Sum and Difference Identities

Learning Objectives

In this section, you will:

• Use sum and difference formulas for cosine.
• Use sum and difference formulas for sine.
• Use sum and difference formulas for tangent.
• Use sum and difference formulas for cofunctions.
• Use sum and difference formulas to verify identities.

Figure 1 Mount McKinley, in Denali National Park, Alaska, rises 20,237 feet (6,168 m) above sea level. It is the highest peak in North America. (credit: Daniel A. Leifheit, Flickr)

How can the height of a mountain be measured? What about the distance from Earth to the sun? Like many seemingly impossible problems, we rely on mathematical formulas to find the answers. The trigonometric identities, commonly used in mathematical proofs, have had real-world applications for centuries, including their use in calculating long distances.

The trigonometric identities we will examine in this section can be traced to a Persian astronomer who lived around 950 AD, but the ancient Greeks discovered these same formulas much earlier and stated them in terms of chords. These are special equations or postulates, true for all values input to the equations, and with innumerable applications.

In this section, we will learn techniques that will enable us to solve problems such as the ones presented above. The formulas that follow will simplify many trigonometric expressions and equations. Keep in mind that, throughout this section, the term formula is used synonymously with the word identity.

Using the Sum and Difference Formulas for Cosine

Finding the exact value of the sine, cosine, or tangent of an angle is often easier if we can rewrite the given angle in terms of two angles that have known trigonometric values. We can use the special angles, which we can review in the unit circle shown in Figure 2.

Figure 2 The Unit Circle

We will begin with the sum and difference formulas for cosine, so that we can find the cosine of a given angle if we can break it up into the sum or difference of two of the special angles. See Table 1.

Sum formula for cosine	cos(α+β)=cosαcosβ−sinαsinβcos(𝛼+𝛽)=cos𝛼cos𝛽−sin𝛼sin𝛽
Difference formula for cosine	cos(α−β)=cosαcosβ+sinαsinβcos(𝛼−𝛽)=cos𝛼cos𝛽+sin𝛼sin𝛽

First, we will prove the difference formula for cosines. Let’s consider two points on the unit circle. See Figure 3. Point P𝑃 is at an angle α𝛼 from the positive x-axis with coordinates (cosα,sinα)(cos𝛼,sin𝛼) and point Q𝑄 is at an angle of β𝛽 from the positive x-axis with coordinates (cosβ,sinβ).(cos𝛽,sin𝛽). Note the measure of angle POQ𝑃𝑂𝑄 is α−β.𝛼−𝛽.

Label two more points: A𝐴 at an angle of (α−β)(𝛼−𝛽) from the positive x-axis with coordinates (cos(α−β),sin(α−β));(cos(𝛼−𝛽),sin(𝛼−𝛽)); and point B𝐵 with coordinates (1,0).(1,0). Triangle POQ𝑃𝑂𝑄 is a rotation of triangle AOB𝐴𝑂𝐵 and thus the distance from P𝑃 to Q𝑄 is the same as the distance from A𝐴 to B.𝐵.

Figure 3

We can find the distance from P𝑃 to Q𝑄 using the distance formula.

dPQ==(cosα−cosβ)2+(sinα−sinβ)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−√cos2α−2cosαcosβ+cos2β+sin2α−2sinαsinβ+sin2β−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√𝑑𝑃𝑄=(cos𝛼−cos𝛽)2+(sin𝛼−sin𝛽)2=cos2𝛼−2cos𝛼cos𝛽+cos2𝛽+sin2𝛼−2sin𝛼sin𝛽+sin2𝛽

Then we apply the Pythagorean identity and simplify.

===(cos2α+sin2α)+(cos2β+sin2β)−2cosαcosβ−2sinαsinβ−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√1+1−2cosαcosβ−2sinαsinβ−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2−2cosαcosβ−2sinαsinβ−−−−−−−−−−−−−−−−−−−−−−−−−−−√=(cos2𝛼+sin2𝛼)+(cos2𝛽+sin2𝛽)−2cos𝛼cos𝛽−2sin𝛼sin𝛽=1+1−2cos𝛼cos𝛽−2sin𝛼sin𝛽=2−2cos𝛼cos𝛽−2sin𝛼sin𝛽

Similarly, using the distance formula we can find the distance from A𝐴 to B.𝐵.

dAB==(cos(α−β)−1)2+(sin(α−β)−0)2−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√cos2(α−β)−2cos(α−β)+1+sin2(α−β)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√𝑑𝐴𝐵=(cos(𝛼−𝛽)−1)2+(sin(𝛼−𝛽)−0)2=cos2(𝛼−𝛽)−2cos(𝛼−𝛽)+1+sin2(𝛼−𝛽)

Applying the Pythagorean identity and simplifying we get:

===(cos2(α−β)+sin2(α−β))−2cos(α−β)+1−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√1−2cos(α−β)+1−−−−−−−−−−−−−−−−√2−2cos(α−β)−−−−−−−−−−−−−√=(cos2(𝛼−𝛽)+sin2(𝛼−𝛽))−2cos(𝛼−𝛽)+1=1−2cos(𝛼−𝛽)+1=2−2cos(𝛼−𝛽)

Because the two distances are the same, we set them equal to each other and simplify.

2−2cosαcosβ−2sinαsinβ−−−−−−−−−−−−−−−−−−−−−−−−−−−√2−2cosαcosβ−2sinαsinβ==2−2cos(α−β)−−−−−−−−−−−−−√2−2cos(α−β)2−2cos𝛼cos𝛽−2sin𝛼sin𝛽=2−2cos(𝛼−𝛽)2−2cos𝛼cos𝛽−2sin𝛼sin𝛽=2−2cos(𝛼−𝛽)

Finally we subtract 22 from both sides and divide both sides by −2.−2.

cosαcosβ+sinαsinβ=cos(α−β)  cos𝛼cos𝛽+sin𝛼sin𝛽=cos(𝛼−𝛽)  

Thus, we have the difference formula for cosine. We can use similar methods to derive the cosine of the sum of two angles.

SUM AND DIFFERENCE FORMULAS FOR COSINE

These formulas can be used to calculate the cosine of sums and differences of angles.

cos(α+β)=cosαcosβ−sinαsinβcos(𝛼+𝛽)=cos𝛼cos𝛽−sin𝛼sin𝛽

cos(α−β)=cosαcosβ+sinαsinβcos(𝛼−𝛽)=cos𝛼cos𝛽+sin𝛼sin𝛽

HOW TO

Given two angles, find the cosine of the difference between the angles.

1. Write the difference formula for cosine.
2. Substitute the values of the given angles into the formula.
3. Simplify.

EXAMPLE 1

Finding the Exact Value Using the Formula for the Cosine of the Difference of Two Angles

Using the formula for the cosine of the difference of two angles, find the exact value of cos(5π4−π6).cos(5𝜋4−𝜋6).

Solution

Begin by writing the formula for the cosine of the difference of two angles. Then substitute the given values.

cos(α−β)cos(5π4−π6)=====cosαcosβ+sinαsinβcos(5π4)cos(π6)+sin(5π4)sin(π6)(−2√2)(3√2)−(2√2)(12)−6√4−2√4−6√−2√4cos(𝛼−𝛽)=cos𝛼cos𝛽+sin𝛼sin𝛽cos(5𝜋4−𝜋6)=cos(5𝜋4)cos(𝜋6)+sin(5𝜋4)sin(𝜋6)=(−22)(32)−(22)(12)=−64−24=−6−24

Keep in mind that we can always check the answer using a graphing calculator in radian mode.

TRY IT #1

Find the exact value of cos(π3−π4).cos(𝜋3−𝜋4).

EXAMPLE 2

Finding the Exact Value Using the Formula for the Sum of Two Angles for Cosine

Find the exact value of cos(75°).cos(75°).

Solution

As 75°=45°+30°,75°=45°+30°, we can evaluate cos(75°)cos(75°) as cos(45°+30°).cos(45°+30°).

cos(α+β)cos(45°+30°)=====cosαcosβ−sinαsinβcos(45°)cos(30°)−sin(45°)sin(30°)2√2(3√2)−2√2(12)6√4−2√46√−2√4cos(𝛼+𝛽)=cos𝛼cos𝛽−sin𝛼sin𝛽cos(45°+30°)=cos(45°)cos(30°)−sin(45°)sin(30°)=22(32)−22(12)=64−24=6−24

Keep in mind that we can always check the answer using a graphing calculator in degree mode.

Analysis

Note that we could have also solved this problem using the fact that 75°=135°−60°.75°=135°−60°.

cos(α−β)cos(135°−60°)=====cosαcosβ+sinαsinβcos(135°)cos(60°)+sin(135°)sin(60°)(−2√2)(12)+(2√2)(3√2)(−2√4)+(6√4)(6√−2√4)cos(𝛼−𝛽)=cos𝛼cos𝛽+sin𝛼sin𝛽cos(135°−60°)=cos(135°)cos(60°)+sin(135°)sin(60°)=(−22)(12)+(22)(32)=(−24)+(64)=(6−24)

TRY IT #2

Find the exact value of cos(105°).cos(105°).

Using the Sum and Difference Formulas for Sine

The sum and difference formulas for sine can be derived in the same manner as those for cosine, and they resemble the cosine formulas.

SUM AND DIFFERENCE FORMULAS FOR SINE

These formulas can be used to calculate the sines of sums and differences of angles.

sin(α+β)=sinαcosβ+cosαsinβsin(𝛼+𝛽)=sin𝛼cos𝛽+cos𝛼sin𝛽

sin(α−β)=sinαcosβ−cosαsinβsin(𝛼−𝛽)=sin𝛼cos𝛽−cos𝛼sin𝛽

HOW TO

Given two angles, find the sine of the difference between the angles.

1. Write the difference formula for sine.
2. Substitute the given angles into the formula.
3. Simplify.

EXAMPLE 3

Using Sum and Difference Identities to Evaluate the Difference of Angles

Use the sum and difference identities to evaluate the difference of the angles and show that part a equals part b.

1. ⓐ sin(45°−30°)sin(45°−30°)
2. ⓑ sin(135°−120°)sin(135°−120°)

Solution

1. ⓐ Let’s begin by writing the formula and substitute the given angles.sin(α−β)sin(45°−30°)==sinαcosβ−cosαsinβsin(45°)cos(30°)−cos(45°)sin(30°)sin(𝛼−𝛽)=sin𝛼cos𝛽−cos𝛼sin𝛽sin(45°−30°)=sin(45°)cos(30°)−cos(45°)sin(30°)Next, we need to find the values of the trigonometric expressions.sin(45°)=2–√2,cos(30°)=3–√2,cos(45°)=2–√2,sin(30°)=12sin(45°)=22,cos(30°)=32,cos(45°)=22,sin(30°)=12Now we can substitute these values into the equation and simplify.sin(45°−30°)==2√2(3√2)−2√2(12)6√−2√4sin(45°−30°)=22(32)−22(12)=6−24
2. ⓑ Again, we write the formula and substitute the given angles.sin(α−β)sin(135°−120°)==sinαcosβ−cosαsinβsin(135°)cos(120°)−cos(135°)sin(120°)sin(𝛼−𝛽)=sin𝛼cos𝛽−cos𝛼sin𝛽sin(135°−120°)=sin(135°)cos(120°)−cos(135°)sin(120°)Next, we find the values of the trigonometric expressions.sin(135°)=2–√2,cos(120°)=−12,cos(135°)=2–√2,sin(120°)=3–√2sin(135°)=22,cos(120°)=−12,cos(135°)=22,sin(120°)=32Now we can substitute these values into the equation and simplify.sin(135°−120°)===2√2(−12)−(−2√2)(3√2)−2√+6√46√−2√4sin(135°−120°)=22(−12)−(−22)(32)=−2+64=6−24

EXAMPLE 4

Finding the Exact Value of an Expression Involving an Inverse Trigonometric Function

Find the exact value of sin(cos−112+sin−135).sin(cos−112+sin−135). Then check the answer with a graphing calculator.

Solution

The pattern displayed in this problem is sin(α+β).sin(𝛼+𝛽). Let α=cos−112𝛼=cos−112 and β=sin−135.𝛽=sin−135. Then we can write

cosαsinβ==12,0≤α≤π35,−π2≤β≤π2cos𝛼=12,0≤𝛼≤𝜋sin𝛽=35,−𝜋2≤𝛽≤𝜋2

We will use the Pythagorean identities to find sinαsin𝛼 and cosβ.cos𝛽.

sinαcosβ========1−cos2α−−−−−−−−√1−14−−−−−√34−−√3√21−sin2β−−−−−−−√1−925−−−−−√1625−−√45sin𝛼=1−cos2𝛼=1−14=34=32cos𝛽=1−sin2𝛽=1−925=1625=45

Using the sum formula for sine,

sin(cos−112+sin−135)====sin(α+β)sinαcosβ+cosαsinβ3√2⋅45+12⋅3543√+310sin(cos−112+sin−135)=sin(𝛼+𝛽)=sin𝛼cos𝛽+cos𝛼sin𝛽=32⋅45+12⋅35=43+310

Using the Sum and Difference Formulas for Tangent

Finding exact values for the tangent of the sum or difference of two angles is a little more complicated, but again, it is a matter of recognizing the pattern.

Finding the sum of two angles formula for tangent involves taking quotient of the sum formulas for sine and cosine and simplifying. Recall, tanx=sinxcosx,cosx≠0.tan𝑥=sin𝑥cos𝑥,cos𝑥≠0.

Let’s derive the sum formula for tangent.

tan(α+β)======sin(α+β)cos(α+β)sinαcosβ+cosαsinβcosαcosβ−sinαsinβsinαcosβ+cosαsinβcosαcosβcosαcosβ−sinαsinβcosαcosβsinαcosβcosαcosβ+cosαsinβcosαcosβcosαcosβcosαcosβ−sinαsinβcosαcosβsinαcosα+sinβcosβ1−sinαsinβcosαcosβtanα+tanβ1−tanαtanβDivide the numerator and denominator by cosαcosβ.tan(𝛼+𝛽)=sin(𝛼+𝛽)cos(𝛼+𝛽)=sin𝛼cos𝛽+cos𝛼sin𝛽cos𝛼cos𝛽−sin𝛼sin𝛽=sin𝛼cos𝛽+cos𝛼sin𝛽cos𝛼cos𝛽cos𝛼cos𝛽−sin𝛼sin𝛽cos𝛼cos𝛽Divide the numerator and denominator by cos𝛼cos𝛽.=sin𝛼cos𝛽cos𝛼cos𝛽+cos𝛼sin𝛽cos𝛼cos𝛽cos𝛼cos𝛽cos𝛼cos𝛽−sin𝛼sin𝛽cos𝛼cos𝛽=sin𝛼cos𝛼+sin𝛽cos𝛽1−sin𝛼sin𝛽cos𝛼cos𝛽=tan𝛼+tan𝛽1−tan𝛼tan𝛽

We can derive the difference formula for tangent in a similar way.

SUM AND DIFFERENCE FORMULAS FOR TANGENT

The sum and difference formulas for tangent are:

tan(α+β)=tanα+tanβ1−tanαtanβtan(𝛼+𝛽)=tan𝛼+tan𝛽1−tan𝛼tan𝛽

tan(α−β)=tanα−tanβ1+tanαtanβtan(𝛼−𝛽)=tan𝛼−tan𝛽1+tan𝛼tan𝛽

HOW TO

Given two angles, find the tangent of the sum of the angles.

1. Write the sum formula for tangent.
2. Substitute the given angles into the formula.
3. Simplify.

EXAMPLE 5

Finding the Exact Value of an Expression Involving Tangent

Find the exact value of tan(π6+π4).tan(𝜋6+𝜋4).

Solution

Let’s first write the sum formula for tangent and then substitute the given angles into the formula.

tan(α+β)tan(π6+π4)==tanα+tanβ1−tanαtanβtan(π6)+tan(π4)1−(tan(π6))(tan(π4))tan(𝛼+𝛽)=tan𝛼+tan𝛽1−tan𝛼tan𝛽tan(𝜋6+𝜋4)=tan(𝜋6)+tan(𝜋4)1−(tan(𝜋6))(tan(𝜋4))

Next, we determine the individual function values within the formula:

tan(π6)=13–√,tan(π4)=1tan(𝜋6)=13,tan(𝜋4)=1

So we have

tan(π6+π4)====13√+11−(13√)(1)1+3√3√3√−13√1+3√3√(3√3√−1)3√+13√−1tan(𝜋6+𝜋4)=13+11−(13)(1)=1+333−13=1+33(33−1)=3+13−1

TRY IT #3

Find the exact value of tan(2π3+π4).tan(2𝜋3+𝜋4).

EXAMPLE 6

Finding Multiple Sums and Differences of Angles

Given sinα=35,0<α<π2,cosβ=−513,π<β<3π2,sin𝛼=35,0<𝛼<𝜋2,cos𝛽=−513,𝜋<𝛽<3𝜋2, find

1. ⓐ sin(α+β)sin(𝛼+𝛽)
2. ⓑ cos(α+β)cos(𝛼+𝛽)
3. ⓒ tan(α+β)tan(𝛼+𝛽)
4. ⓓ tan(α−β)tan(𝛼−𝛽)

Solution

We can use the sum and difference formulas to identify the sum or difference of angles when the ratio of sine, cosine, or tangent is provided for each of the individual angles. To do so, we construct what is called a reference triangle to help find each component of the sum and difference formulas.

• ⓐ To find sin(α+β),sin(𝛼+𝛽), we begin with sinα=35sin𝛼=35 and 0<α<π2.0<𝛼<𝜋2. The side opposite α𝛼 has length 3, the hypotenuse has length 5, and α𝛼 is in the first quadrant. See Figure 4. Using the Pythagorean Theorem, we can find the length of side a:𝑎:a2+32a2a===52164𝑎2+32=52𝑎2=16𝑎=4Figure 4Since cosβ=−513cos𝛽=−513 and π<β<3π2,𝜋<𝛽<3𝜋2, the side adjacent to β𝛽 is −5,−5, the hypotenuse is 13, and β𝛽 is in the third quadrant. See Figure 5. Again, using the Pythagorean Theorem, we have(−5)2+a225+a2a2a====132169144±12(−5)2+𝑎2=13225+𝑎2=169𝑎2=144𝑎=±12Since β𝛽 is in the third quadrant, a=–12.𝑎=–12.Figure 5The next step is finding the cosine of α𝛼 and the sine of β.𝛽. The cosine of α𝛼 is the adjacent side over the hypotenuse. We can find it from the triangle in Figure 5: cosα=45.cos𝛼=45. We can also find the sine of β𝛽 from the triangle in Figure 5, as opposite side over the hypotenuse: sinβ=−1213.sin𝛽=−1213. Now we are ready to evaluate sin(α+β).sin(𝛼+𝛽).sin(α+β)====sinαcosβ+cosαsinβ(35)(−513)+(45)(−1213)−1565−4865−6365sin(𝛼+𝛽)=sin𝛼cos𝛽+cos𝛼sin𝛽=(35)(−513)+(45)(−1213)=−1565−4865=−6365
• ⓑ We can find cos(α+β)cos(𝛼+𝛽) in a similar manner. We substitute the values according to the formula.cos(α+β)====cosαcosβ−sinαsinβ(45)(−513)−(35)(−1213)−2065+36651665cos(𝛼+𝛽)=cos𝛼cos𝛽−sin𝛼sin𝛽=(45)(−513)−(35)(−1213)=−2065+3665=1665
• ⓒ For tan(α+β),tan(𝛼+𝛽), if sinα=35sin𝛼=35 and cosα=45,cos𝛼=45, thentanα=3545=34tan𝛼=3545=34If sinβ=−1213sin𝛽=−1213 and cosβ=−513,cos𝛽=−513, thentanβ=−1213−513=125tan𝛽=−1213−513=125Then,tan(α+β)====tanα+tanβ1−tanαtanβ34+1251−34(125)  6320−1620−6316tan(𝛼+𝛽)=tan𝛼+tan𝛽1−tan𝛼tan𝛽=34+1251−34(125)=  6320−1620=−6316
• ⓓ To find tan(α−β),tan(𝛼−𝛽), we have the values we need. We can substitute them in and evaluate.tan(α−β)====tanα−tanβ1+tanαtanβ34−1251+34(125)−33205620−3356tan(𝛼−𝛽)=tan𝛼−tan𝛽1+tan𝛼tan𝛽=34−1251+34(125)=−33205620=−3356

Analysis

A common mistake when addressing problems such as this one is that we may be tempted to think that α𝛼 and β𝛽 are angles in the same triangle, which of course, they are not. Also note that

tan(α+β)=sin(α+β)cos(α+β)tan(𝛼+𝛽)=sin(𝛼+𝛽)cos(𝛼+𝛽)

Using Sum and Difference Formulas for Cofunctions

Now that we can find the sine, cosine, and tangent functions for the sums and differences of angles, we can use them to do the same for their cofunctions. You may recall from Right Triangle Trigonometry that, if the sum of two positive angles is π2,𝜋2, those two angles are complements, and the sum of the two acute angles in a right triangle is π2,𝜋2, so they are also complements. In Figure 6, notice that if one of the acute angles is labeled as θ,𝜃, then the other acute angle must be labeled (π2−θ).(𝜋2−𝜃).

Notice also that sinθ=cos(π2−θ),sin𝜃=cos(𝜋2−𝜃), which is opposite over hypotenuse. Thus, when two angles are complementary, we can say that the sine of θ𝜃 equals the cofunction of the complement of θ.𝜃. Similarly, tangent and cotangent are cofunctions, and secant and cosecant are cofunctions.

Figure 6

From these relationships, the cofunction identities are formed. Recall that you first encountered these identities in The Unit Circle: Sine and Cosine Functions.

COFUNCTION IDENTITIES

The cofunction identities are summarized in Table 2.

sinθ=cos(π2−θ)sin𝜃=cos(𝜋2−𝜃)	cosθ=sin(π2−θ)cos𝜃=sin(𝜋2−𝜃)
tanθ=cot(π2−θ)tan𝜃=cot(𝜋2−𝜃)	cotθ=tan(π2−θ)cot𝜃=tan(𝜋2−𝜃)
secθ=csc(π2−θ)sec𝜃=csc(𝜋2−𝜃)	cscθ=sec(π2−θ)csc𝜃=sec(𝜋2−𝜃)

Notice that the formulas in the table may also be justified algebraically using the sum and difference formulas. For example, using

cos(α−β)=cosαcosβ+sinαsinβ,cos(𝛼−𝛽)=cos𝛼cos𝛽+sin𝛼sin𝛽,

we can write

cos(π2−θ)===cosπ2cosθ+sinπ2sinθ(0)cosθ+(1)sinθsinθcos(𝜋2−𝜃)=cos𝜋2cos𝜃+sin𝜋2sin𝜃=(0)cos𝜃+(1)sin𝜃=sin𝜃

EXAMPLE 7

Finding a Cofunction with the Same Value as the Given Expression

Write tanπ9tan𝜋9 in terms of its cofunction.

Solution

The cofunction of tanθ=cot(π2−θ).tan𝜃=cot(𝜋2−𝜃). Thus,

tan(π9)===cot(π2−π9)cot(9π18−2π18)cot(7π18)tan(𝜋9)=cot(𝜋2−𝜋9)=cot(9𝜋18−2𝜋18)=cot(7𝜋18)

TRY IT #4

Write sinπ7sin𝜋7 in terms of its cofunction.

Using the Sum and Difference Formulas to Verify Identities

Verifying an identity means demonstrating that the equation holds for all values of the variable. It helps to be very familiar with the identities or to have a list of them accessible while working the problems. Reviewing the general rules presented earlier may help simplify the process of verifying an identity.

HOW TO

Given an identity, verify using sum and difference formulas.

1. Begin with the expression on the side of the equal sign that appears most complex. Rewrite that expression until it matches the other side of the equal sign. Occasionally, we might have to alter both sides, but working on only one side is the most efficient.
2. Look for opportunities to use the sum and difference formulas.
3. Rewrite sums or differences of quotients as single quotients.
4. If the process becomes cumbersome, rewrite the expression in terms of sines and cosines.

EXAMPLE 8

Verifying an Identity Involving Sine

Verify the identity sin(α+β)+sin(α−β)=2sinαcosβ.sin(𝛼+𝛽)+sin(𝛼−𝛽)=2sin𝛼cos𝛽.

Solution

We see that the left side of the equation includes the sines of the sum and the difference of angles.

sin(α+β)sin(α−β)==sinαcosβ+cosαsinβsinαcosβ−cosαsinβsin(𝛼+𝛽)=sin𝛼cos𝛽+cos𝛼sin𝛽sin(𝛼−𝛽)=sin𝛼cos𝛽−cos𝛼sin𝛽

We can rewrite each using the sum and difference formulas.

sin(α+β)+sin(α−β)==sinαcosβ+cosαsinβ+sinαcosβ−cosαsinβ2sinαcosβsin(𝛼+𝛽)+sin(𝛼−𝛽)=sin𝛼cos𝛽+cos𝛼sin𝛽+sin𝛼cos𝛽−cos𝛼sin𝛽=2sin𝛼cos𝛽

We see that the identity is verified.

EXAMPLE 9

Verifying an Identity Involving Tangent

Verify the following identity.

sin(α−β)cosαcosβ=tanα−tanβsin(𝛼−𝛽)cos𝛼cos𝛽=tan𝛼−tan𝛽

Solution

We can begin by rewriting the numerator on the left side of the equation.

sin(α−β)cosαcosβ====sinαcosβ−cosαsinβcosαcosβsinαcosβcosαcosβ−cosαsinβcosαcosβsinαcosα−sinβcosβtanα−tanβRewrite using a common denominator.Cancel.Rewrite in terms of tangent.sin(𝛼−𝛽)cos𝛼cos𝛽=sin𝛼cos𝛽−cos𝛼sin𝛽cos𝛼cos𝛽=sin𝛼cos𝛽cos𝛼cos𝛽−cos𝛼sin𝛽cos𝛼cos𝛽Rewrite using a common denominator.=sin𝛼cos𝛼−sin𝛽cos𝛽Cancel.=tan𝛼−tan𝛽Rewrite in terms of tangent.

We see that the identity is verified. In many cases, verifying tangent identities can successfully be accomplished by writing the tangent in terms of sine and cosine.

TRY IT #5

Verify the identity: tan(π−θ)=−tanθ.tan(𝜋−𝜃)=−tan𝜃.

EXAMPLE 10

Using Sum and Difference Formulas to Solve an Application Problem

Let L1𝐿1 and L2𝐿2 denote two non-vertical intersecting lines, and let θ𝜃 denote the acute angle between L1𝐿1 and L2.𝐿2. See Figure 7. Show that

tanθ=m2−m11+m1m2tan𝜃=𝑚2−𝑚11+𝑚1𝑚2

where m1𝑚1 and m2𝑚2 are the slopes of L1𝐿1 and L2𝐿2 respectively. (Hint: Use the fact that tanθ1=m1tan𝜃1=𝑚1 and tanθ2=m2.tan𝜃2=𝑚2. )

Figure 7

Solution

Using the difference formula for tangent, this problem does not seem as daunting as it might.

tanθ===tan(θ2−θ1)tanθ2−tanθ11+tanθ1tanθ2m2−m11+m1m2tan𝜃=tan(𝜃2−𝜃1)=tan𝜃2−tan𝜃11+tan𝜃1tan𝜃2=𝑚2−𝑚11+𝑚1𝑚2

EXAMPLE 11

Investigating a Guy-wire Problem

For a climbing wall, a guy-wire R𝑅 is attached 47 feet high on a vertical pole. Added support is provided by another guy-wire S𝑆 attached 40 feet above ground on the same pole. If the wires are attached to the ground 50 feet from the pole, find the angle α𝛼 between the wires. See Figure 8.

Figure 8

Solution

Let’s first summarize the information we can gather from the diagram. As only the sides adjacent to the right angle are known, we can use the tangent function. Notice that tanβ=4750,tan𝛽=4750, and tan(β−α)=4050=45.tan(𝛽−𝛼)=4050=45. We can then use difference formula for tangent.

tan(β−α)=tanβ−tanα1+tanβtanαtan(𝛽−𝛼)=tan𝛽−tan𝛼1+tan𝛽tan𝛼

Now, substituting the values we know into the formula, we have

454(1+4750tanα)==4750−tanα1+4750tanα5(4750−tanα)45=4750−tan𝛼1+4750tan𝛼4(1+4750tan𝛼)=5(4750−tan𝛼)

Use the distributive property, and then simplify the functions.

4(1)+4(4750)tanα4+3.76tanα5tanα+3.76tanα8.76tanαtanαtan−1(0.07991)====≈≈5(4750)−5tanα4.7−5tanα0.70.70.07991.0797414(1)+4(4750)tan𝛼=5(4750)−5tan𝛼4+3.76tan𝛼=4.7−5tan𝛼5tan𝛼+3.76tan𝛼=0.78.76tan𝛼=0.7tan𝛼≈0.07991tan−1(0.07991)≈.079741

Now we can calculate the angle in degrees.

α≈0.079741(180π)≈4.57°𝛼≈0.079741(180𝜋)≈4.57°

Analysis

Occasionally, when an application appears that includes a right triangle, we may think that solving is a matter of applying the Pythagorean Theorem. That may be partially true, but it depends on what the problem is asking and what information is given.