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Sum and Difference Identities

By the end of this section, you will be able to:

  • 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.

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 below.

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 the table below.

Sum formula for cosinecos(α+β)=cosαcosβsinαsinβ\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta
Difference formula for cosinecos(αβ)=cosαcosβ+sinαsinβ\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta

First, we will prove the difference formula for cosines. Let’s consider two points on the unit circle. Point PP is at an angle α\alpha from the positive xx-axis with coordinates (cosα,sinα)(\cos\alpha,\sin\alpha) and point QQ is at an angle of β\beta from the positive xx-axis with coordinates (cosβ,sinβ)(\cos\beta,\sin\beta). Note the measure of angle POQPOQ is αβ\alpha-\beta.

Label two more points: AA at an angle of (αβ)(\alpha-\beta) from the positive xx-axis with coordinates (cos(αβ),sin(αβ))(\cos(\alpha-\beta),\sin(\alpha-\beta)); 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 BB.

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

dPQ=(cosαcosβ)2+(sinαsinβ)2=cos2α2cosαcosβ+cos2β+sin2α2sinαsinβ+sin2β \begin{array}{lrcl} & d_{PQ} &=& \sqrt{(\cos\alpha-\cos\beta)^2+(\sin\alpha-\sin\beta)^2} \\[4pt] & &=& \sqrt{\cos^2\alpha-2\cos\alpha\cos\beta+\cos^2\beta+\sin^2\alpha-2\sin\alpha\sin\beta+\sin^2\beta} \end{array}

Then we apply the Pythagorean Identity and simplify.

=(cos2α+sin2α)+(cos2β+sin2β)2cosαcosβ2sinαsinβ=1+12cosαcosβ2sinαsinβ=22cosαcosβ2sinαsinβ \begin{array}{lrcl} & &=& \sqrt{(\cos^2\alpha+\sin^2\alpha)+(\cos^2\beta+\sin^2\beta)-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta} \\[4pt] & &=& \sqrt{1+1-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta} \\[4pt] & &=& \sqrt{2-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta} \end{array}

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

dAB=(cos(αβ)1)2+(sin(αβ)0)2=cos2(αβ)2cos(αβ)+1+sin2(αβ) \begin{array}{lrcl} & d_{AB} &=& \sqrt{(\cos(\alpha-\beta)-1)^2+(\sin(\alpha-\beta)-0)^2} \\[4pt] & &=& \sqrt{\cos^2(\alpha-\beta)-2\cos(\alpha-\beta)+1+\sin^2(\alpha-\beta)} \end{array}

Applying the Pythagorean Identity and simplifying we get:

=(cos2(αβ)+sin2(αβ))2cos(αβ)+1=12cos(αβ)+1=22cos(αβ) \begin{array}{lrcl} & &=& \sqrt{(\cos^2(\alpha-\beta)+\sin^2(\alpha-\beta))-2\cos(\alpha-\beta)+1} \\[4pt] & &=& \sqrt{1-2\cos(\alpha-\beta)+1} \\[4pt] & &=& \sqrt{2-2\cos(\alpha-\beta)} \end{array}

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

22cosαcosβ2sinαsinβ=22cos(αβ)22cosαcosβ2sinαsinβ=22cos(αβ) \begin{array}{lrcl} & \sqrt{2-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta} &=& \sqrt{2-2\cos(\alpha-\beta)} \\[4pt] & 2-2\cos\alpha\cos\beta-2\sin\alpha\sin\beta &=& 2-2\cos(\alpha-\beta) \end{array}

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

cosαcosβ+sinαsinβ=cos(αβ)\cos\alpha\cos\beta+\sin\alpha\sin\beta=\cos(\alpha-\beta)

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)=cosαcosβsinαsinβ\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\betacos(αβ)=cosαcosβ+sinαsinβ\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta

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. Using the formula for the cosine of the difference of two angles, find the exact value of cos(5π4π6)\cos\left(\tfrac{5\pi}{4}-\tfrac{\pi}{6}\right).

Solution. Use the formula for the cosine of the difference of two angles. We have

cos(αβ)=cosαcosβ+sinαsinβcos(5π4π6)=cos(5π4)cos(π6)+sin(5π4)sin(π6)=(22)(32)(22)(12)=6424=624 \begin{array}{lrcl} & \cos(\alpha-\beta) &=& \cos\alpha\cos\beta+\sin\alpha\sin\beta \\[4pt] & \cos\left(\tfrac{5\pi}{4}-\tfrac{\pi}{6}\right) &=& \cos\left(\tfrac{5\pi}{4}\right)\cos\left(\tfrac{\pi}{6}\right)+\sin\left(\tfrac{5\pi}{4}\right)\sin\left(\tfrac{\pi}{6}\right) \\[4pt] & &=& \left(-\tfrac{\sqrt2}{2}\right)\left(\tfrac{\sqrt3}{2}\right)-\left(\tfrac{\sqrt2}{2}\right)\left(\tfrac12\right) \\[4pt] & &=& -\tfrac{\sqrt6}{4}-\tfrac{\sqrt2}{4} \\[4pt] & &=& \tfrac{-\sqrt6-\sqrt2}{4} \end{array}

Find the exact value ofcos(π3π4)\cos\left(\tfrac{\pi}{3}-\tfrac{\pi}{4}\right).

Example. Find the exact value of cos(75)\cos(75^\circ).

Solution. As 75=45+3075^\circ=45^\circ+30^\circ, we can evaluate cos(75)\cos(75^\circ) as cos(45+30)\cos(45^\circ+30^\circ). Thus,

cos(45+30)=cos(45)cos(30)sin(45)sin(30)=22(32)22(12)=6424=624 \begin{array}{lrcl} & \cos(45^\circ+30^\circ) &=& \cos(45^\circ)\cos(30^\circ)-\sin(45^\circ)\sin(30^\circ) \\[4pt] & &=& \tfrac{\sqrt2}{2}\left(\tfrac{\sqrt3}{2}\right)-\tfrac{\sqrt2}{2}\left(\tfrac12\right) \\[4pt] & &=& \tfrac{\sqrt6}{4}-\tfrac{\sqrt2}{4} \\[4pt] & &=& \tfrac{\sqrt6-\sqrt2}{4} \end{array}

Find the exact value ofcos(105)\cos(105^\circ).

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+cosαsinβ\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\betasin(αβ)=sinαcosβcosαsinβ\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta

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

sin(4530)\sin(45^\circ-30^\circ)sin(135120)\sin(135^\circ-120^\circ)

Solution. ⓐ Let’s begin by writing the formula and substitute the given angles.

sin(αβ)=sinαcosβcosαsinβsin(4530)=sin(45)cos(30)cos(45)sin(30) \begin{array}{lrcl} & \sin(\alpha-\beta) &=& \sin\alpha\cos\beta-\cos\alpha\sin\beta \\[4pt] & \sin(45^\circ-30^\circ) &=& \sin(45^\circ)\cos(30^\circ)-\cos(45^\circ)\sin(30^\circ) \end{array}

Next, we need to find the values of the trigonometric expressions.

sin(45)=22, cos(30)=32, cos(45)=22, sin(30)=12\sin(45^\circ)=\tfrac{\sqrt2}{2},\ \cos(30^\circ)=\tfrac{\sqrt3}{2},\ \cos(45^\circ)=\tfrac{\sqrt2}{2},\ \sin(30^\circ)=\tfrac12

Now we can substitute these values into the equation and simplify.

sin(4530)=22(32)22(12)=624 \begin{array}{lrcl} & \sin(45^\circ-30^\circ) &=& \tfrac{\sqrt2}{2}\left(\tfrac{\sqrt3}{2}\right)-\tfrac{\sqrt2}{2}\left(\tfrac12\right) \\[4pt] & &=& \tfrac{\sqrt6-\sqrt2}{4} \end{array}

ⓑ Again, we write the formula and substitute the given angles.

sin(αβ)=sinαcosβcosαsinβsin(135120)=sin(135)cos(120)cos(135)sin(120) \begin{array}{lrcl} & \sin(\alpha-\beta) &=& \sin\alpha\cos\beta-\cos\alpha\sin\beta \\[4pt] & \sin(135^\circ-120^\circ) &=& \sin(135^\circ)\cos(120^\circ)-\cos(135^\circ)\sin(120^\circ) \end{array}

Next, we find the values of the trigonometric expressions.

sin(135)=22, cos(120)=12, cos(135)=22, sin(120)=32\sin(135^\circ)=\tfrac{\sqrt2}{2},\ \cos(120^\circ)=-\tfrac12,\ \cos(135^\circ)=-\tfrac{\sqrt2}{2},\ \sin(120^\circ)=\tfrac{\sqrt3}{2}

Now we can substitute these values into the equation and simplify.

sin(135120)=22(12)(22)(32)=2+64=624 \begin{array}{lrcl} & \sin(135^\circ-120^\circ) &=& \tfrac{\sqrt2}{2}\left(-\tfrac12\right)-\left(-\tfrac{\sqrt2}{2}\right)\left(\tfrac{\sqrt3}{2}\right) \\[4pt] & &=& \tfrac{-\sqrt2+\sqrt6}{4} \\[4pt] & &=& \tfrac{\sqrt6-\sqrt2}{4} \end{array}

Example. Find the exact value of sin(cos112+sin135)\sin\left(\cos^{-1}\tfrac12+\sin^{-1}\tfrac35\right).

Solution. The pattern displayed in this problem is sin(α+β)\sin(\alpha+\beta). Let α=cos112\alpha=\cos^{-1}\tfrac12 and β=sin135\beta=\sin^{-1}\tfrac35. Then we can write

cosα=12, 0απsinβ=35, π2βπ2 \begin{array}{l} \cos\alpha=\tfrac12,\ 0\le\alpha\le\pi \\ \sin\beta=\tfrac35,\ -\tfrac{\pi}{2}\le\beta\le\tfrac{\pi}{2} \end{array}

We will use the Pythagorean Identities to find sinα\sin\alpha and cosβ\cos\beta.

sinα=1cos2α=114=34=32cosβ=1sin2β=1925=1625=45 \begin{array}{lrcl} & \sin\alpha &=& \sqrt{1-\cos^2\alpha} \\[4pt] & &=& \sqrt{1-\tfrac14} \\[4pt] & &=& \sqrt{\tfrac34} \\[4pt] & &=& \tfrac{\sqrt3}{2} \end{array} \qquad \begin{array}{lrcl} & \cos\beta &=& \sqrt{1-\sin^2\beta} \\[4pt] & &=& \sqrt{1-\tfrac{9}{25}} \\[4pt] & &=& \sqrt{\tfrac{16}{25}} \\[4pt] & &=& \tfrac45 \end{array}

Using the sum formula for sine,

sin(cos112+sin135)=sin(α+β)=sinαcosβ+cosαsinβ=3245+1235=43+310 \begin{array}{lrcl} & \sin\left(\cos^{-1}\tfrac12+\sin^{-1}\tfrac35\right) &=& \sin(\alpha+\beta) \\[4pt] & &=& \sin\alpha\cos\beta+\cos\alpha\sin\beta \\[4pt] & &=& \tfrac{\sqrt3}{2}\cdot\tfrac45+\tfrac12\cdot\tfrac35 \\[4pt] & &=& \tfrac{4\sqrt3+3}{10} \end{array}

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 the quotient of the sum formulas for sine and cosine and simplifying. Recall, tanx=sinxcosx, cosx0\tan x=\tfrac{\sin x}{\cos x},\ \cos x\ne0.

Let’s derive the sum formula for tangent.

tan(α+β)=sin(α+β)cos(α+β)=sinαcosβ+cosαsinβcosαcosβsinαsinβDivide the numerator and denominator by cosαcosβ.=sinαcosβ+cosαsinβcosαcosβcosαcosβsinαsinβcosαcosβ=sinαcosβcosαcosβ+cosαsinβcosαcosβcosαcosβcosαcosβsinαsinβcosαcosβ=sinαcosα+sinβcosβ1sinαsinβcosαcosβ=tanα+tanβ1tanαtanβ \begin{array}{lrcl} & \tan(\alpha+\beta) &=& \tfrac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} \\[4pt] & &=& \tfrac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta-\sin\alpha\sin\beta} \\[4pt] \text{Divide the numerator and denominator by }\cos\alpha\cos\beta. & &=& \cfrac{\tfrac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta}}{\tfrac{\cos\alpha\cos\beta-\sin\alpha\sin\beta}{\cos\alpha\cos\beta}} \\[10pt] & &=& \cfrac{\tfrac{\sin\alpha\cos\beta}{\cos\alpha\cos\beta}+\tfrac{\cos\alpha\sin\beta}{\cos\alpha\cos\beta}}{\tfrac{\cos\alpha\cos\beta}{\cos\alpha\cos\beta}-\tfrac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}} \\[10pt] & &=& \cfrac{\tfrac{\sin\alpha}{\cos\alpha}+\tfrac{\sin\beta}{\cos\beta}}{1-\tfrac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}} \\[10pt] & &=& \tfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} \end{array}

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+tanβ1tanαtanβ\tan(\alpha+\beta)=\tfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}tan(αβ)=tanαtanβ1+tanαtanβ\tan(\alpha-\beta)=\tfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}

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. Find the exact value of tan(π6+π4)\tan\left(\tfrac{\pi}{6}+\tfrac{\pi}{4}\right).

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

tan(α+β)=tanα+tanβ1tanαtanβtan(π6+π4)=tan(π6)+tan(π4)1(tan(π6))(tan(π4)) \begin{array}{lrcl} & \tan(\alpha+\beta) &=& \tfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} \\[4pt] & \tan\left(\tfrac{\pi}{6}+\tfrac{\pi}{4}\right) &=& \tfrac{\tan\left(\tfrac{\pi}{6}\right)+\tan\left(\tfrac{\pi}{4}\right)}{1-\left(\tan\left(\tfrac{\pi}{6}\right)\right)\left(\tan\left(\tfrac{\pi}{4}\right)\right)} \end{array}

Next, we determine the individual tangents within the formula:

tan(π6)=13,tan(π4)=1\tan\left(\tfrac{\pi}{6}\right)=\tfrac{1}{\sqrt3},\qquad\tan\left(\tfrac{\pi}{4}\right)=1

So we have

tan(π6+π4)=13+11(13)(1)=1+33313=1+33(331)=3+131 \begin{array}{lrcl} & \tan\left(\tfrac{\pi}{6}+\tfrac{\pi}{4}\right) &=& \cfrac{\tfrac{1}{\sqrt3}+1}{1-\left(\tfrac{1}{\sqrt3}\right)(1)} \\[10pt] & &=& \cfrac{\tfrac{1+\sqrt3}{\sqrt3}}{\tfrac{\sqrt3-1}{\sqrt3}} \\[10pt] & &=& \tfrac{1+\sqrt3}{\sqrt3}\left(\tfrac{\sqrt3}{\sqrt3-1}\right) \\[4pt] & &=& \tfrac{\sqrt3+1}{\sqrt3-1} \end{array}

Find the exact value oftan(2π3+π4)\tan\left(\tfrac{2\pi}{3}+\tfrac{\pi}{4}\right).

Example. Given sinα=35, 0<α<π2, cosβ=513, π<β<3π2\sin\alpha=\tfrac35,\ 0<\alpha<\tfrac{\pi}{2},\ \cos\beta=-\tfrac{5}{13},\ \pi<\beta<\tfrac{3\pi}{2}, find

sin(α+β)\sin(\alpha+\beta)cos(α+β)\cos(\alpha+\beta)tan(α+β)\tan(\alpha+\beta)tan(αβ)\tan(\alpha-\beta)

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(\alpha+\beta), we begin with sinα=35\sin\alpha=\tfrac35 and 0<α<π20<\alpha<\tfrac{\pi}{2}. The side opposite α\alpha has length 3, the hypotenuse has length 5, and α\alpha is in the first quadrant. See the figure below. Using the Pythagorean Theorem, we can find the length of side aa:

a2+32=52a2=16a=4 \begin{array}{lrcl} & a^2+3^2 &=& 5^2 \\[4pt] & a^2 &=& 16 \\[4pt] & a &=& 4 \end{array}

Since cosβ=513\cos\beta=-\tfrac{5}{13} and π<β<3π2\pi<\beta<\tfrac{3\pi}{2}, the side adjacent to β\beta is 5-5, the hypotenuse is 13, and β\beta is in the third quadrant. See the figure below. Again, using the Pythagorean Theorem, we have

(5)2+a2=13225+a2=169a2=144a=±12 \begin{array}{lrcl} & (-5)^2+a^2 &=& 13^2 \\[4pt] & 25+a^2 &=& 169 \\[4pt] & a^2 &=& 144 \\[4pt] & a &=& \pm12 \end{array}

Since β\beta is in the third quadrant, a=12a=-12.

The next step is finding the cosine of α\alpha and the sine of β\beta. The cosine of α\alpha is the adjacent side over the hypotenuse. We can find it from the triangle above: cosα=45\cos\alpha=\tfrac45. We can also find the sine of β\beta from the triangle above, as opposite side over the hypotenuse: sinβ=1213\sin\beta=-\tfrac{12}{13}. Now we are ready to evaluate sin(α+β)\sin(\alpha+\beta).

sin(α+β)=sinαcosβ+cosαsinβ=(35)(513)+(45)(1213)=15654865=6365 \begin{array}{lrcl} & \sin(\alpha+\beta) &=& \sin\alpha\cos\beta+\cos\alpha\sin\beta \\[4pt] & &=& \left(\tfrac35\right)\left(-\tfrac{5}{13}\right)+\left(\tfrac45\right)\left(-\tfrac{12}{13}\right) \\[4pt] & &=& -\tfrac{15}{65}-\tfrac{48}{65} \\[4pt] & &=& -\tfrac{63}{65} \end{array}

ⓑ We can find cos(α+β)\cos(\alpha+\beta) in a similar manner. We substitute the values according to the formula.

cos(α+β)=cosαcosβsinαsinβ=(45)(513)(35)(1213)=2065+3665=1665 \begin{array}{lrcl} & \cos(\alpha+\beta) &=& \cos\alpha\cos\beta-\sin\alpha\sin\beta \\[4pt] & &=& \left(\tfrac45\right)\left(-\tfrac{5}{13}\right)-\left(\tfrac35\right)\left(-\tfrac{12}{13}\right) \\[4pt] & &=& -\tfrac{20}{65}+\tfrac{36}{65} \\[4pt] & &=& \tfrac{16}{65} \end{array}

ⓒ For tan(α+β)\tan(\alpha+\beta), if sinα=35\sin\alpha=\tfrac35 and cosα=45\cos\alpha=\tfrac45, then

tanα=3545=34\tan\alpha=\tfrac{\tfrac35}{\tfrac45}=\tfrac34

If sinβ=1213\sin\beta=-\tfrac{12}{13} and cosβ=513\cos\beta=-\tfrac{5}{13}, then

tanβ=1213513=125\tan\beta=\tfrac{-\tfrac{12}{13}}{-\tfrac{5}{13}}=\tfrac{12}{5}

Then,

tan(α+β)=tanα+tanβ1tanαtanβ=34+125134(125)=63201620=6316 \begin{array}{lrcl} & \tan(\alpha+\beta) &=& \tfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} \\[10pt] & &=& \cfrac{\tfrac34+\tfrac{12}{5}}{1-\tfrac34\left(\tfrac{12}{5}\right)} \\[10pt] & &=& \cfrac{\tfrac{63}{20}}{-\tfrac{16}{20}} \\[10pt] & &=& -\tfrac{63}{16} \end{array}

ⓓ To find tan(αβ)\tan(\alpha-\beta), we have the values we need. We can substitute them in and evaluate.

tan(αβ)=tanαtanβ1+tanαtanβ=341251+34(125)=33205620=3356 \begin{array}{lrcl} & \tan(\alpha-\beta) &=& \tfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta} \\[10pt] & &=& \cfrac{\tfrac34-\tfrac{12}{5}}{1+\tfrac34\left(\tfrac{12}{5}\right)} \\[10pt] & &=& \cfrac{-\tfrac{33}{20}}{\tfrac{56}{20}} \\[10pt] & &=& -\tfrac{33}{56} \end{array}

Analysis. A common mistake when addressing problems such as this one is that we may be tempted to think that α\alpha and β\beta 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(\alpha+\beta)=\tfrac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)}

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\tfrac{\pi}{2}, those two angles are complements, and the sum of the two acute angles in a right triangle is π2\tfrac{\pi}{2}, so they are also complements. In the figure below, notice that if one of the acute angles is labeled as θ\theta, then the other acute angle must be labeled (π2θ)\left(\tfrac{\pi}{2}-\theta\right).

Notice also that sinθ=cos(π2θ)\sin\theta=\cos\left(\tfrac{\pi}{2}-\theta\right): opposite over hypotenuse. Thus, when two angles are complementary, we can say that the sine of θ\theta equals the cofunction of the complement of θ\theta. Similarly, tangent and cotangent are cofunctions, and secant and cosecant are cofunctions.

From these relationships, the cofunction identities are formed.

Cofunction Identities.

sinθ=cos(π2θ)\sin\theta=\cos\left(\tfrac{\pi}{2}-\theta\right)cosθ=sin(π2θ)\cos\theta=\sin\left(\tfrac{\pi}{2}-\theta\right)
tanθ=cot(π2θ)\tan\theta=\cot\left(\tfrac{\pi}{2}-\theta\right)cotθ=tan(π2θ)\cot\theta=\tan\left(\tfrac{\pi}{2}-\theta\right)
secθ=csc(π2θ)\sec\theta=\csc\left(\tfrac{\pi}{2}-\theta\right)cscθ=sec(π2θ)\csc\theta=\sec\left(\tfrac{\pi}{2}-\theta\right)

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

cos(αβ)=cosαcosβ+sinαsinβ,\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta,

we can write

cos(π2θ)=cosπ2cosθ+sinπ2sinθ=(0)cosθ+(1)sinθ=sinθ \begin{array}{lrcl} & \cos\left(\tfrac{\pi}{2}-\theta\right) &=& \cos\tfrac{\pi}{2}\cos\theta+\sin\tfrac{\pi}{2}\sin\theta \\[4pt] & &=& (0)\cos\theta+(1)\sin\theta \\[4pt] & &=& \sin\theta \end{array}

Example. Write tanπ9\tan\tfrac{\pi}{9} in terms of its cofunction.

Solution. The cofunction of tanθ=cot(π2θ)\tan\theta=\cot\left(\tfrac{\pi}{2}-\theta\right). Thus,

tan(π9)=cot(π2π9)=cot(9π182π18)=cot(7π18) \begin{array}{lrcl} & \tan\left(\tfrac{\pi}{9}\right) &=& \cot\left(\tfrac{\pi}{2}-\tfrac{\pi}{9}\right) \\[4pt] & &=& \cot\left(\tfrac{9\pi}{18}-\tfrac{2\pi}{18}\right) \\[4pt] & &=& \cot\left(\tfrac{7\pi}{18}\right) \end{array}

Writesinπ7\sin\tfrac{\pi}{7}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 from Simplifying and Verifying Trigonometric Identities 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. Verify the identity sin(α+β)+sin(αβ)=2sinαcosβ\sin(\alpha+\beta)+\sin(\alpha-\beta)=2\sin\alpha\cos\beta.

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

sin(α+β)=sinαcosβ+cosαsinβsin(αβ)=sinαcosβcosαsinβ \begin{array}{l} \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta \\ \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta \end{array}

We can rewrite each using the sum and difference formulas.

sin(α+β)+sin(αβ)=sinαcosβ+cosαsinβ+sinαcosβcosαsinβ=2sinαcosβ \begin{array}{lrcl} & \sin(\alpha+\beta)+\sin(\alpha-\beta) &=& \sin\alpha\cos\beta+\cos\alpha\sin\beta+\sin\alpha\cos\beta-\cos\alpha\sin\beta \\[4pt] & &=& 2\sin\alpha\cos\beta \end{array}

We see that the identity is verified.

Example. Verify the following identity.

sin(αβ)cosαcosβ=tanαtanβ\tfrac{\sin(\alpha-\beta)}{\cos\alpha\cos\beta}=\tan\alpha-\tan\beta

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

sin(αβ)cosαcosβ=sinαcosβcosαsinβcosαcosβRewrite using a common denominator.=sinαcosβcosαcosβcosαsinβcosαcosβCancel.=sinαcosαsinβcosβRewrite in terms of tangent.=tanαtanβ \begin{array}{lrcl} \text{} & \tfrac{\sin(\alpha-\beta)}{\cos\alpha\cos\beta} &=& \tfrac{\sin\alpha\cos\beta-\cos\alpha\sin\beta}{\cos\alpha\cos\beta} \\[8pt] \text{Rewrite using a common denominator.} & &=& \tfrac{\sin\alpha\cos\beta}{\cos\alpha\cos\beta}-\tfrac{\cos\alpha\sin\beta}{\cos\alpha\cos\beta} \\[8pt] \text{Cancel.} & &=& \tfrac{\sin\alpha}{\cos\alpha}-\tfrac{\sin\beta}{\cos\beta} \\[8pt] \text{Rewrite in terms of tangent.} & &=& \tan\alpha-\tan\beta \end{array}

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.

Which of the following is equivalent totan(πθ)\tan(\pi-\theta)for everyθ\thetain its domain?

Example. Let L1L_1 and L2L_2 denote two non-vertical intersecting lines, and let θ\theta denote the acute angle between L1L_1 and L2L_2. See the figure below. Show that

tanθ=m1m21+m1m2\tan\theta=\tfrac{m_1-m_2}{1+m_1m_2}

where m1m_1 and m2m_2 are the slopes of L1L_1 and L2L_2 respectively. (Hint: Use the fact that tanθ1=m1\tan\theta_1=m_1 and tanθ2=m2\tan\theta_2=m_2.)

Source note. The source states this goal as tanθ=m2m11+m1m2\tan\theta=\tfrac{m_2-m_1}{1+m_1m_2}, but its own solution — reproduced below — sets θ=θ1θ2\theta=\theta_1-\theta_2 and derives m1m21+m1m2\tfrac{m_1-m_2}{1+m_1m_2}, the negative of that statement, and its figure (recreated here) draws θ1>θ2\theta_1>\theta_2, which matches the derivation. This page states the goal the solution actually proves; the mathematics of the derivation is unchanged.

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

tanθ=tan(θ1θ2)=tanθ1tanθ21+tanθ1tanθ2=m1m21+m1m2 \begin{array}{lrcl} & \tan\theta &=& \tan(\theta_1-\theta_2) \\[4pt] & &=& \tfrac{\tan\theta_1-\tan\theta_2}{1+\tan\theta_1\tan\theta_2} \\[4pt] & &=& \tfrac{m_1-m_2}{1+m_1m_2} \end{array}

Example. 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 α\alpha between the wires. See the figure below.

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\tan\beta=\tfrac{47}{50}, and tan(βα)=4050=45\tan(\beta-\alpha)=\tfrac{40}{50}=\tfrac45. We can then use the difference formula for tangent.

tan(βα)=tanβtanα1+tanβtanα\tan(\beta-\alpha)=\tfrac{\tan\beta-\tan\alpha}{1+\tan\beta\tan\alpha}

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

45=4750tanα1+4750tanα4(1+4750tanα)=5(4750tanα) \begin{array}{lrcl} & \tfrac45 &=& \cfrac{\tfrac{47}{50}-\tan\alpha}{1+\tfrac{47}{50}\tan\alpha} \\[10pt] & 4\left(1+\tfrac{47}{50}\tan\alpha\right) &=& 5\left(\tfrac{47}{50}-\tan\alpha\right) \end{array}

Use the distributive property, and then simplify the functions.

4(1)+4(4750)tanα=5(4750)5tanα4+3.76tanα=4.75tanα5tanα+3.76tanα=0.78.76tanα=0.7tanα0.07991tan1(0.07991)0.079741 \begin{array}{lrcl} & 4(1)+4\left(\tfrac{47}{50}\right)\tan\alpha &=& 5\left(\tfrac{47}{50}\right)-5\tan\alpha \\[4pt] & 4+3.76\tan\alpha &=& 4.7-5\tan\alpha \\[4pt] & 5\tan\alpha+3.76\tan\alpha &=& 0.7 \\[4pt] & 8.76\tan\alpha &=& 0.7 \\[4pt] & \tan\alpha &\approx& 0.07991 \\[4pt] & \tan^{-1}(0.07991) &\approx& 0.079741 \end{array}

Now we can calculate the angle in degrees.

α0.079741(180π)4.57\alpha\approx0.079741\left(\tfrac{180}{\pi}\right)\approx4.57^\circ

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.

Key equations

Sum formula for cosinecos(α+β)=cosαcosβsinαsinβ\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta
Difference formula for cosinecos(αβ)=cosαcosβ+sinαsinβ\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta
Sum formula for sinesin(α+β)=sinαcosβ+cosαsinβ\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta
Difference formula for sinesin(αβ)=sinαcosβcosαsinβ\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta
Sum formula for tangenttan(α+β)=tanα+tanβ1tanαtanβ\tan(\alpha+\beta)=\tfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}
Difference formula for tangenttan(αβ)=tanαtanβ1+tanαtanβ\tan(\alpha-\beta)=\tfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}
Cofunction identitiessinθ=cos(π2θ)cosθ=sin(π2θ)tanθ=cot(π2θ)cotθ=tan(π2θ)secθ=csc(π2θ)cscθ=sec(π2θ)\begin{array}{l}\sin\theta=\cos\left(\tfrac{\pi}{2}-\theta\right) \\ \cos\theta=\sin\left(\tfrac{\pi}{2}-\theta\right) \\ \tan\theta=\cot\left(\tfrac{\pi}{2}-\theta\right) \\ \cot\theta=\tan\left(\tfrac{\pi}{2}-\theta\right) \\ \sec\theta=\csc\left(\tfrac{\pi}{2}-\theta\right) \\ \csc\theta=\sec\left(\tfrac{\pi}{2}-\theta\right) \end{array}

Key concepts

  • The sum formula for cosines states that the cosine of the sum of two angles equals the product of the cosines of the angles minus the product of the sines of the angles. The difference formula for cosines states that the cosine of the difference of two angles equals the product of the cosines of the angles plus the product of the sines of the angles.
  • The sum and difference formulas can be used to find the exact values of the sine, cosine, or tangent of an angle.
  • The sum formula for sines states that the sine of the sum of two angles equals the product of the sine of the first angle and cosine of the second angle plus the product of the cosine of the first angle and the sine of the second angle. The difference formula for sines states that the sine of the difference of two angles equals the product of the sine of the first angle and cosine of the second angle minus the product of the cosine of the first angle and the sine of the second angle.
  • The sum and difference formulas for sine and cosine can also be used for inverse trigonometric functions.
  • The sum formula for tangent states that the tangent of the sum of two angles equals the sum of the tangents of the angles divided by 1 minus the product of the tangents of the angles. The difference formula for tangent states that the tangent of the difference of two angles equals the difference of the tangents of the angles divided by 1 plus the product of the tangents of the angles.
  • The Pythagorean Theorem along with the sum and difference formulas can be used to find multiple sums and differences of angles.
  • The cofunction identities apply to complementary angles and pairs of reciprocal functions.
  • Sum and difference formulas are useful in verifying identities.
  • Application problems are often easier to solve by using sum and difference formulas.

Practice

Use sum and difference formulas for cosine.

Find the exact value:cos(π12)\cos\left(\tfrac{\pi}{12}\right).

Rewritecos(x+2π3)\cos\left(x+\tfrac{2\pi}{3}\right)in terms ofsinx\sin xandcosx\cos x.

Use sum and difference formulas for sine.

Find the exact value:sin(11π12)\sin\left(\tfrac{11\pi}{12}\right).

Rewritesin(x3π4)\sin\left(x-\tfrac{3\pi}{4}\right)in terms ofsinx\sin xandcosx\cos x.

Use sum and difference formulas for tangent.

Find the exact value:tan(19π12)\tan\left(\tfrac{19\pi}{12}\right).

Simplifytan(32x)tan(75x)1+tan(32x)tan(75x)\cfrac{\tan\left(\tfrac32 x\right)-\tan\left(\tfrac75 x\right)}{1+\tan\left(\tfrac32 x\right)\tan\left(\tfrac75 x\right)}to a single trigonometric function ofxx.

Use sum and difference formulas for cofunctions.

Simplifysec(π2θ)\sec\left(\tfrac{\pi}{2}-\theta\right).

Simplifytan(π2x)\tan\left(\tfrac{\pi}{2}-x\right).

Use sum and difference formulas to verify identities.

True or false:tan(uv)=tanutanv1+tanutanv\tan(u-v)=\tfrac{\tan u-\tan v}{1+\tan u\tan v}for everyu,vu,vin the domain.

Use a graph to determine whetherf(x)=sin(2x)f(x)=\sin(2x)andg(x)=2sinxcosxg(x)=2\sin x\cos xare the same function or different functions.


This section is adapted from Precalculus 2e, Section 7.2: Sum and Difference Identities by Jay Abramson and OpenStax, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: omitted the decorative photograph of Denali (Figure 1). Recreated all seven instructional figures as accessible spec-first SVGs: the unit circle of special angles (Figure 2), reusing the coordinate-labeled unit circle already authored for Section 5.2 with its degree/radian ray labels omitted, since the (cosine, sine) coordinates alone carry this figure’s reason for being shown here (deriving the sum and difference formulas), and every special angle’s degree and radian measure is transcribed in full in Sections 5.2 and 5.3; the difference-formula proof diagram (Figure 3), rebuilt with representative angles α=140\alpha=140^\circ, β=20\beta=20^\circ standing in for the source’s generic α,β\alpha,\beta, since the proof depends only on the two points’ being distinct and in general position, not on their specific measures; the two reference-triangle figures for Example 6 (Figures 4–5); the cofunction right triangle (Figure 6); the two-line application diagram (Figure 7); and the guy-wire diagram (Figure 8), with the source’s separate dashed horizontal reference ray for angles α\alpha and β\beta replaced by the solid 5050-foot ground segment itself, which is both the same reference direction and the segment the problem already measures. Every retained Try It became a real fillin or multiplechoice component. Two Try Its whose printed subject is itself a single trigonometric application of a cofunction- or supplement-related argument — “write sinπ7\sin\tfrac{\pi}{7} in terms of its cofunction” and “verify tan(πθ)=tanθ\tan(\pi-\theta)=-\tan\theta” — were authored as multiplechoice rather than fillin: since both the printed subject and the keyed answer are exactly one trigonometric application of the variable, the grader’s single-trig-function token cannot refuse a learner who simply retypes the prompt, so the response mode changed instead of the token; the corresponding cofunction Practice items (sec(π2θ)\sec\left(\tfrac{\pi}{2}-\theta\right), tan(π2x)\tan\left(\tfrac{\pi}{2}-x\right)) received the same adaptation. One Extension “prove or disprove” item and one Graphical “same or different” item, adapted into the closing Practice block’s verify-identities group, were changed from free-response proofs to multiplechoice for the same reason a proof has no gradable free-response answer; their keyed alternative is the source’s own printed solution (“True” / “They are the same”). Omitted the “Access these online resources” media links. The two-line application example’s stated goal is corrected from the source’s m2m11+m1m2\tfrac{m_2-m_1}{1+m_1m_2} to the m1m21+m1m2\tfrac{m_1-m_2}{1+m_1m_2} its own solution derives and its own figure supports, with a visible Source note beside the statement (the source’s stated goal is the negative of what its solution proves). The five source Try Its (three fillin, two multiplechoice) were transcribed as in-page practice following their examples; ten further end-of-section exercises — two per objective, covering exact-value evaluation, rewriting in terms of sinx\sin x and cosx\cos x, a tangent-difference simplification, cofunction simplification, and identity/graph-comparison recognition — were adapted into the closing Practice block, one group per objective.