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n! = n · (n − 1) · (n − 2) · · · 2 · 1 with 0! = 1 as a special case. (We will study factorials in greater detail in Section 9.4.) The world famous Fibonacci Numbers are defined recursively and are explored in the exercises. While none of the sequences worked out to be the sequence in (1), they do give us some insight ... |
. bk = (−1)k 2k + 1, k ≥ 0 4, − 27 16,... 2We’re basically talking about the ‘countably infinite’ subsets of the real number line when we do this. 3Sequences which are both arithmetic and geometric are discussed in the Exercises. 9.1 Sequences 655 Solution. A good rule of thumb to keep in mind when working with sequence... |
4 5 and − 1 = − 1 = − 3 3. As we saw in Example 9.1.1, the sequence {2n − 1}∞ n=1 generates the odd numbers: 1, 3, 5, 7,.... Computing the first few differences, we find a2 − a1 = 2, a3 − a2 = 2, and a4 − a3 = 2. This suggests that the sequence is arithmetic. To verify this, we find an+1 − an = (2(n + 1) − 1) − (2n − 1) =... |
difference d, the way we get from one term to the next is by adding d. Hence, the terms of the sequence are: a, a + d, a + 2d, a + 3d,.... We see that to reach the nth term, we add d to a exactly (n − 1) times, which is what the formula says. The derivation of the formula for geometric series follows similarly. Here, w... |
10 n−1 10 2. As the reader can verify, this sequence is neither arithmetic nor geometric. In an attempt to find a pattern, we rewrite the second term with a denominator to make all the terms appear as fractions. We have 2 7,.... If we associate the negative ‘−’ of the last two −3, 2 1, 2 terms with the denominators we ... |
reader to show that this checks out. = (−2)n−1 13, −8 1, −2 7, 4 While the last problem in Example 9.1.3 was neither geometric nor arithmetic, it did resolve into a combination of these two kinds of sequences. If handed the sequence 2, 5, 10, 17,..., we would be hard-pressed to find a formula for an if we restrict our ... |
the formula for Ak, the amount in the account after k compounding periods, is Ak = P 1 + r, k ≥ 1. We now spot this as a geometric sequence with first term P 1 + r. In retirement planning, it is seldom n the case that an investor deposits a set amount of money into an account and waits for it to grow. Usually, addition... |
d; if it is geometric, find the common ratio r. 14. {3n − 5}∞ n=1 16. 1 3, 1 6, 1 12, 1 24,... 15. an = n2 + 3n + 2, n ≥ 1 3 1 5 17. n−1∞ n=1 18. 17, 5, −7, −19,... 19. 2, 22, 222, 2222,... 20. 0.9, 9, 90, 900,... 21. an = n! 2, n ≥ 0. In Exercises 22 - 30, find an explicit formula for the nth term of the given sequence... |
30 days. So, for example, you get two pennies on the second day and four pennies on the third day. How many pennies do you get on the 30th day? What is the total dollar value of the gift you have received? 35. Research the terms ‘arithmetic mean’ and ‘geometric mean.’ With the help of your classmates, show that a give... |
2, n ≥ 1 27. (−1)n−1x2n−1 2n−1, n ≥ 1 28. an = 10n−1 10n, n ≥ 1 29. an = (n + 2)3, n ≥ 1 30. an = 1+(−1)n−1 2, n ≥ 1 9.2 Summation Notation 661 9.2 Summation Notation In the previous section, we introduced sequences and now we shall present notation and theorems concerning the sum of terms of a sequence. We begin with ... |
definitions. For example, summation notation allows us to define polynomials as functions of the form f (x) = n k=0 akxk for real numbers ak, k = 0, 1,... n. The reader is invited to compare this with what is given in Definition 3.1. Summation notation is particularly useful when talking about matrix operations. For exam... |
n, but not the variable x, with the values 1 through 5 and adding the resulting terms. 9.2 Summation Notation 663 5 n=1 (−1)n+1 n (x − 1)n = (−1)1+1 1 (x − 1)1 + (−1)2+1 2 (x − 1)2 + (−1)3+1 3 (x − 1)3 + (−1)1+4 4 = (x − 1) − (x − 1)4 + (x − 1)2 2 + (−1)1+5 5 (x − 1)3 3 (x − 1)5 − (x − 1)4 4 + (x − 1)5 5 2. The key to... |
)n−1 for our terms, and we find the lower and upper limits of summation to be n = 1 and n = 117, respectively. Thus 117 = 117 n=1 (−1)n−1 n (c) Thanks to Example 9.1.3, we know that one formula for the nth term is an = 9 10n for n ≥ 1. This gives us a formula for the summation as well as a lower limit of summation. To d... |
and geometric sequences. Given an arithmetic sequence ak = a + (k − 1)d for k ≥ 1, we let S denote the sum of the first n terms. To derive a formula for S, we write it out in two different ways S = S = (a + (n − 1)d) + (a + (n − 2)d) +... + (a + d) + a +... + (a + (n − 2)d) + (a + (n − 1)d) (a + d) + a If we add these t... |
a and arn to cancel out and we get S − rS = a − arn. Factoring, we get S(1 − r) = a (1 − rn). Assuming r = 1, we can divide both sides by the quantity (1 − r) to obtain S = a 1 − rn 1 − r If we distribute a through the numerator, we get a − arn = a1 − an+1 which yields the formula S = a1 − an+1 1 − r In the case when ... |
= r n denote the interest rate per period. Suppose we wish to make ongoing deposits of P dollars at the end of each compounding period. Let Ak denote the amount in the account after k compounding periods. Then A1 = P, because we have made our first deposit at the end of the first compounding period and no interest has b... |
i)2 +... + 1 (1 + i)k−1 = 1 Hence, we get 1 − 1 (1 + i)1 + i) 1 − (1 + i)−k i Ak = P (1 + i)k−1 (1 + i) 1 − (1 + i)−k i P (1 + i)k − 1 i = If we let t be the number of years this investment strategy is followed, then k = nt, and we get the formula for the future value of an ordinary annuity. Equation 9.3. Futu... |
logs of both sides of the equation. 100000 = 50 (1 + 0.005)12t − 1 0.005 10 = (1.005)12t − 1 (1.005)12t = 11 ln (1.005)12t = ln(11) 12t ln(1.005) = ln(11) ln(11) t = 12 ln(1.005) ≈ 40.06 This means that it takes just over 40 years for the investment to grow to $100,000. Comparing this with our answer to part 1, we see... |
1 − 1 Theorem 9.2. Geometric Series: Given the sequence ak = ark−1 for k ≥ 1, where |r| < 1, a + ar + ar2 +... = ∞ ark−1 = k=1 If |r| ≥ 1, the sum a + ar + ar2 +... is not defined. a 1 − r The justification of the result in Theorem 9.2 comes from taking the formula in Equation 9.2 for the sum of the first n terms of a ge... |
5)2 + 1 6 (x − 5)3 + 1 8 (x − 5)4 In Exercises 17 - 28, use the formulas in Equation 9.2 to find the sum. 17. 20. 10 n=1 5n + 3 n 10 n=1 1 2 18. 21. 20 n=1 2n − 1 n 5 n=1 3 2 19. 22. 15 k=0 3 − k k 5 k=0 1 4 2 23. 1 + 4 + 7 +... + 295 24. 4 + 2 + 0 − 2 −... − 146 25. 1 + 3 + 9 +... + 2187 26 256 27 256 28. 10 n=1 −2n +... |
(1.0001)k−1 (b) k=1 ∞ (−1)k−1 (c) k=1 8When in doubt, write them out! 672 Sequences and the Binomial Theorem 9.2.2 Answers 1. 213 5. 17 2 9. 13. 5 (3k + 5) k=1 5 k=1 k + 1 k 17. 305 21. 633 32 25. 3280 29. 7 9 2. 341 280 6. 0 10. 14. 8 k=1 20 k=3 (−1)k−1k (−1)k ln(k) 18. 400 22. 26. 30. 1365 512 255 256 13 99 3. 63 7.... |
adding d. In symbols, a, a + d, a + 2d, a + 3d, a + 4d +... The pattern suggested here is that to reach the nth term, we start with a and add d to it exactly n − 1 times, which lead us to our formula an = a + (n − 1)d for n ≥ 1. But how do we prove this to be the case? We have the following. The Principle of Mathemati... |
need to show is that whenever P (k) is true, it follows that P (k + 1) is true. In other words, we assume P (k) is true (this is called the ‘induction hypothesis’) and deduce that P (k + 1) is also true. Assuming P (k) to be true seems to invite disaster - after all, isn’t this essentially what we’re trying to prove i... |
+ 1) − 1)d. Hence, P (k + 1) is true. In essence, by showing that P (k + 1) must always be true when P (k) is true, we are showing that the formula P (1) can be used to get the formula P (2), which in turn can be used to derive the formula P (3), which in turn can be used to establish the formula P (4), and so on. Thu... |
Kai. 9.3 Mathematical Induction 675 This shows the base case P (1) is true. Next we assume P (k) is true, that is, we assume k (a + (j − 1)d) = j=1 k 2 (2a + (k − 1)d) and attempt to use this to show P (k + 1) is true. Namely, we must show k+1 (a + (j − 1)d) = j=1 k + 1 2 (2a + (k + 1 − 1)d) To see how we can use P (k... |
The base case P (1) is (z)1 = z1, which reduces to z = z which is true. We now assume P (k) is true, that is, we assume (z)k = zk and attempt to show that P (k + 1) is true. Since (z)k+1 = (z)k z, we can use the induction hypothesis and 676 Sequences and the Binomial Theorem write (z)k = zk. Hence, (z)k+1 = (z)k z = z... |
n > 100n for all n ≥ 6. 4. To prove this determinant property, we use induction on n, where we take P (n) to be that the property we wish to prove is true for all n × n matrices. For the base case, we note that if A is a 1 × 1 matrix, then A = [a] so A = [ca]. By definition, det(A) = a and det(A) = ca so we have det(A) ... |
this notation. 9.3 Mathematical Induction 677 it is needed. While mathematically more elegant, it is less intuitive, and we stand by our approach because of its pedagogical value.) Case 2: The row R being replaced is the not the first row of A. By definition, det(A) = n p=1 1pC a 1p, where in this case, a 1p = a1p, sinc... |
n > 12 4. 3n ≥ n3 for n ≥ 4 5. Use the Product Rule for Absolute Value to show |xn| = |x|n for all real numbers x and all natural numbers n ≥ 1 6. Use the Product Rule for Logarithms to show log (xn) = n log(x) for all real numbers x > 0 and all natural numbers n ≥ 1. an a 0 0 0 b 0 bn n = for n ≥ 1. 7. 8. Prove Equat... |
k + 1) + 6(k + 1)) 6 (k + 1) 2k2 + 7k + 6 6 (k + 1)(k + 2)(2k + 3k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 (k + 1)(k + 2)(2k + 3) 6 By induction, n j=1 j2 = n(n + 1)(2n + 1) 6 is true for all natural numbers n ≥ 1... |
1 log = log xk · x = log xk + log(x) = k log(x) + log(x) = (k + 1) log(x) Hence, log xk+1 = (k + 1) log(x). By induction log (xn) = n log(x) is true for all x > 0 and all natural numbers n ≥ 1. 9. Let A be an n × n lower triangular matrix. We proceed to prove the det(A) is the product of the entries along the main diag... |
diagonal consists of all 1’s. Hence, det (In) = 1, as required. 9.4 The Binomial Theorem 681 9.4 The Binomial Theorem In this section, we aim to prove the celebrated Binomial Theorem. Simply stated, the Binomial Theorem is a formula for the expansion of quantities (a + b)n for natural numbers n. In Elementary and Inte... |
. Informally, n! = n · (n − 1) · (n − 2) · · · 2 · 1 with 0! = 1 as our ‘base case.’ Our first example familiarizes us with some of the basic computations involving factorials. Example 9.4.1. 1. Simplify the following expressions. (a) 3! 2! 0! (b) 7! 5! (c) 1000! 998! 2! (d) (k + 2)! (k − 1)!, k ≥ 1 2. Prove n! > 3n for... |
ator. To see this, we begin by writing out (k + 2)! starting with (k + 2) and multiplying it by the numbers which precede it until we reach (k − 1): (k + 2)! = (k + 2)(k + 1)(k)(k − 1)!. As a result, we have (k + 2)! (k − 1)! = (k + 2)(k + 1)(k)(k − 1)! (k − 1)! = (k + 2)(k + 1)(k) (k − 1)! (k − 1)! = k(k + 1)(k + 2) T... |
the wild often involve counting arrangements. For example, if you have fifty songs on your mp3 player and wish arrange these songs in a playlist in which the order of the n! → 0 as n → ∞.2 2This fact is far more important than you could ever possibly imagine. 9.4 The Binomial Theorem 683 songs matters, it turns out tha... |
the possibilities. We now state anf prove a theorem which is crucial to the proof of the Binomial Theorem. Theorem 9.3. For natural numbers n and j with n ≥ j = The proof of Theorem 9.3 is purely computational and uses the definition of binomial coefficients, the recursive property of factorials and common denominators. ... |
s consider the specific case of n = 4. According to the theorem, we have 9.4 The Binomial Theorem 685 (a + b)4 = 4 a4−jbj 4 j j=0 4 0 4 0 = = a4−0b0 + a4−1b1 + 4 1 a4−2b2 + 4 2 a4−3b3 + a4−4b4 4 4 a4 + a3b + 4 1 4 2 a2b2 + 4 3 ab3 + 4 3 4 4 b4 We forgo the simplification of the coefficients in order to note the pattern in ... |
b Now we assume that P (k) is true. That is, we assume that we can expand (a + b)k using the formula given in Theorem 9.4 and attempt to show that P (k + 1) is true. 4and a fair amount of tenacity and attention to detail. 686 Sequences and the Binomial Theorem (a + b)k+1 = (a + b)(a + b)k k = (a + b) k j ak−jbj j=0 = ... |
powers of a and b, they do so for different values of j. To resolve this, we need to shift the index on the second summation so that the index j starts at j = 1 instead of j = 0 and we make use of Theorem 9.1 in the process. 9.4 The Binomial Theorem 687 k−1 j=0 k j ak−jbj+1 = k−1+1 k j − 1 j=0+1 ak−(j−1)b(j−1)+1 = k k ... |
1(−2)1 + 4 1 4 2 x4−2(−2)2 + 4 3 x4−3(−2)3 + x4−4(−2)4 4 4 = x4 − 8x3 + 24x2 − 32x + 16 2. At first this problem seem misplaced, but we can write 2.13 = (2 + 0.1)3. Identifying a = 2, b = 0.1 = 1 10 and n = 3, we get 3 23−j 3 j j 1 10 2 + 3 1 10 = = j=0 3 0 23−0 0 1 10 3 1 + 23−1 1 1 10 3 2 + 23−2 2 1 10 3 3 + 23−3 3 1 ... |
and ends with 1 and the middle numbers are generated using Theorem 9.3. Below we attempt to demonstrate this building process to generate the first five rows of Pascal’s Triangle −−−−−−→ −−−−−−→ −−−−−−→ 690 Sequences and the Binomial Theorem To see how we can use Pascal’s Triangle to expedite the Binomial Theorem, suppo... |
Exercises 14 - 17, use Pascal’s Triangle to simplify the given power of a complex number. 14. (1 + 2i)4 √ 16. 3 i 3 2 + 1 2 15. −1 + i √ √ 33 √ 4 i 2 2 − 2 2 17. In Exercises 18 - 22, use the Binomial Theorem to find the indicated term. 18. The term containing x3 in the expansion (2x − y)5 19. The term containing x117 ... |
x + y23 13. x − x−14 = 1 27 x3 + 1 3 x2y2 + xy4 + y6 = x4 − 4x2 + 6 − 4x−2 + x−4 14. −7 − 24i 15. 8 16. i 17. −1 18. 80x3y2 19. 236x117 20. −24x 7 2 21. −40x−7 22. 70 Chapter 10 Foundations of Trigonometry 10.1 Angles and their Measure This section begins our study of Trigonometry and to get started, we recall some ba... |
degrees are denoted by the familiar ‘◦’ symbol. One complete revolution as shown below is 360◦, and parts of a revolution are measured proportionately.2 Thus half of a revolution (a straight angle) measures 1 2 (360◦) = 180◦, a quarter of a revolution (a right angle) measures 1 4 (360◦) = 90◦ and so on. One revolution... |
into sixty minutes, and in turn, each minute is divided equally into sixty seconds.5 In symbols, we write 1◦ = 60 and 1 = 60, from which it follows that 1◦ = 3600. To convert a measure of 42.125◦ to the DMS system, we start by noting that 42.125◦ = 42◦ + 0.125◦. Converting the partial amount of degrees to minutes, we ... |
Convert α to the DMS system. Round your answer to the nearest second. 2. Convert β to decimal degrees. Round your answer to the nearest thousandth of a degree. 3. Sketch α and β. 4. Find a supplementary angle for α. 5. Find a complementary angle for β. Solution. 1. To convert α to the DMS system, we start with 111.371... |
angle γ so that β + γ = 90◦. We get γ = 90◦ − β = 90◦ − 37◦2817. While we could reach for the calculator to obtain an approximate answer, we choose instead to do a bit of sexagesimal7 arithmetic. We first rewrite 90◦ = 90◦00 = 89◦600 = 89◦5960. In essence, we are ‘borrowing’ 1◦ = 60 from the degree place, and then borr... |
angle with measure 450◦ we start with an initial side, rotate counter-clockwise one complete revolution (to take care of the ‘first’ 360◦) then continue with an additional 90◦ counter-clockwise rotation, as seen below. 450◦ To further connect angles with the Algebra which has come before, we shall often overlay an angl... |
, at least one of which is positive and one of which is negative. 1. α = 60◦ 2. β = −225◦ 3. γ = 540◦ 4. φ = −750◦ Solution. 1. To graph α = 60◦, we draw an angle with its initial side on the positive x-axis and rotate counter-clockwise 60◦ 6 of a revolution. We see that α is a Quadrant I angle. To find angles which are... |
k, where k is an integer. Working through the arithmetic, we find three such angles: 180◦, −180◦ and 900◦. 4. The Greek letter φ is pronounced ‘fee’ or ‘fie’ and since φ is negative, we begin our rotation clockwise from the positive x-axis. Two full revolutions account for 720◦, with just 30◦ or 1 12 of a revolution to ... |
which is explored in the Exercises. Since the diameter of a circle is twice its radius, we can quickly rearrange the equation in Definition 10.1 to get a formula more useful for our purposes, namely: 2π = C r 10.1 Angles and their Measure 701 This tells us that for any circle, the ratio of its circumference to its radi... |
word ‘radians’ to denote these dimensionless units as needed. For instance, we say one revolution measures ‘2π radians,’ half of a revolution measures ‘π radians,’ and so forth. As with degree measure, the distinction between the angle itself and its measure is often blurred in practice, so when we write ‘θ = π 2 radi... |
be a Quadrant II angle. To find coterminal angles, we proceed as before using 2π = 6π 3 · k for integer values of k. We obtain 2π 3 + 6π 3, − 10π 3, and compute θ = − 4π 2π = 2 3 as coterminal angles. 3 and 8π 13The authors are well aware that we are now identifying radians with real numbers. We will justify this short... |
2 in standard position. It is worth mentioning that we could have plotted the angles in Example 10.1.3 by first converting them to degree measure and following the procedure set forth in Example 10.1.2. While converting back and forth from degrees and radians is certainly a good skill to have, it is best that you learn... |
θ = s. In order to identify real numbers with oriented angles, we make good use of this fact by essentially ‘wrapping’ the real number line around the Unit Circle and associating to each real number t an oriented arc on the Unit Circle with initial point (1, 0). s 1 s r = Viewing the vertical line x = 1 as another rea... |
around the unit circle. Since π ≈ 3.14 and π 2 ≈ 1.57, we find that rotating 2 radians clockwise from the point (1, 0) lands us in Quadrant III. To more accurately place the endpoint, we proceed as we did in Example 10.1.1, successively halving the angle measure until we find 5π 8 ≈ 1.96 which tells us our arc extends j... |
v has units of length have v = displacement time and conveys two ideas: the direction in which the object is moving and how fast the position of the object is changing. The contribution of direction in the quantity v is either to make it positive (in the case of counter-clockwise motion) or negative (in the case of cl... |
ians. The supposed contradiction in units is resolved by remembering that radians are a dimensionless quantity and angles in radian measure are identified with real numbers so that the units length·radians time. We are long overdue for an example. reduce to the units length time time Example 10.1.5. Assuming that the su... |
the period of the motion is 24 hours, or one day. The concepts of frequency and period help frame the equation v = rω in a new light. That is, if ω is fixed, points which are farther from the center of rotation need to travel faster to maintain the same angular frequency since they have farther to travel to make one re... |
the angle from degree measure into radian measure, giving the exact value in terms of π. 29. 0◦ 33. −315◦ 30. 240◦ 34. 150◦ 31. 135◦ 35. 45◦ 32. −270◦ 36. −225◦ In Exercises 37 - 44, convert the angle from radian measure into degree measure. 37. π 41. π 3 38. − 2π 3 42. 5π 3 39. 7π 6 43. − π 6 40. 44. 11π 6 π 2 710 Fo... |
is 136 feet. (Remember this from Exercise 17 in Section 7.2?) It completes two revolutions in 2 minutes and 7 seconds.20 Assuming the riders are at the edge of the circle, how fast are they traveling in miles per hour? 56. Consider the circle of radius r pictured below with central angle θ, measured in radians, and su... |
� 12. 405◦ is a Quadrant I angle coterminal with 45◦ and −3154 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −2 −3 −4 13. −270◦ lies on the positive y-axis 14. coterminal with 90◦ and −630◦ is a Quadrant II angle 5π 6 coterminal with 17π 6 and − 7π 4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x −4 −3 −2 −1 −1 −2 −3 −4 10.1... |
minal with 19π 6 and − 5π 4 −3 −2 −1 −1 −2 −3 −4 23. − 5π 3 is a Quadrant I angle 24. 3π lies on the negative x-axis coterminal with y 4 3 2 1 π 3 and − 11π 3 coterminal with π and −4 −3 −2 −1 −1 −2 −3 −4 1 2 3 4 x −4 −3 −2 −1 −1 −2 −3 −4 25. −2π lies on the positive x-axis 26. − π 4 is a Quadrant IV angle coterminal w... |
π 3 square units 58. 6250π square units 60. π 2 square units 62. 38.025π ≈ 119.46 square units 10.2 The Unit Circle: Cosine and Sine 717 10.2 The Unit Circle: Cosine and Sine In Section 10.1.1, we introduced circular motion and derived a formula which describes the linear velocity of an object moving on a circular path... |
axis. The point on the Unit Circle that lies on the negative x-axis is (−1, 0) which means cos(−π) = −1 and sin(−π) = 0. 1The etymology of the name ‘sine’ is quite colorful, and the interested reader is invited to research it; the ‘co’ in ‘cosine’ is explained in Section 10.4. 718 Foundations of Trigonometry y 1 θ = 27... |
the Unit Circle, we drop a perpendicular line segment from P to the x-axis to form = 1 a 30◦ − 60◦ − 90◦ right triangle. After a bit of Geometry3 we find y = 1 2. Since P (x, y) lies on the Unit Circle, we substitute y = 1 4, or x = ± 2 into x2 + y2 = 1 to get x2 = 3 2. Here, x > 0 so x = cos π 2 so sin x, y) x 1 P (x,... |
θ) + sin2(θ) = 1. The moniker ‘Pythagorean’ brings to mind the Pythagorean Theorem, from which both the Distance Formula and the equation for a circle are ultimately derived.5 The word ‘Identity’ reminds us that, regardless of the angle θ, the equation in Theorem 10.1 is always true. If one of cos(θ) or sin(θ) is known... |
2(θ) = 1, we find cos(θ) = 0. Another tool which helps immensely in determining cosines and sines of angles is the symmetry inherent in the Unit Circle. Suppose, for instance, we wish to know the cosine and sine of θ = 5π 6. We plot θ in standard position below and, as usual, let P (x, y) denote the point on the termina... |
angle 722 Foundations of Trigonometry Reference angle α for a Quadrant III angle Reference angle α for a Quadrant IV angle We have just outlined the proof of the following theorem. Theorem 10.2. Reference Angle Theorem. Suppose α is the reference angle for θ. Then cos(θ) = ± cos(α) and sin(θ) = ± sin(α), where the cho... |
so the Reference Angle Theorem gives: cos 11π 6 2 and sin 11π = − sin π 6 = cos = 225◦ 45 = 11π 6 Finding cos (225◦) and sin (225◦) Finding cos 11π 6 and sin 11π 6 3. To plot θ = − 5π 4, we rotate clockwise an angle of 5π 4 from the positive x-axis. The terminal side of θ, therefore, lies in Quadrant II making an angl... |
Points on the Unit Circle 6For once, we have something convenient about using radian measure in contrast to the abstract theoretical nonsense about using them as a ‘natural’ way to match oriented angles with real numbers! 10.2 The Unit Circle: Cosine and Sine 725 The next example summarizes all of the important ideas ... |
this. 726 Foundations of Trigonometry Visualizing θ = π + α θ has reference angle α (b) Rewriting θ = 2π − α as θ = 2π + (−α), we can plot θ by visualizing one complete revolution counter-clockwise followed by a clockwise revolution, or ‘backing up,’ of α radians. We see that α is θ’s reference angle, and since θ is a... |
� 2 + α Using symmetry to determine Q(x, y) 728 Foundations of Trigonometry Our next example asks us to solve some very basic trigonometric equations.8 Example 10.2.5. Find all of the angles which satisfy the given equation. 1. cos(θ) = 1 2 2. sin(θ) = − 1 2 3. cos(θ) = 0. Solution. Since there is no context in the pro... |
The Unit Circle: Cosine and Sine 729 In Quadrant III, one solution is 7π multiples of 2π: θ = 7π are of the form θ = 11π 6, so we capture all Quadrant III solutions by adding integer so all the solutions here 6 + 2πk. In Quadrant IV, one solution is 11π 6 + 2πk for integers k. 6 3. The angles with cos(θ) = 0 are quadr... |
position. Let Q(x, y) be the point on the terminal side of θ which lies on the circle x2 + y2 = r2, and let P (x, y) be the point on the terminal side of θ which lies on the Unit Circle. Now consider dropping perpendiculars from P and Q to create two right triangles, ∆OP A and ∆OQB. These triangles are similar,10 thus... |
2 = 1, so Theorem 10.3 reduces to Example 10.2.6. 1. Suppose that the terminal side of an angle θ, when plotted in standard position, contains the point Q(4, −2). Find sin(θ) and cos(θ). 2. In Example 10.1.5 in Section 10.1, we approximated the radius of the earth at 41.628◦ north latitude to be 2960 miles. Justify thi... |
10.3, the location of the object Q(x, y) on the circle is found using the equations x = r cos(θ) = r cos(ωt) and y = r sin(θ) = r sin(ωt). Hence, at time t, the object is at the point (r cos(ωt), r sin(ωt)). We have just argued the following. Equation 10.3. Suppose an object is traveling in a circular path of radius r... |
triangle opposite θ; and the remaining side of length c (the side opposite the 11You may have been exposed to this in High School. 10.2 The Unit Circle: Cosine and Sine 733 right angle) is called the hypotenuse. We now imagine drawing this triangle in Quadrant I so that the angle θ is in standard position with the adj... |
√ 3 3 3 3 734 Foundations of Trigonometry could use the Pythagorean Theorem to find the missing side and solve (7)2 + b2 = Alternatively, we could use Theorem 10.4, namely that sin (30◦) = b b = c sin (30◦) = 14 3 √ 3 3. The triangle with all of its data is recorded below. 2 = 7 · 1 √ 3 c. Choosing the latter, we find 2... |
�). Since cos(t) and sin(t) represent x- and y-coordinates, respectively, of points on the Unit Circle, they both take on all of the values between −1 an 1, inclusive. In other words, the range of f (t) = cos(t) and of g(t) = sin(t) is the interval [−1, 1]. To summarize: 10.2 The Unit Circle: Cosine and Sine 735 Theore... |
θ = 14. θ = π 4 3π 4 4π 3 23π 6 18. θ = − π 6 3. θ = π 3 7. θ = π 11. θ = 3π 2 15. θ = − 13π 2 19. θ = 10π 3 4. θ = 8. θ = 12. θ = π 2 7π 6 5π 3 16. θ = − 43π 6 20. θ = 117π In Exercises 21 - 30, use the results developed throughout the section to find the requested value. 21. If sin(θ) = − 7 25 with θ in Quadrant IV, ... |
0 36. cos(θ) = −1 39. cos(θ) = −1.001 37. sin(θ) = −1 38. cos(θ) = In Exercises 40 - 48, solve the equation for t. (See the comments following Theorem 10.5.) 40. cos(t) = 0 41. sin(t) = − 43. sin(t) = − 1 2 44. cos(t) = 1 2 √ 2 2 42. cos(t) = 3 45. sin(t) = −2 46. cos(t) = 1 47. sin(t) = 1 48. cos(t) = − √ 2 2 In Exer... |
enuse has length 10, how long is the side adjacent to θ? 64. If θ = 37.5◦ and the side opposite θ has length 306, how long is the side adjacent to θ? In Exercises 65 - 68, let θ be the angle in standard position whose terminal side contains the given point then compute cos(θ) and sin(θ). 65. P (−7, 24) 66. Q(3, 4) 67. ... |
, the object was at the point (r, 0). If this is not the case, we can adjust the equations of motion by introducing a ‘time delay.’ If t0 > 0 is the first time the object passes through the point (r, 0), show, with the help of your classmates, the equations of motion are x = r cos(ω(t − t0)) and y = r sin(ω(t − t0)). 74... |
− 2 11 25. If sin(θ) = − 2 3 26. If cos(θ) = 28 53 with θ in Quadrant IV, then cos(θ) = with θ in Quadrant I, then sin(θ) = with θ in Quadrant II, then cos(θ) = − 12 13. with θ in Quadrant III, then sin(θ) = − 24 25.. √ 65 9 √ 117 11. √ 5 3. with θ in Quadrant III, then cos(θ) = − with θ in Quadrant IV, then sin(θ) = ... |
= (2k + 1)π for any integer k. 37. sin(θ) = −1 when θ = 38. cos(θ) = √ 3 2 when θ = 3π 2 π 6 + 2πk for any integer k. + 2πk or θ = 11π 6 + 2πk for any integer k. 39. cos(θ) = −1.001 never happens 40. cos(t) = 0 when t = π 2 + πk for any integer k. 41. sin(t) = − √ 2 2 when t = 5π 4 + 2πk or t = 7π 4 + 2πk for any inte... |
c2 − 62 ≈ 5.402 59. The hypotenuse has length 4 cos(12◦) ≈ 4.089. 60. The side adjacent to θ has length 5280 cos(78.123◦) ≈ 1086.68. 61. The hypotenuse has length 117.42 sin(59◦) ≈ 136.99. 62. The side opposite θ has length 10 sin(5◦) ≈ 0.872. 63. The side adjacent to θ has length 10 cos(5◦) ≈ 9.962. 64. The hypotenus... |
Fundamental Identities In section 10.2, we defined cos(θ) and sin(θ) for angles θ using the coordinate values of points on the Unit Circle. As such, these functions earn the moniker circular functions.1 It turns out that cosine and sine are just two of the six commonly used circular functions which we define below. Defin... |
residing in a right triangle so we could just as easily call them trigonometric functions. In later sections, you will find that we do indeed use the phrase ‘trigonometric function’ interchangeably with the term ‘circular function’. 10.3 The Six Circular Functions and Fundamental Identities 745 y = 1 x which gives y = ... |
��nition 10.2, it is customary to rephrase the remaining four circular functions in terms of cosine and sine. The following theorem is a result of simply replacing x with cos(θ) and y with sin(θ) in Definition 10.2. Theorem 10.6. Reciprocal and Quotient Identities: sec(θ) = 1 cos(θ), provided cos(θ) = 0; if cos(θ) = 0, ... |
3 radians is not one of the ‘common angles’ from Section 10.2, we resort to the calculator for a decimal approximation. Ensuring that the calculator is in radian mode, we find cot(3) = cos(3) sin(3) ≈ −7.015. = 0 and sin(θ) = sin 3π 2 = −1. Attempting 4. If θ is coterminal with 3π to compute tan(θ) = sin(θ) 2, then cos... |
Since π < θ < 3π 2, θ is a Quadrant III angle. This means sin(θ) < 0, so our final answer is sin(θ) = − 3 = 1. Solving, we get sin2(θ) = 9 10 √ √ 10 10. While the Reciprocal and Quotient Identities presented in Theorem 10.6 allow us to always reduce problems involving secant, cosecant, tangent and cotangent to problems... |
) = 2 Solution. 2. tan(θ) = √ 3 3. cot(θ) = −1. 1. To solve sec(θ) = 2, we convert to cosines and get 1 cos(θ) = 2 or cos(θ) = 1 same equation we solved in Example 10.2.5, number 1, so we know the answer is: θ = π or θ = 5π 3 + 2πk for integers k. 2. This is the exact 3 + 2πk √ √ = 2. From the table of common values, w... |
one negative.) Hence, our solutions lie in Quadrants II and IV. Our Quadrant II solution is θ = 3π 4 + 2πk, and for Quadrant IV, we get θ = 7π 4 +2πk for integers k. Can these lists be combined? Indeed they can - one such way to capture all the solutions is: θ = 3π 4 + πk for integers k We have already seen the import... |
(θ) = 0. Common Alternate Forms: csc2(θ) − cot2(θ) = 1 csc2(θ) − 1 = cot2(θ) Trigonometric identities play an important role in not just Trigonometry, but in Calculus as well. We’ll use them in this book to find the values of the circular functions of an angle and solve equations and inequalities. In Calculus, they are ... |
Expanding the left hand side of the equation gives: (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = sec2(θ) − tan2(θ). According to Theorem 10.8, sec2(θ) − tan2(θ) = 1. Putting it all together, (sec(θ) − tan(θ))(sec(θ) + tan(θ)) = sec2(θ) − tan2(θ) = 1. 4. While both sides of our last identity contain fractions, the left side aff... |
to another way, earn more partial credit if this were an exam question! 10.3 The Six Circular Functions and Fundamental Identities 751 At this point, it is worth pausing to remind ourselves of our goal. We wish to transform this expression into 6 sec(θ) tan(θ). Using a reciprocal and quotient identity, we find 6 sec(θ)... |
.8. This is exactly the √ same kind of phenomenon that occurs when we multiply expressions such as 1 − 2 or 3 − 4i by 3 + 4i. (Can you recall instances from Algebra where we did such things?) For this reason, the quantities (1 − cos(θ)) and (1 + cos(θ)) are called ‘Pythagorean Conjugates.’ Below is a list of other comm... |
identities. Nevertheless, a summary of some strategies which may be helpful (depending on the situation) is provided below and ample practice is provided for you in the Exercises. Try working on the more complicated side of the identity. Strategies for Verifying Identities Use the Reciprocal and Quotient Identities in... |
(θ) = −4. Find the values of the five remaining circular functions of θ. Solution. 1. Since x = 3 and y = −4, r = x2 + y2 = (3)2 + (−4)2 = √ cos(θ) = 3 5, sin(θ) = − 4 5, sec(θ) = 5 3, csc(θ) = − 5 4, tan(θ) = − 4 25 = 5. Theorem 10.9 tells us 3 and cot(θ) = − 3 4. 2. In order to use Theorem 10.9, we need to find a point... |
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