9.2 Sum and Difference Identities
Figure 1 Mount McKinley, in Denali National Park, Alaska, rises 20,310 feet (6,190 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.
9.2.1 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 | \(\cos\left( {\alpha + \beta} \right) = \cos\;\alpha\;\cos\;\beta - \sin\;\alpha\;\sin\;\beta\) |
| Difference formula for cosine | \(\cos\left( {\alpha - \beta} \right) = \cos\;\alpha\;\cos\;\beta + \sin\;\alpha\;\sin\;\beta\) |
Table 1
First, we will prove the difference formula for cosines. Let’s consider two points on the unit circle. See Figure 3. Point \(P\) is at an angle \(\alpha\) from the positive \(x\)-axis with coordinates \(\left( {\cos\;\alpha,\sin\;\alpha} \right)\) and point \(Q\) is at an angle of \(\beta\) from the positive \(x\)-axis with coordinates \(\left( {\cos\;\beta,\sin\;\beta} \right).\) Note the measure of angle \(POQ\) is \(\alpha - \beta.\)
Label two more points: \(A\) at an angle of \(\left( {\alpha - \beta} \right)\) from the positive \(x\)-axis with coordinates \(\left( {\cos\left( {\alpha - \beta} \right),\sin\left( {\alpha - \beta} \right)} \right);\) and point \(B\) with coordinates \(\left( {1,0} \right).\) Triangle \(POQ\) is a rotation of triangle \(AOB\) and thus the distance from \(P\) to \(Q\) is the same as the distance from \(A\) to \(B.\)
Figure 3
We can find the distance from \(P\) to \(Q\) using the distance formula.
\[\begin{array}{ccl} d_{PQ} & = & \sqrt{{(\cos\;\alpha - \cos\;\beta)}^{2} + {(\sin\;\alpha - \sin\;\beta)}^{2}} \\ & = & \sqrt{\cos^{2}\alpha - 2\;\cos\;\alpha\;\cos\;\beta + \cos^{2}\beta + \sin^{2}\alpha - 2\;\sin\;\alpha\;\sin\;\beta + \sin^{2}\beta} \end{array}\]
Then we apply the Pythagorean identity and simplify.
\[\begin{array}{cl} = & \sqrt{\left( {\cos^{2}\alpha + \sin^{2}\alpha} \right) + \left( {\cos^{2}\beta + \sin^{2}\beta} \right) - 2\;\cos\;\alpha\;\cos\;\beta - 2\;\sin\;\alpha\;\sin\;\beta} \\ = & \sqrt{1 + 1 - 2\;\cos\;\alpha\;\cos\;\beta - 2\;\sin\;\alpha\;\sin\;\beta} \\ = & \sqrt{2 - 2\;\cos\;\alpha\;\cos\;\beta - 2\;\sin\;\alpha\;\sin\;\beta} \end{array}\]
Similarly, using the distance formula we can find the distance from \(A\) to \(B.\)
\[\begin{array}{ccl} d_{AB} & = & \sqrt{{(\cos(\alpha - \beta) - 1)}^{2} + {(\sin(\alpha - \beta) - 0)}^{2}} \\ & = & \sqrt{\cos^{2}(\alpha - \beta) - 2\;\cos(\alpha - \beta) + 1 + \sin^{2}(\alpha - \beta)} \end{array}\]
Applying the Pythagorean identity and simplifying we get:
\[\begin{array}{cl} = & \sqrt{\left( {\cos^{2}(\alpha - \beta) + \sin^{2}(\alpha - \beta)} \right) - 2\;\cos(\alpha - \beta) + 1} \\ = & \sqrt{1 - 2\;\cos(\alpha - \beta) + 1} \\ = & \sqrt{2 - 2\;\cos(\alpha - \beta)} \end{array}\]
Because the two distances are the same, we set them equal to each other and simplify.
\[\begin{array}{rcl} \sqrt{2 - 2\;\cos\;\alpha\;\cos\;\beta - 2\;\sin\;\alpha\;\sin\;\beta} & = & \sqrt{2 - 2\;\cos(\alpha - \beta)} \\ {2 - 2\;\cos\;\alpha\;\cos\;\beta - 2\;\sin\;\alpha\;\sin\;\beta} & = & {2 - 2\;\cos(\alpha - \beta)} \end{array}\]
Finally we subtract \(2\) from both sides and divide both sides by \(-2.\)
\[\cos\;\alpha\;\cos\;\beta + \sin\;\alpha\;\sin\;\beta = \cos(\alpha - \beta)\text{~~}\]
Thus, we have the difference formula for cosine. We can use similar methods to derive the cosine of the sum of two angles.
9.2.2 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.
9.2.3 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, \(\tan\; x = \frac{\sin\; x}{\cos\; x},\cos\; x \neq 0.\)
Let’s derive the sum formula for tangent.
\[\begin{array}{cclc} {\tan(\alpha + \beta)} & = & \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} & \\ & = & \frac{\sin\;\alpha\;\cos\;\beta + \cos\;\alpha\;\sin\;\beta}{\cos\;\alpha\;\cos\;\beta - \sin\;\alpha\;\sin\;\beta} & \\ & = & \frac{\frac{\sin\;\alpha\;\cos\;\beta + \cos\;\alpha\;\sin\;\beta}{\cos\;\alpha\;\cos\;\beta}}{\frac{\cos\;\alpha\;\cos\;\beta - \sin\;\alpha\;\sin\;\beta}{\cos\;\alpha\;\cos\;\beta}} & {\text{Divide~the~numerator~and~denominator~by~cos}\;\alpha\;\text{cos}\;\beta.} \\ & = & \frac{\frac{\sin\;\alpha\cancel{\;\cos\;\beta}}{\cos\;\alpha\cancel{\;\cos\;\beta}} + \frac{\cancel{\cos\;\alpha}\;\sin\;\beta}{\cancel{\cos\;\alpha}\;\cos\;\beta}}{\frac{\cancel{\cos\;\alpha}\cancel{\;\cos\;\beta}}{\cancel{\cos\;\alpha}\;\cancel{\cos\;\beta}} - \frac{\sin\;\alpha\;\sin\;\beta}{\cos\;\alpha\;\cos\;\beta}} & \\ & = & \frac{\frac{\sin\;\alpha}{\cos\;\alpha} + \frac{\sin\;\beta}{\cos\;\beta}}{1 - \frac{\sin\;\alpha\;\sin\;\beta}{\cos\;\alpha\;\cos\;\beta}} & \\ & = & \frac{\tan\;\alpha + \tan\;\beta}{1 - \tan\;\alpha\;\tan\;\beta} & \end{array}\]
We can derive the difference formula for tangent in a similar way.
Given \(\sin\;\alpha = \frac{3}{5},0 < \alpha < \frac{\pi}{2},\cos\;\beta = - \frac{5}{13},\pi < \beta < \frac{3\pi}{2},\) find
ⓐ \(\sin\left( {\alpha + \beta} \right)\)
ⓑ \(\cos\left( {\alpha + \beta} \right)\)
ⓒ \(\tan\left( {\alpha + \beta} \right)\)
ⓓ \(\tan\left( {\alpha - \beta} \right)\)
Solution (click to reveal)
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 \(\sin\left( {\alpha + \beta} \right),\) we begin with \(\sin\;\alpha = \frac{3}{5}\) and \(0 < \alpha < \frac{\pi}{2}.\) The side opposite \(\alpha\) has length 3, the hypotenuse has length 5, and \(\alpha\) is in the first quadrant. See Figure 4. Using the Pythagorean Theorem, we can find the length of side \(a\text{:}\)
\[\begin{array}{rcl} {a^{2} + 3^{2}} & = & 5^{2} \\ a^{2} & = & 16 \\ a & = & 4 \end{array}\]
Figure 4
Since \(\cos\;\beta = - \frac{5}{13}\) and \(\pi < \beta < \frac{3\pi}{2},\) the side adjacent to \(\beta\) is \(-5,\) the hypotenuse is 13, and \(\beta\) is in the third quadrant. See Figure 5. Again, using the Pythagorean Theorem, we have
\[\begin{array}{rcl} {{(-5)}^{2} + a^{2}} & = & 13^{2} \\ {25 + a^{2}} & = & 169 \\ a^{2} & = & 144 \\ a & = & {\pm 12} \end{array}\]
Since \(\beta\) is in the third quadrant, \(a = –12.\)
Figure 5
The next step is finding the cosine of \(\alpha\) and the sine of \(\beta.\) The cosine of \(\alpha\) is the adjacent side over the hypotenuse. We can find it from the triangle in Figure 5: \(\cos\;\alpha = \frac{4}{5}.\) We can also find the sine of \(\beta\) from the triangle in Figure 5, as opposite side over the hypotenuse: \(\sin\;\beta = - \frac{12}{13}.\) Now we are ready to evaluate \(\sin\left( {\alpha + \beta} \right).\)
\[\begin{array}{ccl} {\sin(\alpha + \beta)} & = & {\sin\;\alpha\;\cos\;\beta + \cos\;\alpha\;\sin\;\beta} \\ & = & {\left( \frac{3}{5} \right)\left( {- \frac{5}{13}} \right) + \left( \frac{4}{5} \right)\left( {- \frac{12}{13}} \right)} \\ & = & {- \frac{15}{65} - \frac{48}{65}} \\ & = & {- \frac{63}{65}} \end{array}\]
- ⓑ We can find \(\cos\left( {\alpha + \beta} \right)\) in a similar manner. We substitute the values according to the formula.
\[\begin{array}{ccl} {\cos(\alpha + \beta)} & = & {\cos\;\alpha\;\cos\;\beta - \sin\;\alpha\;\sin\;\beta} \\ & = & {\left( \frac{4}{5} \right)\left( {- \frac{5}{13}} \right) - \left( \frac{3}{5} \right)\left( {- \frac{12}{13}} \right)} \\ & = & {- \frac{20}{65} + \frac{36}{65}} \\ & = & \frac{16}{65} \end{array}\]
- ⓒ For \(\tan\left( {\alpha + \beta} \right),\) if \(\sin\;\alpha = \frac{3}{5}\) and \(\cos\;\alpha = \frac{4}{5},\) then
\[\tan\;\alpha = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}\]
If \(\sin\;\beta = - \frac{12}{13}\) and \(\cos\;\beta = - \frac{5}{13},\) then
\[\tan\;\beta = \frac{\frac{- 12}{13}}{\frac{- 5}{13}} = \frac{12}{5}\]
Then,
\[\begin{array}{ccl} {\tan(\alpha + \beta)} & = & \frac{\tan\;\alpha + \tan\;\beta}{1 - \tan\;\alpha\;\tan\;\beta} \\ & = & \frac{\frac{3}{4} + \frac{12}{5}}{1 - \frac{3}{4}\left( \frac{12}{5} \right)} \\ & = & \frac{\text{~~}\frac{63}{20}}{- \frac{16}{20}} \\ & = & {- \frac{63}{16}} \end{array}\]
- ⓓ To find \(\tan\left( {\alpha - \beta} \right),\) we have the values we need. We can substitute them in and evaluate.
\[\begin{array}{ccl} {\tan(\alpha - \beta)} & = & \frac{\tan\;\alpha - \tan\;\beta}{1 + \tan\;\alpha\;\tan\;\beta} \\ & = & \frac{\frac{3}{4} - \frac{12}{5}}{1 + \frac{3}{4}\left( \frac{12}{5} \right)} \\ & = & \frac{- \frac{33}{20}}{\frac{56}{20}} \\ & = & {- \frac{33}{56}} \end{array}\]
A common mistake when addressing problems such as this one is that we may be tempted to think that \(\alpha\) and \(\beta\) are angles in the same triangle, which of course, they are not. Also note that
\[\tan\left( {\alpha + \beta} \right) = \frac{\sin\left( {\alpha + \beta} \right)}{\cos\left( {\alpha + \beta} \right)}\]
9.2.4 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 \(\frac{\pi}{2},\) those two angles are complements, and the sum of the two acute angles in a right triangle is \(\frac{\pi}{2},\) so they are also complements. In Figure 6, notice that if one of the acute angles is labeled as \(\theta,\) then the other acute angle must be labeled \(\left( {\frac{\pi}{2} - \theta} \right).\)
Notice also that \(\sin\;\theta = \cos\left( {\frac{\pi}{2} - \theta} \right),\) which is opposite over hypotenuse. Thus, when two angles are complementary, we can say that the sine of \(\theta\) equals the cofunction of the complement of \(\theta.\) 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.
Notice that the formulas in the table may also be justified algebraically using the sum and difference formulas. For example, using
\[\cos\left( {\alpha - \beta} \right) = \cos\;\alpha\cos\;\beta + \sin\;\alpha\sin\;\beta,\]
we can write
\[\begin{array}{ccl} {\cos\left( {\frac{\pi}{2} - \theta} \right)} & = & {\cos\;\frac{\pi}{2}\;\cos\;\theta + \sin\;\frac{\pi}{2}\;\sin\;\theta} \\ & = & {(0)\cos\;\theta + (1)\sin\;\theta} \\ & = & {\sin\;\theta} \end{array}\]
9.2.5 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.
Let \(L_{1}\) and \(L_{2}\) denote two non-vertical intersecting lines, and let \(\theta\) denote the acute angle between \(L_{1}\) and \(L_{2}.\) See Figure 7. Show that
\[\tan\;\theta = \frac{m_{2} - m_{1}}{1 + m_{1}m_{2}}\]
where \(m_{1}\) and \(m_{2}\) are the slopes of \(L_{1}\) and \(L_{2}\) respectively. (Hint: Use the fact that \(\tan\;\theta_{1} = m_{1}\) and \(\tan\;\theta_{2} = m_{2}.\) )
Figure 7
Solution (click to reveal)
Using the difference formula for tangent, this problem does not seem as daunting as it might.
\[\begin{array}{ccl} {\tan\;\theta} & = & {\tan\left( {\theta_{2} - \theta_{1}} \right)} \\ & = & \frac{\tan\;\theta_{2} - \tan\;\theta_{1}}{1 + \tan\;\theta_{1}\tan\;\theta_{2}} \\ & = & \frac{m_{2} - m_{1}}{1 + m_{1}m_{2}} \end{array}\]
For a climbing wall, a guy-wire \(R\) is attached 47 feet high on a vertical pole. Added support is provided by another guy-wire \(S\) 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 \(\alpha\) between the wires. See Figure 8.
Figure 8
Solution (click to reveal)
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 \(\tan\;\beta = \frac{47}{50},\) and \(\tan\left( {\beta - \alpha} \right) = \frac{40}{50} = \frac{4}{5}.\) We can then use difference formula for tangent.
\[\tan\left( {\beta - \alpha} \right) = \frac{\tan\;\beta - \tan\;\alpha}{1 + \tan\;\beta\tan\;\alpha}\]
Now, substituting the values we know into the formula, we have
\[\begin{array}{rcl} \frac{4}{5} & = & \frac{\frac{47}{50} - \tan\;\alpha}{1 + \frac{47}{50}\tan\;\alpha} \\ {4\left( {1 + \frac{47}{50}\tan\;\alpha} \right)} & = & {5\left( {\frac{47}{50} - \tan\;\alpha} \right)} \end{array}\]
Use the distributive property, and then simplify the functions.
\[\begin{array}{rcl} {4(1) + 4\left( \frac{47}{50} \right)\tan\;\alpha} & = & {5\left( \frac{47}{50} \right) - 5\;\tan\;\alpha} \\ {4 + 3.76\;\tan\;\alpha} & = & {4.7 - 5\;\tan\;\alpha} \\ {5\;\tan\;\alpha + 3.76\;\tan\;\alpha} & = & 0.7 \\ {8.76\;\tan\;\alpha} & = & 0.7 \\ {\tan\;\alpha} & \approx & 0.07991 \\ {\tan^{- 1}(0.07991)} & \approx & .079741 \end{array}\]
Now we can calculate the angle in degrees.
\[\alpha \approx 0.079741\left( \frac{180}{\pi} \right) \approx 4.57{^\circ}\]
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.
Section Exercises
Verbal
1. Explain the basis for the cofunction identities and when they apply.
Solution (click to reveal)
The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures \(x,\) the second angle measures \(\frac{\pi}{2} - x.\) Then \(\sin x = \cos\left( {\frac{\pi}{2} - x} \right).\) The same holds for the other cofunction identities. The key is that the angles are complementary.
2. Is there only one way to evaluate \(\cos\left( \frac{5\pi}{4} \right)?\) Explain how to set up the solution in two different ways, and then compute to make sure they give the same answer.
3. Explain to someone who has forgotten the even-odd properties of sinusoidal functions how the addition and subtraction formulas can determine this characteristic for \(f(x) = \sin(x)\) and \(g(x) = \cos(x).\) (Hint: \(0 - x = - x\) )
Solution (click to reveal)
\(\sin\left( {- x} \right) = - \sin x,\) so \(\sin x\) is odd. \(\cos\left( {- x} \right) = \cos\left( {0 - x} \right) = \cos x,\) so \(\cos x\) is even.
Algebraic
For the following exercises, find the exact value.
4. \(\cos\left( \frac{7\pi}{12} \right)\)
5. \(\cos\left( \frac{\pi}{12} \right)\)
Solution (click to reveal)
\(\frac{\sqrt{2} + \sqrt{6}}{4}\)
6. \(\sin\left( \frac{5\pi}{12} \right)\)
7. \(\sin\left( \frac{11\pi}{12} \right)\)
Solution (click to reveal)
\(\frac{\sqrt{6} - \sqrt{2}}{4}\)
8. \(\tan\left( {- \frac{\pi}{12}} \right)\)
9. \(\tan\left( \frac{19\pi}{12} \right)\)
Solution (click to reveal)
\(- 2 - \sqrt{3}\)
For the following exercises, rewrite in terms of \(\sin\; x\) and \(\cos\; x.\)
10. \(\sin\left( {x + \frac{11\pi}{6}} \right)\)
11. \(\sin\left( {x - \frac{3\pi}{4}} \right)\)
Solution (click to reveal)
\(- \frac{\sqrt{2}}{2}\sin x - \frac{\sqrt{2}}{2}\cos x\)
12. \(\cos\left( {x - \frac{5\pi}{6}} \right)\)
13. \(\cos\left( {x + \frac{2\pi}{3}} \right)\)
Solution (click to reveal)
\(- \frac{1}{2}\cos x - \frac{\sqrt{3}}{2}\sin x\)
For the following exercises, simplify the given expression.
14. \(\csc\left( {\frac{\pi}{2} - t} \right)\)
15. \(\sec\left( {\frac{\pi}{2} - \theta} \right)\)
Solution (click to reveal)
\(\csc\theta\)
16. \(\cot\left( {\frac{\pi}{2} - x} \right)\)
17. \(\tan\left( {\frac{\pi}{2} - x} \right)\)
Solution (click to reveal)
\(\cot x\)
18. \(\sin(2x)\;\cos(5x) - \sin(5x)\;\cos(2x)\)
19. \(\frac{\tan\left( {\frac{3}{2}x} \right) - \tan\left( {\frac{7}{5}x} \right)}{1 + \tan\left( {\frac{3}{2}x} \right)\tan\left( {\frac{7}{5}x} \right)}\)
Solution (click to reveal)
\(\tan\left( \frac{x}{10} \right)\)
For the following exercises, find the requested information.
20. Given that \(\sin\; a = \frac{2}{3}\) and \(\cos\; b = - \frac{1}{4},\) with \(a\) and \(b\) both in the interval \(\left\lbrack {\frac{\pi}{2},\pi} \right),\) find \(\sin(a + b)\) and \(\cos(a - b).\)
21. Given that \(\sin\; a = \frac{4}{5},\) and \(\cos\; b = \frac{1}{3},\) with \(a\) and \(b\) both in the interval \(\left\lbrack {0,\frac{\pi}{2}} \right),\) find \(\sin(a - b)\) and \(\cos(a + b).\)
Solution (click to reveal)
\(\begin{array}{rclcl} {\sin(a - b)} & = & {\left( \frac{4}{5} \right)\left( \frac{1}{3} \right) - \left( \frac{3}{5} \right)\left( \frac{2\sqrt{2}}{3} \right)} & = & \frac{4 - 6\sqrt{2}}{15} \\ {\cos(a + b)} & = & {\left( \frac{3}{5} \right)\left( \frac{1}{3} \right) - \left( \frac{4}{5} \right)\left( \frac{2\sqrt{2}}{3} \right)} & = & \frac{3 - 8\sqrt{2}}{15} \end{array}\)
For the following exercises, find the exact value of each expression.
22. \(\sin\left( {\cos^{- 1}(0) - \cos^{- 1}\left( \frac{1}{2} \right)} \right)\)
23. \(\cos\left( {\cos^{- 1}\left( \frac{\sqrt{2}}{2} \right) + \sin^{- 1}\left( \frac{\sqrt{3}}{2} \right)} \right)\)
Solution (click to reveal)
\(\frac{\sqrt{2} - \sqrt{6}}{4}\)
24. \(\tan\left( {\sin^{- 1}\left( \frac{1}{2} \right) - \cos^{- 1}\left( \frac{1}{2} \right)} \right)\)
Graphical
For the following exercises, simplify the expression, and then graph both expressions as functions to verify the graphs are identical. Confirm your answer using a graphing calculator.
25. \(\cos\left( {\frac{\pi}{2} - x} \right)\)
Solution (click to reveal)
\(\sin x\)
26. \(\sin(\pi - x)\)
27. \(\tan\left( {\frac{\pi}{3} + x} \right)\)
Solution (click to reveal)
\(\cot\left( {\frac{\pi}{6} - x} \right)\)
28. \(\sin\left( {\frac{\pi}{3} + x} \right)\)
29. \(\tan\left( {\frac{\pi}{4} - x} \right)\)
Solution (click to reveal)
\(\cot\left( {\frac{\pi}{4} + x} \right)\)
30. \(\cos\left( {\frac{7\pi}{6} + x} \right)\)
31. \(\sin\left( {\frac{\pi}{4} + x} \right)\)
Solution (click to reveal)
\(\frac{\sin x}{\sqrt{2}} + \frac{\cos x}{\sqrt{2}}\)
32. \(\cos\left( {\frac{5\pi}{4} + x} \right)\)
For the following exercises, use a graph to determine whether the functions are the same or different. If they are the same, show why. If they are different, replace the second function with one that is identical to the first. (Hint: think \(2x = x + x.\) )
33. \(f(x) = \sin\left( {4x} \right) - \sin\left( {3x} \right)\cos\; x,g(x) = \sin\; x\;\cos\left( {3x} \right)\)
Solution (click to reveal)
They are the same.
34. \(f(x) = \cos\left( {4x} \right) + \sin\; x\;\sin\left( {3x} \right),g(x) = - \cos\; x\;\cos\left( {3x} \right)\)
35. \(f(x) = \sin\left( {3x} \right)\cos\left( {6x} \right),g(x) = - \sin\left( {3x} \right)\cos\left( {6x} \right)\)
Solution (click to reveal)
They are the different, try \(g(x) = \sin\left( {9x} \right) - \cos\left( {3x} \right)\sin\left( {6x} \right).\)
36. \(f(x) = \sin(4x),g(x) = \sin(5x)\cos\; x - \cos(5x)\sin\; x\)
37. \(f(x) = \sin(2x),g(x) = 2\;\sin\; x\;\cos\; x\)
Solution (click to reveal)
They are the same.
38. \(f(\theta) = \cos\left( {2\theta} \right),g(\theta) = \cos^{2}\theta - \sin^{2}\theta\)
39. \(f(\theta) = \tan(2\theta),g(\theta) = \frac{\tan\;\theta}{1 + \tan^{2}\theta}\)
Solution (click to reveal)
They are the different, try \(g(\theta) = \frac{2\;\tan\theta}{1 - \tan^{2}\theta}.\)
40. \(f(x) = \sin(3x)\sin\; x,g(x) = \sin^{2}(2x)\cos^{2}x - \cos^{2}(2x)\sin^{2}x\)
41. \(f(x) = \tan( - x),g(x) = \frac{\tan\; x - \tan(2x)}{1 - \tan\; x\;\tan(2x)}\)
Solution (click to reveal)
They are different, try \(g(x) = \frac{\tan x - \tan\left( {2x} \right)}{1 + \tan x\tan\left( {2x} \right)}.\)
Technology
For the following exercises, find the exact value algebraically, and then confirm the answer with a calculator to the fourth decimal point.
42. \(\sin(75{^\circ})\)
43. \(\sin(195{^\circ})\)
Solution (click to reveal)
\(- \frac{\sqrt{3} - 1}{2\sqrt{2}},\text{or~} - 0.2588\)
44. \(\cos(165{^\circ})\)
45. \(\cos(345{^\circ})\)
Solution (click to reveal)
\(\frac{1 + \sqrt{3}}{2\sqrt{2}},\) or 0.9659
46. \(\tan(-15{^\circ})\)
Extensions
For the following exercises, prove the identities provided.
47. \(\tan(x + \frac{\pi}{4}) = \frac{\tan\; x + 1}{1 - \tan\; x}\)
Solution (click to reveal)
\(\begin{array}{rcc} {\tan\left( {x + \frac{\pi}{4}} \right)} & = & \\ \frac{\tan x + \tan\left( \frac{\pi}{4} \right)}{1 - \tan x\tan\left( \frac{\pi}{4} \right)} & = & \\ \frac{\tan x + 1}{1 - \tan x(1)} & = & \frac{\tan x + 1}{1 - \tan x} \end{array}\)
48. \(\frac{\tan(a + b)}{\tan(a - b)} = \frac{\sin\; a\;\cos\; a + \sin\; b\;\cos\; b}{\sin\; a\;\cos\; a - \sin\; b\;\cos\; b}\)
49. \(\frac{\cos(a + b)}{\cos\; a\;\cos\; b} = 1 - \tan\; a\;\tan\; b\)
Solution (click to reveal)
\(\begin{array}{rcc} \frac{\cos(a + b)}{\cos a\cos b} & = & \\ {\frac{\cos a\cos b}{\cos a\cos b} - \frac{\sin a\sin b}{\cos a\cos b}} & = & {1 - \tan a\tan b} \end{array}\)
50. \(\cos\left( {x + y} \right)\cos\left( {x - y} \right) = \cos^{2}x - \sin^{2}y\)
51. \(\frac{\cos(x + h) - \cos\; x}{h} = \cos\; x\frac{\cos\; h - 1}{h} - \sin\; x\frac{\sin\; h}{h}\)
Solution (click to reveal)
\(\begin{array}{rcc} \frac{\cos(x + h) - \cos x}{h} & = & \\ \frac{\cos x\cosh - \sin x\sinh - \cos x}{h} & = & \\ \frac{\cos x(\cosh - 1) - \sin x\sinh}{h} & = & {\cos x\frac{\cos h - 1}{h} - \sin x\frac{\sin h}{h}} \end{array}\)
For the following exercises, prove or disprove the statements.
52. \(\tan(u + v) = \frac{\tan\; u + \tan\; v}{1 - \tan\; u\;\tan\; v}\)
53. \(\tan(u - v) = \frac{\tan\; u - \tan\; v}{1 + \tan\; u\;\tan\; v}\)
Solution (click to reveal)
True
54. \(\frac{\tan\left( {x + y} \right)}{1 + \tan\; x\;\tan\; x} = \frac{\tan\; x + \tan\; y}{1 - \tan^{2}x\;\tan^{2}y}\)
55. If \(\alpha,\beta,\) and \(\gamma\) are angles in the same triangle, then prove or disprove \(\sin\left( {\alpha + \beta} \right) = \sin\;\gamma.\)
Solution (click to reveal)
True. Note that \(\,\sin\left( {\alpha + \beta} \right) = \sin\left( {\pi - \gamma} \right)\,\) and expand the right hand side.
56. If \(\alpha,\beta,\) and \(y\) are angles in the same triangle, then prove or disprove \(\tan\;\alpha + \tan\;\beta + \tan\;\gamma = \tan\;\alpha\;\tan\;\beta\;\tan\;\gamma\)