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Mathematics LibreTexts

7.2 Addition and Subtraction Identities

In this section, we begin expanding our repertoire of trigonometric identities. 

Identities

The sum and difference identities

\[ \cos (\alpha - \beta) = \cos (\alpha) \cos (\beta) + \sin (\alpha) \sin (\beta)\]

\[ \cos (\alpha + \beta) = \cos (\alpha) \cos (\beta) - \sin (\alpha) \sin (\beta)\]

\[ \sin (\alpha - \beta) = \sin (\alpha) \cos (\beta) + \cos (\alpha) \sin (\beta)\]

\[ \sin (\alpha + \beta) = \sin (\alpha) \cos (\beta) - \cos (\alpha) \sin (\beta)\]

We will prove the difference of angles identity for cosine. The rest of the identities can be derived from this one.

Proof: Difference of Angles identity for Cosine

Consider two points on a unit circle:

  • \(P\) at an angle of \(\alpha\) from the positive \(x\) axis with coordinates \(\cos(\alpha),\sin(\alpha)\)
  • \(Q\) at an angle of \(\beta\) with coordinates \(\cos(\beta),\sin(\beta)\)

Notice the measure of angle \(POQ\) is \(\alpha-\beta\). Label two more points:

  • \(C\) at an angle of \(\alpha-\beta\), with coordinates \(\cos(\alpha-\beta),\sin(\alpha-\beta)\),
  • \(D\) at the point (1, 0).

Notice that the distance from \(C\) to \(D\) is the same as the distance from \(P\) to \(Q\) because triangle \(COD\) is a rotation of triangle \(POQ\).

Using the distance formula to find the distance from \(P\) to \(Q\) yields

\[ \sqrt{ \left(\cos(\alpha)-\cos(\beta) \right)^2 + \left(\sin(\alpha)-\sin(\beta) \right)^2 } \]

Expanding this

\[ \sqrt{ \cos^2(\alpha)-2\cos(\alpha)\cos(\beta) + \cos^2(\beta) + \sin^2(\alpha)-2\sin(\alpha)\sin(\beta) + \sin^2(\beta)  } \]

Applying the Pythagorean Identity and simplifying

\[\sqrt{2-2\cos(\alpha)\cos(\beta)-2\sin(\alpha)\sin(\beta)}\]

Similarly, using the distance formula to find the distance from \(C\) to \(D\)

\[ \sqrt{ (\cos(\alpha-\beta-1)^2+ (\sin(\alpha-\beta)-0)^2}\]

Expanding this

\[ \sqrt{-\cos^2(\alpha-\beta)-2\cos(\alpha-\beta)+1+\sin^2(\alpha-\beta)}\]

Applying the Pythagorean Identity and simplifying

\[ \sqrt{-2\cos(\alpha-\beta)+2}\]

Since the two distances are the same we set these two formulas equal to each other and simplify

Establishing the identity

Try it Now: 1

By writing \(\cos(\alpha+\beta)\) as \(\cos(\alpha-(-\beta)\), show the sum of angles identity for cosine follows from the difference of angles identity proven above

The sum and difference of angles identities are often used to rewrite expressions in other forms, or to rewrite an angle in terms of simpler angles.

Example 1

Find the exact value of \(cos(75°\).

SOLUTION

Since , we can evaluate  as

                            Apply the cosine sum of angles identity

        Evaluate

                                      Simply

Try it Now: 2

Find the exact value of \(\sin\left(\dfrac{\pi}{12}\right)\).

Example 2

Rewrite \(\sin\left(x-\dfrac{\pi}{4} right)\) in terms of \(\sin(x)\) and \(\cos(x)\).

SOLUTION

\[\sin \left(x-\dfrac{\pi}{4} right)\]

Use the difference of angles identity for sine

\[= \sin(x)\cos \left(\dfrac{\pi}{4} \right)-\cos(x) \sin \left(\dfrac{\pi}{4} \right)\]

Evaluate the cosine and sine and rearrange

\[=\dfrac{\sqrt{2}}{2} \sin(x) - \dfrac{\sqrt{2}}{2}\cos(x)\]

Additionally, these identities can be used to simplify expressions or prove new identities

Example 3

Prove .

SOLUTION

As with any identity, we need to first decide which side to begin with. Since the left side involves sum and difference of angles, we might start there

                                         Apply the sum and difference of angle identities

        

Since it is not immediately obvious how to proceed, we might start on the other side, and see if the path is more apparent.

                                Rewriting the tangents using the tangent identity

                           Multiplying the top and bottom by cos(a)cos(b)

  Distriuting and simplifying

         From above, we recognize this

 

                                                 Establishing the identity

These identities can also be used to solve equations.

Example 4

Solve .

SOLUTION

By recognizing the left side of the equation as the result of the difference of angles identity for cosine, we can simplify the equation

           Apply the difference of angles identity

                                              Use the negative angle identity

Since this is a special cosine value we recognize from the unit circle, we can quickly write the answers:

, where k is an integer

Combining Waves of Equal Period

A sinusoidal function of the form  can be rewritten using the sum of angles identity.

Example 5

Rewrite  as a sum of sine and cosine.

SOLUTION

Using the sum of angles identity

        Evaluate the sine and cosine

                    Distribute and simplify

Notice that the result is a stretch of the sine added to a different stretch of the cosine, but both have the same horizontal compression, which results in the same period.

We might ask now whether this process can be reversed – can a combination of a sine and cosine of the same period be written as a single sinusoidal function?  To explore this, we will look in general at the procedure used in the example above.

 

                                     Use the sum of angles identity

                        Distribute the A

           Rearrange the terms a bit

Based on this result, if we have an expression of the form , we could rewrite it as a single sinusoidal function if we can find values A and C so that

, which will require that:

  which can be rewritten as  

To find A,

                    Apply the Pythagorean Identity and simplify

Rewriting a Sum of Sine and Cosine as a Single Sine

To rewrite  as

, , and

We can use either of the last two equations to solve for possible values of C. Since there will usually be two possible solutions, we will need to look at both to determine which quadrant C is in and determine which solution for C satisfies both equations.

Example 6

Rewrite  as a single sinusoidal function.

SOLUTION

Using the formulas above, , so A = 8. 

Solving for C,

, so  or

However, since , the angle that works for both is

 

Combining these results gives us the expression

Try it Now: 3

Rewrite  as a single sinusoidal function.

Rewriting a combination of sine and cosine of equal periods as a single sinusoidal function provides an approach for solving some equations.

Example 7

Solve  to find two positive solutions.

SOLUTION

To approach this, since the sine and cosine have the same period, we can rewrite them as a single sinusoidal function. 

, so A = 5

, so  or

Since , a positive value, we need the angle in the first quadrant, C = 0.927.

Using this, our equation becomes

                         Divide by 5

                          Make the substitution u = 2x + 0.927

                                        The inverse gives a first solution

                       By symmetry, the second solution is

                      A third solution is

                               

Undoing the substitution, we can find two positive solutions for x.

   or              or        

                                                      

                                                          

Since the first of these is negative, we eliminate it and keep the two positive solutions,  and .

The Product-to-Sum and Sum-to-Product Identities

Identities

The Product-to-Sum Identities

We will prove the first of these, using the sum and difference of angles identities from the beginning of the section.  The proofs of the other two identities are similar and are left as an exercise.

Proof of Product-to-Sum Identity for \(\sin(\alpha) \cos(\beta)\)

Recall the sum and difference of angles identities from earlier

Adding these two equations, we obtain

Dividing by 2, we establish the identity

Example 8

Write  as a sum or difference.

SOLUTION

Using the product-to-sum identity for a product of sines

                               If desired, apply the negative angle identity

                                 Distribute

Try it Now: 4

Evaluate \(\cos \left( \dfrac{11\pi}{12}  \cos \left( \dfrac{\pi}{12} \right)\).

Identities

The Sum-to-Product Identities

We will again prove one of these and leave the rest as an exercise. 

Proof of Sum-to-product Identity for Sine Function

We begin with the product-to-sum identity

We define two new variables:

Adding these equations yields , giving

Subtracting the equations yields , or

Substituting these expressions into the product-to-sum identity above,

Multiply by 2 on both sides

Establishing the identity

Example 9

Evaluate .

SOLUTION

Using the sum-to-product identity for the difference of cosines,

                         Simplify

 

                                            Evaluate

Example 10

Prove the identity .

SOLUTION

Since the left side seems more complicated, we can start there and simplify.

                             Using the sum-to-product identities

       Simplify

                            Simplify further

                                         Rewrite as a tangent

                                        

Establishing the identity

Try it Now: 5

Notice that, using the negative angle identity, .  Use this along with the sum of sines identity to prove the sum-to-product identity for .

Example 11

Solve  for all solutions with .

SOLUTION

In an equation like this, it is not immediately obvious how to proceed. One option would be to combine the two sine functions on the left side of the equation.  Another would be to move the cosine to the left side of the equation, and combine it with one of the sines.  For no particularly good reason, we’ll begin by combining the sines on the left side of the equation and see how things work out.

                       Apply the sum to product identity on the left

   Simplify

                    Apply the negative angle identity

                      Rearrange the equation to be 0 on one side

                 Factor out the cosine

                        

Using the Zero Product Theorem we know that at least one of the two factors must be zero.  The first factor, , has period , so the solution interval of  represents one full cycle of this function.

                                                  Substitute

                                                    On one cycle, this has solutions

 or                                          Undo the substitution

 

, so

, so

The second factor, , has period of , so the solution interval  contains two complete cycles of this function.

                                         Isolate the sine

                                               

                                                    On one cycle, this has solutions

 or                                          On the second cycle, the solutions are

 or   Undo the substitution

, so

, so

, so

, so

Altogether, we found six solutions on , which we can confirm by looking at the graph.

Important Topics of This Section

  • The sum and difference identities
  • Combining waves of equal periods
  • Product-to-sum identities
  • Sum-to-product identities
  • Completing proofs

 

Try it Now Answers

  1.                                        Use negative angle identity for sine

                                            Use sum-to-product identity for sine

                    Eliminate the parenthesis

                               Establishing the identity

Section 7.2 Exercises

 

Find an exact value for each of the following.

1.                  2.                3.                 4.               

5.                 6.                  7.                 8.

 

Rewrite in terms of  and .

9.          10.          11.         12.

 

Simplify each expression.

13.            14.           15.            16.

 

Rewrite the product as a sum.

17.                                  18.

19.                                      20.

 

Rewrite the sum as a product.

21.                                      22.

23.                                      24.

 

25. Given  and , with a and b both in the interval :

            a. Find                                b. Find

 

26. Given  and , with a and b both in the interval :

            a. Find                                b. Find

 

Solve each equation for all solutions.

27.

28.

29.

30.

Solve each equation for all solutions.

31.

32.

33.

34.

 

 

Rewrite as a single function of the form .

35.                                    36.

37.                                38.

 

Solve for the first two positive solutions.

39.                             40.

41.                           42.

 

Simplify.

43.                                                 44.

 

Prove the identity.

44.

45.

46.

47.

48.

49.

50.

51.

52.

Contributors

  • David Lippman (Pierce College)
  • Melonie Rasmussen (Pierce College)