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Trigonometry Through Wayfinding and Navigation Across the Pacific

Kamuela E Yong

Section 3.4 Product-to-Sum and Sum-to-Product Formulas

In this section, we will learn how to convert sums of trigonometric functions to products of trigonometric functions, and vice versa. These techniques provide us with tools to simplify expressions and solve equations.

Subsection 3.4.1 Product to Sum Formulas

Definition 3.4.1. Product to Sum Formulas.

\begin{align*} \sin\alpha\cos\beta \amp = \frac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha-\beta)] \\ \cos\alpha\sin\beta \amp = \frac{1}{2}[\sin(\alpha+\beta)-\sin(\alpha-\beta)] \\ \cos\alpha\cos\beta \amp = \frac{1}{2}[\cos(\alpha+\beta)+\cos(\alpha-\beta)] \\ \sin\alpha\sin\beta \amp = \frac{1}{2}[\cos(\alpha-\beta)-\cos(\alpha+\beta)] \end{align*}

Proof.

We add the addition and subtraction formulas for cosine:
\begin{align*} \cos(\alpha+\beta)\amp = \cos\alpha\cos\beta-\sin\alpha\sin\beta \\ +\cos(\alpha-\beta)\amp = \cos\alpha\cos\beta+\sin\alpha\sin\beta \\ \hline \\ \cos(\alpha+\beta)+\cos(\alpha-\beta) \amp = 2\cos\alpha\cos\beta \end{align*}
Dividing both sides by 2, we get
\begin{equation*} \frac{1}{2}[\cos(\alpha+\beta)+\cos(\alpha-\beta)]=\cos\alpha\cos\beta \end{equation*}
Next, we add the addition and subtraction formulas for sine:
\begin{align*} \sin(\alpha+\beta)\amp = \sin\alpha\cos\beta+\cos\alpha\sin\beta \\ +\sin(\alpha-\beta)\amp = \sin\alpha\cos\beta-\cos\alpha\sin\beta \\ \hline \\ \sin(\alpha+\beta)+\sin(\alpha-\beta) \amp = 2\sin\alpha\cos\beta \end{align*}
Dividing both sides by 2, we get
\begin{equation*} \frac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha-\beta)]=\sin\alpha\cos\beta \end{equation*}
Next, we subtract the addition and subtraction formulas for sine:
\begin{align*} \sin(\alpha+\beta)\amp = \sin\alpha\cos\beta+\cos\alpha\sin\beta \\ -\sin(\alpha-\beta)\amp = \sin\alpha\cos\beta-\cos\alpha\sin\beta \\ \hline \\ \sin(\alpha+\beta)-\sin(\alpha-\beta) \amp = 2\sin\beta\cos\alpha \end{align*}
Dividing both sides by 2, we get
\begin{equation*} \frac{1}{2}[\sin(\alpha+\beta)-\sin(\alpha-\beta)]=\sin\beta\cos\alpha \end{equation*}
Finally, we subtract the addition and subtraction formulas for cosine:
\begin{align*} \cos(\alpha-\beta)\amp = \cos\alpha\cos\beta+\sin\alpha\sin\beta \\ -\cos(\alpha+\beta)\amp = \cos\alpha\cos\beta-\sin\alpha\sin\beta \\ \hline \\ \cos(\alpha-\beta)-\cos(\alpha+\beta) \amp = 2\sin\alpha\sin\beta \end{align*}
Dividing both sides by 2, we get
\begin{equation*} \frac{1}{2}[\cos(\alpha-\beta)-\cos(\alpha+\beta)]=\sin\alpha\sin\beta \end{equation*}

Example 3.4.2.

Express the product of \(\cos(2x)\cos(5x)\) as a sum or difference of sine and cosine with no products.

Solution.

Using the formula we get
\begin{align*} \cos(2x)\cos(5x)\amp = \frac{1}{2}[\cos(2x+5x)+\cos(2x-5x)]\\ \amp = \frac{1}{2}[\cos(7x)+\cos(-3x)] \end{align*}
This satisfies the requirement of expressing the product of \(\cos(2x)\cos(5x)\) as a sum or difference of sine and cosine with no products. However, we can simplify it further.
Since cosine is an even function, \(\cos(-3x) = \cos(3x)\text{.}\) Thus, we can simplify the expression to:
\begin{equation*} \cos(2x)\cos(5x)=\frac{1}{2}[\cos(7x)+\cos(3x)] \end{equation*}

Example 3.4.3.

Express the product of \(\sin(6\theta)\cos(4\theta)\) as a sum or difference of sine and cosine with no products.

Solution.

Using the formula we get
\begin{align*} \sin(6\theta)\cos(4\theta) \amp = \frac{1}{2}[\sin(6\theta+4\theta)+\sin(6\theta-4\theta)]\\ \amp = \frac{1}{2}[\sin(10\theta)+\cos(2\theta)] \end{align*}

Remark 3.4.4. Negative Angles.

In the previous example, both forms \(\cos(2x)\cos(5x) = \frac{1}{2}[\cos(7x)+\cos(-3x)]\) and \(\cos(2x)\cos(5x) = \frac{1}{2}[\cos(7x)+\cos(3x)]\) are valid representations of the answer. However, it is more common to write the first form (where the angles are positive) because it simplifies the expression and aligns with standard conventions for representing trigonometric identities. Positive angles are often preferred for clarity and consistency in mathematical notation. Negative angles can be transformed into positive angles using the even-odd properties of trigonometric functions (Definition 1.5.22).

Subsection 3.4.2 Sum to Product Formulas

Definition 3.4.5. Sum-to-Product Formula.

\begin{align*} \sin\alpha+\sin\beta\amp = 2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)\\ \sin\alpha-\sin\beta\amp = 2\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)\\ \cos\alpha+\cos\beta\amp = 2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)\\ \cos\alpha-\cos\beta\amp = -2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right) \end{align*}

Proof.

We first let \(\alpha=\frac{u+v}{2}\) and \(\beta=\frac{u-v}{2}\text{.}\) Then
\begin{equation*} \alpha+\beta=\frac{u+v}{2}+\frac{u-v}{2}=\frac{2u}{2}=u \end{equation*}
and
\begin{equation*} \alpha-\beta=\frac{u+v}{2}-\frac{u-v}{2}=\frac{2v}{2}=v \end{equation*}
Substituting these values for \(\alpha\text{,}\) \(\beta\text{,}\) \(\alpha+\beta\text{,}\) and \(\alpha-\beta\) into the Product-to-Sum Formulas, we get
\begin{align*} \sin\left(\frac{u+v}{2}\right)\cos\left(\frac{u-v}{2}\right)\amp = \frac{1}{2}[\sin(u)+\sin(v)]\\ \cos\left(\frac{u+v}{2}\right)\sin\left(\frac{u-v}{2}\right)\amp = \frac{1}{2}[\sin(u)-\sin(v)]\\ \cos\left(\frac{u+v}{2}\right)\cos\left(\frac{u-v}{2}\right)\amp = \frac{1}{2}[\cos(u)+\cos(v)]\\ \sin\left(\frac{u+v}{2}\right)\sin\left(\frac{u-v}{2}\right)\amp = \frac{1}{2}[\cos(v)-\cos(u)] \end{align*}
Multiplying both sides by 2 and substituting \(u\) with \(\alpha\) and \(v\) with \(\beta\text{,}\) we arrive at the Sum-to-Product Formula, where we negate the last equation.

Example 3.4.6.

Express the sum \(\sin(8x)+\sin(2x)\) as a product of sines or cosines.

Solution.

Using the formula we get
\begin{align*} \sin(8x)+\sin(2x) \amp = 2\sin\left(\frac{8x+2x}{2}\right)\cos\left(\frac{8x-2x}{2}\right) \\ \amp = 2\sin\left(\frac{10x}{2}\right)\cos\left(\frac{6x}{2}\right) \\ \amp = 2\sin\left(5x\right)\cos\left(3x\right) \end{align*}

Example 3.4.7.

Express the difference \(\cos(3t)-\cos(5t)\) as a product of sines or cosines.

Solution.

Using the formula we get
\begin{align*} \cos(3t)-\cos(5t) \amp = -2\sin\left(\frac{3t+5t}{2}\right)\sin\left(\frac{3t-5t}{2}\right) \\ \amp = -2\sin\left(\frac{8t}{2}\right)\sin\left(\frac{-2t}{2}\right) \\ \amp = -2\sin\left(4t\right)\sin\left(-t\right) \\ \amp = 2\sin\left(4t\right)\sin\left(t\right) \end{align*}

Exercises 3.4.3 Exercises

Exercise Group.

Express each product as a sum or difference of sine and cosine.
1.
\(\sin(3x)\cos(5x)\)
Answer.
\(\frac{1}{2}[\sin(8x)+\sin(-2x)]=\frac{1}{2}[\sin(8x)-\sin(2x)]\)
2.
\(\sin(7t)\cos(-2t)\)
Answer.
\(\frac{1}{2}[\sin(5t)+\sin(9t)]\)
3.
\(\cos(-3t)\sin(7t)\)
Answer.
\(\frac{1}{2}[\sin(4t)-\sin(-10t)]=\frac{1}{2}[\sin(4t)+\sin(10t)]\)
4.
\(\cos(9\theta)\sin(6\theta)\)
Answer.
\(\frac{1}{2}[\cos(15\theta)-\cos(3\theta)]\)
5.
\(\cos(-4x)\cos(6x)\)
Answer.
\(\frac{1}{2}[\cos(2x)+\cos(-10x)]=\frac{1}{2}[\cos(2x)+\cos(10x)]\)
6.
\(\cos(2\theta)\cos(4\theta)\)
Answer.
\(\frac{1}{2}[\cos(6\theta)+\cos(-2\theta)]=\frac{1}{2}[\cos(6\theta)+\cos(2\theta)]\)
7.
\(\sin(-\theta)\sin(8\theta)\)
Answer.
\(\frac{1}{2}[\cos(-9\theta)-\cos(7\theta)]=\frac{1}{2}[\cos(9\theta)-\cos(7\theta)]\)
8.
\(\sin(6t)\sin(-3t)\)
Answer.
\(\frac{1}{2}[\cos(9t)-\cos(3t)]\)

Exercise Group.

Express each sum or difference as a product.
9.
\(\sin(-2\theta)+\sin(-9\theta)\)
Answer.
\(2\sin\left(-\frac{11\theta}{2}\right)\cos\left(\frac{7\theta}{2}\right)=-2\sin\left(\frac{11\theta}{2}\right)\cos\left(\frac{7\theta}{2}\right)\)
10.
\(\sin(5x)+\sin(7x)\)
Answer.
\(2\sin\left(6x\right)\cos\left(-x\right)=2\sin\left(6x\right)\cos\left(x\right)\)
11.
\(\sin(4\theta)-\sin(-7\theta)\)
Answer.
\(2\cos\left(\frac{11\theta}{2}\right)\sin\left(-\frac{3\theta}{2}\right)=-2\cos\left(\frac{11\theta}{2}\right)\sin\left(\frac{3\theta}{2}\right)\)
12.
\(\sin(3x)-\sin(-4x)\)
Answer.
\(2\cos\left(\frac{7x}{2}\right)\sin\left(-\frac{x}{2}\right)=-2\cos\left(\frac{7x}{2}\right)\sin\left(\frac{x}{2}\right)\)
13.
\(\cos(8\theta)+\cos(-5\theta)\)
Answer.
\(2\cos\left(\frac{3\theta}{2}\right)\cos\left(\frac{13\theta}{2}\right)\)
14.
\(\cos(9t)+\cos(2t)\)
Answer.
\(2\cos\left(\frac{11t}{2}\right)\cos\left(\frac{7t}{2}\right)\)
15.
\(\cos(6\theta)-\cos(8\theta)\)
Answer.
\(-2\sin\left(7\theta\right)\sin\left(-\theta\right)=2\sin\left(7\theta\right)\sin\left(\theta\right)\)
16.
\(\cos(6t)-\cos(-3t)\)
Answer.
\(-2\sin\left(\frac{9t}{2}\right)\sin\left(\frac{3t}{2}\right)\)

Exercise Group.

Find the exact value of each expression.
17.
\(\sin(195^{\circ})\cos(105^{\circ})\)
Answer.
\(\frac{1}{2}\left(-\frac{\sqrt{3}}{2}+1\right)\)
18.
\(\cos(225^{\circ})\cos(195^{\circ})\)
Answer.
\(\frac{\sqrt{3}}{4}-\frac{1}{4}\)
19.
\(\sin(195^{\circ})-\sin(75^{\circ})\)
Answer.
\(-\frac{\sqrt{6}}{2}\)
20.
\(\cos(165^{\circ})-\cos(105^{\circ})\)
Answer.
\(-\frac{\sqrt{2}}{2}\)
21.
\(\sin(285^{\circ})+\sin(195^{\circ})\)
Answer.
\(-\frac{\sqrt{6}}{2}\)
22.
\(\cos(255^{\circ})+\cos(15^{\circ})\)
Answer.
\(\frac{\sqrt{2}}{2}\)

Exercise Group.

Verify the identity
23.
\(\sin\theta+\sin(3\theta)=4\sin\theta\cos^2\theta\)
24.
\(\cos(3\theta)+\cos\theta=2\left(\cos^3\theta-\sin^2\theta\cos\theta\right)\)
25.
\(6\cos(5\theta)\sin(6\theta)=3\sin(11\theta)+3\sin(\theta)\)
26.
\(\frac{\sin\theta+\sin(3\theta)}{2\sin(2\theta)}=\cos\theta\)
27.
\(\frac{\cos\theta+\cos(3\theta)}{2\cos(2\theta)}=\cos\theta\)
28.
\(\frac{\cos\theta-\cos(3\theta)}{\sin\theta+\sin(3\theta)}=\tan\theta\)
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