \left(\phi\right)\left(2 \cos \left(\phi\right)-\sin \left(\phi\right)\right)\)

Factoring by Grouping

Once you mastered factoring out common expressions from a collection of terms, you moved to factoring by grouping. We only review via example here as we did in the previous topic.

Example \(\PageIndex{6}\)

Factor.

1. \( 10x^2 + 2x - 15xy - 3y \)
2. \( 10\sin^2\left( \theta \right) + 2\sin\left( \theta \right) - 15 \sin\left( \theta \right) \cos\left( \theta \right) - 3 \cos\left( \theta \right) \)

Solutions

1. Grouping terms, we get\[ \begin{array}{rclcl}
   10x^2 + 2x - 15xy - 3y & = & \left( 10x^2 + 2x \right) + \left( -15xy - 3y \right) & \quad & \left( \text{Grouping} \right) \\[6pt] & = & 2x\left( 5x + 1 \right) -3y \left( 5x + 1 \right) & \quad & \left( \text{Factoring out the GCF from each group} \right) \\[6pt] & = & 2x\mathbf{\left( 5x + 1 \right)} -3y \mathbf{\left( 5x + 1 \right)} & \quad & \left( \text{Noticing the result has two common factors} \right) \\[6pt] & = & \left( 5x + 1 \right)\left( 2x - 3y \right) & \quad & \left( \text{Factoring out the common factor of }\left( 5x + 1 \right) \right) \\[6pt] \end{array} \nonumber \]
2. The structure of this expression is almost identical to that of the previous problem. Let \( x = \sin\left( \theta \right) \) and \( y = \cos\left( \theta \right) \) to get\[ \begin{array}{rclcl}
   10\sin^2\left( \theta \right) + 2\sin\left( \theta \right) - 15 \sin\left( \theta \right) \cos\left( \theta \right) - 3 \cos\left( \theta \right) & = & 10x^2 + 2x - 15xy - 3y & \quad & \left( \text{Substitution: }x = \sin\left( \theta \right) \text{ and }y=\cos\left( \theta \right) \right) \\[6pt] & = & \left( 5x + 1 \right)\left( 2x - 3y \right) & \quad & \left( \text{From part (a)} \right) \\[6pt] & = & \left( 5\sin\left( \theta \right) + 1 \right)\left( 2\sin\left( \theta \right) - 3\cos\left( \theta \right) \right) & \quad & \left( \text{"Back-substitute:" }\sin\left( \theta \right) = x \text{ and }\cos\left( \theta \right) = y\right) \\[6pt] \end{array} \nonumber \]

Example \( \PageIndex{ 6b } \) illustrates a common tactic we will use as we move through mathematics. Namely, when an expression (or equation) looks overly complicated, we make it easier to work with by performing substitutions. In Calculus, this will be tied to a technique called \( u \)-Substitutions (where we use the variable \( u \) instead of \( x \)).

Checkpoint \(\PageIndex{6}\)

Factor.

1. \( 12rt + 9r - 20t - 15 \)
2. \( 12\sec\left( \alpha \right)\tan\left( \beta \right) + 9\sec\left( \alpha \right) - 20 \tan\left( \beta \right) - 15 \)

Answers

1. \( \left( 3r - 5 \right)\left( 4t + 3 \right) \)
2. \( \left( 3\sec\left( \alpha \right) - 5 \right)\left( 4\tan\left( \beta \right) + 3 \right) \)

Factoring Trinomials

Despite being more complex than factoring a difference of squares (which we will talk about next), the next factorization technique you learned was factoring trinomials.

Example \( \PageIndex{ 7 } \)

Factor.

1. \(t^2-3 t-10\)
2. \(\tan ^2 \left(\alpha\right)-3 \tan \left(\alpha\right)-10\)

Solutions

1. We look for numbers \(p\) and \(q\) so that \((t+p)(t+q)=t^2-3 t-10\). Their product is \(p q=-10\), and their sum is \(p+q=-3\). By checking the factors of \(-10\) for the correct sum, we find \(p=-5\) and \(q=2\). Thus,\[t^2-3 t-10=(t-5)(t+2)\nonumber \]
2. Now replace \(t\) by \(\tan \left(\alpha\right)\) to find\[\tan ^2 \left(\alpha\right)-3 \tan \left(\alpha\right)-10=\left(\tan \left(\alpha\right)-5\right)\left(\tan \left(\alpha\right)+2\right)\nonumber \]

Checkpoint \( \PageIndex{ 7 } \)

Factor.

1. \(3 z^2-2 z-1\)
2. \(3 \sin ^2 \left(\beta\right)-2 \sin \left(\beta\right)-1\)

Answers

1. \((3 z+1)(z-1)\)
2. \(\left(3 \sin \left(\beta\right)+1\right)\left(\sin \left(\beta\right)-1\right)\)

Factoring Differences of Squares

In addition to factoring out common expressions and factoring trinomials, special factoring formulas for the Difference of Squares and the Sums and Differences of Cubes will be needed. Recall the following theorem from Algebra.

Theorem: Difference of Squares

For any algebraic expressions \( F \) and \( L \), \[ F^2 - L^2 = \left( F - L \right)\left( F + L \right). \nonumber \]

Proof

\[ \begin{array}{rclcl}
\left( F - L \right)\left( F + L \right) & = & F^2 + FL - FL - L^2 & \quad & \left( \text{distributing} \right) \\[6pt]
& = & F^2 - L^2 & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]

Example \(\PageIndex{8}\)

Factor completely.\[ 16 \cos^4\left( \theta \right) - 1 \nonumber \]

Solution

By this point, you (hopefully) understand what we will do.\[ \begin{array}{rclcl}
16\cos^4\left( \theta \right) - 1 & = & 16x^4 - 1 & \quad & \left( \text{Substitute: }x = \cos\left( \theta \right) \right) \\[6pt] & = & \left(4x^2\right)^2 - 1^2 & \quad & \left( \text{Laws of Exponents} \right) \\[6pt] & = & F^2 - L^2 & \quad & \left( \text{Substitute: }F = 4x^2 \text{ and } L = 1 \right) \\[6pt] & = & \left(F - L\right)\left( F + L \right) & \quad & \left( \text{Difference of Squares} \right) \\[6pt] & = & \left(4x^2 - 1\right)\left( 4x^2 + 1 \right) & \quad & \left( \text{Resubstitute: }4x^2 = F \text{ and } 1 = L \right) \\[6pt] & = & \left((2x)^2 - 1^2\right)\left( 4x^2 + 1 \right) & \quad & \left( \text{Laws of Exponents} \right) \\[6pt] & = & \left(\overline{F}^2 - \overline{L}^2\right)\left( 4x^2 + 1 \right) & \quad & \left( \text{Substitute: }\overline{F} = 2x \text{ and } \overline{L} = 1 \right) \\[6pt] & = & \left(\overline{F} - \overline{L}\right)\left(\overline{F} + \overline{L}\right)\left( 4x^2 + 1 \right) & \quad & \left( \text{Difference of Squares} \right) \\[6pt] & = & \left(2x - 1\right)\left(2x + 1\right)\left( 4x^2 + 1 \right) & \quad & \left( \text{Resubstitute: } 2x = \overline{F} \text{ and } 1 = \overline{L} \right) \\[6pt] & = & \left(2\cos\left( \theta \right) - 1\right)\left(2\cos\left( \theta \right) + 1\right)\left( 4\cos^2\left( \theta \right) + 1 \right) & \quad & \left( \text{Resubstitute: } \cos\left( \theta \right) = x \right) \\[6pt] \end{array} \nonumber \]

A couple of notes are in order for Example \( \PageIndex{ 8 } \). First and foremost, while we took our time with substitutions and being meticulous with our work, you need to reach a level of proficiency in mathematics where you do not need to perform those substitutions to do the factoring. That is, you need to reach the point (and very soon) where you see\[ 16\cos^4\left( \theta \right) - 1\nonumber \]and immediately rewrite it (without substitutions) as\[ \left( 4\cos^2\left( \theta \right) - 1\right)\left( 4\cos^2\left( \theta \right) + 1 \right) = \left(2\cos\left( \theta \right) - 1\right)\left(2\cos\left( \theta \right) + 1\right)\left( 4\cos^2\left( \theta \right) + 1 \right). \nonumber \]I cannot express how important it is for you, as the student, to reach that level of mathematical fluency as quickly as possible (truthfully, in a college-level mathematics course, you should already have this level of fluency).

The second note is better mentioned as a cautionary statement.

Caution: \( F^2 + L^2 \) is NOT Factorable

A common mistake of students in lower levels of Algebra is to think that \( F^2 + L^2 \) is factorable - this is not true (unless we are factoring over the complex number system). That is,\[ 4x^2 + 1 \nonumber \]cannot be factored.¹

Checkpoint \(\PageIndex{8}\)

Factor completely.\[ \sin^4\left( \alpha \right) - 81 \cos^4\left( \beta \right) \nonumber \]

Answer

\( \left( \sin\left( \alpha \right) - 3 \cos\left( \beta \right) \right)\left( \sin\left( \alpha \right) + 3 \cos\left( \beta \right) \right) \left( \sin^2\left( \alpha \right) + 9 \cos^2\left( \beta \right) \right) \)

Factoring Sums and Differences of Cubes

We have now reached the final factorization method, which we must recall from Algebra.²

Theorem: Sums and Differences of Cubes

For any algebraic expressions \( F \) and \( L \),\[ \begin{array}{rcl}
F^3 + L^3 & = & \left( F + L \right)\left( F^2 - FL + L^2 \right) \\[6pt] F^3 - L^3 & = & \left( F - L \right)\left( F^2 + FL + L^2 \right) \\[6pt] \end{array} \nonumber \]

Proof

\[ \begin{array}{rclcl}
\left( F + L \right)\left( F^2 - FL + L^2 \right) & = & F^3 - F^2 L + FL^2 + F^2 L - FL^2 + L^3 & \quad & \left( \text{distributing} \right) \\[6pt]
& = & F^3 + L^3 & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]and\[ \begin{array}{rclcl}
\left( F - L \right)\left( F^2 + FL + L^2 \right) & = & F^3 + F^2 L + FL^2 - F^2 L - FL^2 - L^3 & \quad & \left( \text{distributing} \right) \\[6pt]
& = & F^3 - L^3 & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]

Example \(\PageIndex{9}\)

Factor.\[ 27\sin^3\left( A \right) - 8 \nonumber \]

Solution

This time, we will speed things up a bit.\[ \begin{array}{rclcl}
27\sin^3\left( A \right) - 8 & = & 27x^3 - 8 & \quad & \left( \text{Substitute: }x = \sin\left( A \right) \right) \\[6pt] & = & (3x)^3 - 2^3 & \quad & \left( \text{Laws of Exponents} \right) \\[6pt] & = & \left(3x - 2\right)\left( (3x)^2 + (3x)(2) + 2^2 \right) & \quad & \left( \text{Difference of Cubes} \right) \\[6pt] & = & \left(3x - 2\right)\left( (9x^2 + 6x + 4 \right) & \quad & \\[6pt] & = & \left(3\sin\left( A \right) - 2\right)\left( (9\sin^2\left( A \right) + 6\sin\left( A \right) + 4 \right) & \quad & \left( \text{Resubstitute: }\sin\left( A \right) = x \right)\\[6pt] \end{array} \nonumber \]

Checkpoint \(\PageIndex{9}\)

Factor.\[ 125\csc^3\left( \theta \right) + 64\cot^3\left( \phi \right) \nonumber \]

Answer

\( \left( 5\csc\left( \theta \right) + 4\cot\left( \phi \right) \right)\left( 25\csc^2\left( \theta \right) - 20 \csc\left( \theta \right)\cot\left( \phi \right) + 16 \cot^2\left( \phi \right)\right) \)

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Footnotes

¹ Technically, if we are allowing ourselves to factor over the complex number system (i.e. if we are considering numbers like \( 2 + 3i \)), then we can factor sums of squares; however, this will not be the case for us until nearly the end of this textbook. Even then, most instructors might skip that material.

² Believe it or not, other factorization methods exist beyond these basic techniques. For example, in College Algebra (also sometimes referred to as Precalculus), you learn about factoring high-degree polynomials with integer coefficients using the Rational Roots Theorem. There are also critically important methods for factoring expressions involving negative rational exponents - these are encountered frequently in Calculus.