A special case of the above result which you might encounter more often isthe following:

n∑

i=1

c = cn.

Telescoping Sum

n∑

i=m

[f(i+ 1)− f(i)] = f(n+ 1)− f(m)

Proof.

n∑

i=m

[f(i+ 1)− f(i)

]

= [f(m+ 1)− f(m)] + [f(m+ 2)− f(m+ 1)]

+ [f(m+ 3)− f(m+ 2)] + · · ·+ [f(n+ 1)− f(n)]

Note that the terms, f(m+1), f(m+2), . . . , f(n), all cancel out. Hence, we have

n∑

i=m

[f(i+ 1)− f(i)] = f(n+ 1)− f(m). �

Example 2.2.3. Evaluate:30∑

i=1

(4i− 5).

Solution.

30∑

i=1

(4i− 5) =30∑

i=1

4i−30∑

i=1

5

= 430∑

i=1

i−30∑

i=1

5

= 4(30)(31)

2− 5(30)

= 1710 �

Example 2.2.4. Evaluate:

1

1 · 2+

1

2 · 3+

1

3 · 4+ · · ·+

1

99 · 100.

95

Solution.

1

1 · 2+

1

2 · 3+

1

3 · 4+ · · ·+

1

99 · 100

=99∑

i=1

1

i(i+ 1)

=99∑

i=1

i+ 1− i

i(i+ 1)

=99∑

i=1

[i+ 1

i(i+ 1)− i

i(i+ 1)

]

=99∑

i=1

(1

i− 1

i+ 1

)

= −99∑

i=1

(1

i+ 1− 1

i

)

Using f(i) =1

iand the telescoping-sum property, we get

99∑

i=1

1

i(i+ 1)= −

(1

100− 1

1

)

=99

100. �

Example 2.2.5. Derive a formula forn∑

i=1

i2 using a telescoping sum with terms

f(i) = i3.

Solution. The telescoping sum property implies that

n∑

i=1

[i3 − (i− 1)3

]= n3 − 03 = n3.

On the other hand, using expansion and the other properties of summation,we have

n∑

i=1

[i3 − (i− 1)3

]=

n∑

i=1

(i3 − i3 + 3i2 − 3i+ 1)

= 3n∑

i=1

i2 − 3n∑

i=1

i+n∑

i=1

1

= 3n∑

i=1

i2 − 3 ·n(n+ 1)

2+ n.

96

Equating the two results above, we obtain

3n∑

i=1

i2 − 3n(n+ 1)

2+ n = n3

6n∑

i=1

i2 − 3n(n+ 1) + 2n = 2n3

6n∑

i=1

i2 = 2n3 − 2n+ 3n(n+ 1)

= 2n(n2 − 1) + 3n(n+ 1)

= 2n(n− 1)(n+ 1) + 3n(n+ 1)

= n(n+ 1)[2(n− 1) + 3]

= n(n+ 1)(2n+ 1).

Finally, after dividing both sides of the equation by 6, we obtain the desiredformula

n∑

i=1

i2 =n(n+ 1)(2n+ 1)

6. �

Seatwork/Homework 2.2.2

1. Use the properties of sigma notation to evaluate the following summations.

(a)50∑

k=1

(2− 3k) Answer: −3725

(b)n∑

j=1

(1 + 2j) Answer: 2n+ n2

(c)99∑

j=1

1√i+ 1 +

√i

Answer: 9

Solution:99∑

j=1

1√i+ 1 +

√i=

99∑

j=1

1√i+ 1 +

√i·

√i+ 1−

√i√

i+ 1−√i

=99∑

j=1

(√i+ 1−

√i)

=√99 + 1−

√1

= 9

97

2. Ifn∑

i=1

(i+ 1)2 = an3 + bn2 + cn+ d, what is a+ b+ c+ d? Answer: 4

Exercises 2.2

1. Expand each sum.

(a)9∑

i=5

i

x+ iAnswer:

1

x+ 1+

2

x+ 2+

3

x+ 3+

4

x+ 4+

5

x+ 5

(b)6∑

i=0

3√2i Answer: 0 + 3

√2 + 3

√4 + 3

√6 + 2 + 3

√10 + 3

√12

(c)3∑

i=−2

3−i Answer: 9 + 3 + 1 + 1/3 + 1/9 + 1/27

2. Write each expression in sigma notation.

(a) 1 + 22 + 33 + 44 + · · ·+ 1212 Answer:12∑

i=1

ii

(b) (x− 5) + (x− 3) + (x− 1) + (x+ 1) + (x+ 3) + · · ·+ (x+ 15)Teaching Notes

Another possibleanswer for (b) is11∑

i=1

[x+ (2i− 7)].Answer:7∑

i=−3

[x+ (2i+ 1)]

(c) a1 + a4 + a9 + a16 + · · ·+ a81 Answer:9∑

i=1

ai2

3. Evaluate each sum.

(a)120∑

i=1

(4i− 15) Answer: 27240

(b)50∑

i=1

[(5i− 2)(i+ 3)] Answer: 230900

(c)n∑

i=1

(3i− 1)2 Answer:6n3 + 3n2 − 3n+ 2

2

4. If30∑

i=1

f(i) = 70 and30∑

i=1

g(i) = 50, what is the value of30∑

i=1

3g(i)− f(i) + 7

2?

Answer: 145

98

5. If s =100∑

i=1

i, express200∑

i=1

i in terms of s. Answer: 2s+ 100000

6. If s =n∑

i=1

ai, does it follow thatn∑

i=1

a2i = s2?

Answer: No. If s =2∑

i=1

ai = a1 + a2, thenn∑

i=1

a2i = a21 + a22, while s2 =

a21 + 2a1a2 + a22.

7. Derive a formula forn∑

i=1

i3 by using a telescoping sum with terms f(i) = i4.

Answer:n2(n+ 1)2

4

4

Lesson 2.3. Mathematical Induction

Time Frame: 3 one-hour sessions

Learning Outcomes of the Lesson

At the end of the lesson, the student is able to:

(1) illustrate the Principle of Mathematical Induction; and

(2) apply mathematical induction in proving identities.

Lesson Outline

(1) State the principle of mathematical induction

(2) Prove summation identities using mathematical induction

(3) Prove divisibility statements using mathematical induction

(4) Prove inequalities using mathematical induction

Introduction

We have derived and used formulas for the terms of arithmetic and geometricsequences and series. These formulas and many other theorems involving positiveintegers can be proven with the use of a technique called mathematical induction.

99

2.3.1. Proving Summation Identities

There are many mathematical results that can be proven using mathematicalinduction. In this lesson, we will focus on three main categories: summationidentities, divisibility statements, and inequalities.

We first state the Principle of Mathematical Induction, and see how the prin-ciple works in general sense.

The Principle of Mathematical Induction

Let P (n) be a property or statement about an integer n. Supposethat the following conditions can be proven:

(1) P (n0) is true (that is, the statement is true when n = n0).

(2) If P (k) is true for some integer k ≥ n0, then P (k + 1) is true(that is, if the statement is true for n = k, then it is also true forn = k + 1).

Then the statement P (n) is true for all integers n ≥ n0.

The Principle of Mathematical Induction is often compared to climbing aninfinite staircase. First, you need to be able to climb up to the first step. Second,if you are on any step (n = k), you must be able to climb up to the next step(n = k + 1). If you can do these two things, then you will be able to climb upthe infinite staircase.

Part 1 Part 2

Another analogy of the Principle of Mathematical Induction that is used istoppling an infinite line of standing dominoes. You need to give the first dominoa push so that it falls down. Also, the dominoes must be arranged so that if thekth domino falls down, the next domino will also fall down. These two conditionswill ensure that the entire line of dominoes will fall down.

100

https://commons.wikimedia.org/wiki/File:Wallpaper kartu domino.png

By Nara Cute (Own work)[CC BY-SA 4.0 (http://creativecommons.org/licenses/by-sa/4.0)],

via Wikimedia Commons

We now consider some examples on the use of mathematical induction inproving summation identities.

Example 2.3.1. Using mathematical induction, prove that

1 + 2 + 3 + · · ·+ n =n(n+ 1)

2

for all positive integers n.

Solution. We need to establish the two conditions stated in the Principle of Math-ematical Induction.

Part 1. Prove that the identity is true for n = 1.

The left-hand side of the equation consists of one term equal to 1. The right-hand side becomes

1(1 + 1)

2=

2

2= 1.

Hence, the formula is true for n = 1.

Part 2. Assume that the formula is true for n = k ≥ 1:

1 + 2 + 3 + · · ·+ k =k(k + 1)

2.

101

We want to show that the formula is true for n = k + 1; that is,

1 + 2 + 3 + · · ·+ k + (k + 1) =(k + 1)(k + 1 + 1)

2.

Using the formula for n = k and adding k + 1 to both sides of the equation,we get

1 + 2 + 3 + · · ·+ k + (k + 1) =k(k + 1)

2+ (k + 1)

=k(k + 1) + 2(k + 1)

2

=(k + 1)(k + 2)

2

=(k + 1) [(k + 1) + 1]

2

We have proven the two conditions required by the Principle of MathematicalInduction. Therefore, the formula is true for all positive integers n. �

Example 2.3.2. Use mathematical induction to prove the formula for the sumof a geometric series with n terms:

Sn =a1 (1− rn)

1− r, r "= 1.

Solution. Let an be the nth term of a geometric series. From Lesson 2.1, we knowthat an = a1r

n−1. Teaching Notes

The fact thatan = a1rn−1 canalso be proven bymathematicalinduction. Here,however, we simplyrecall a formula inLesson 2.1 becauseour focus in thisexample is theproof of the sum.

Part 1. Prove that the formula is true for n = 1.

a1(1− r1)

1− r= a1 = S1

The formula is true for n = 1.

Part 2. Assume that the formula is true for n = k ≥ 1: Sk =a1(1− rk)

1− r. We

want to prove that it is also true for n = k + 1; that is,

Sk+1 =a1(1− rk+1)

1− r.

We know that

Sk+1 = a1 + a2 + · · ·+ ak︸ ︷︷ ︸

Sk

+ak+1

= Sk + ak+1

102

=a1

(1− rk

)

1− r+ a1r

k

=a1

(1− rk

)+ a1r

k (1− r)

1− r

=a1

(1− rk + rk − rk+1

)

1− r

=a1

(1− rk+1

)

1− r

By the Principle of Mathematical Induction, we have proven that

Sn =a1(1− rn)

1− r

for all positive integers n. �

Example 2.3.3. Using mathematical induction, prove that

12 + 22 + 32 + · · ·+ n2 =n(n+ 1)(2n+ 1)

6

for all positive integers n.

Solution. We again establish the two conditions stated in the Principle of Math-ematical Induction.

Part 11(1 + 1)(2 · 1 + 1)

6=

1 · 2 · 3

6= 1 = 12

The formula is true for n = 1.

Part 2

Assume: 12 + 22 + 32 + · · ·+ k2 =k(k + 1)(2k + 1)

6.

Prove: 12 + 22 + 32 + · · ·+ k2 + (k + 1)2

=(k + 1)(k + 2) [2(k + 1) + 1]

6

=(k + 1)(k + 2)(2k + 3)

6.

12 + 22 + 32 + · · ·+ k2 + (k + 1)2

=k(k + 1)(2k + 1)

6+ (k + 1)2

=k(k + 1)(2k + 1) + 6(k + 1)2

6

103

=(k + 1) [k(2k + 1) + 6(k + 1)]

6

=(k + 1) (2k2 + 7k + 6)

6

=(k + 1)(k + 2)(2k + 3)

6

Therefore, by the Principle of Mathematical Induction,

12 + 22 + 32 + · · ·+ n2 =n(n+ 1)(2n+ 1)

6

for all positive integers n. �

Seatwork/Homework 2.3.1

Using mathematical induction, prove that

1 · 3 + 2 · 4 + 3 · 5 + · · ·+ n(n+ 2) =n(n+ 1)(2n+ 7)

6.

Answer:

Part 11(1 + 1)[2(1) + 7]

6=

2 · 9

6= 3 = 1 · 3

The formula is true for n = 1.

Part 2

Assume: 1 · 3 + 2 · 4 + 3 · 5 + · · ·+ k(k + 2) =k(k + 1)(2k + 7)

6

To show: 1 · 3 + 2 · 4 + · · ·+ k(k + 2) + (k + 1)(k + 3)

=(k + 1)(k + 2) [2(k + 1) + 7]

6

=(k + 1)(k + 2)(2k + 9)

6

1 · 3 + 2 · 4 + · · ·+ k(k + 2) + (k + 1)(k + 3)

=k(k + 1)(2k + 7)

6+ (k + 1)(k + 3)

=(k + 1)

6[k(2k + 7) + 6(k + 3)]

=(k + 1)

6

[2k2 + 13k + 18

]

104

=(k + 1)(k + 2)(2k + 9)

6

Therefore, by the Principle of Math Induction, the formula is true for all positiveintegers n.Teaching Notes

Recall thedefinition of

divisibility: aninteger n is

divisible by aninteger k if n = krfor some integer r.

2.3.2. Proving Divisibility Statements

We now prove some divisibility statements using mathematical induction.

Example 2.3.4. Use mathematical induction to prove that, for every positiveinteger n, 7n − 1 is divisible by 6.

Solution. Similar to what we did in the previous session, we establish the twoconditions stated in the Principle of Mathematical Induction.

Part 1

71 − 1 = 6 = 6 · 1

71 − 1 is divisible by 6.

Part 2

Assume: 7k − 1 is divisible by 6.

To show: 7k+1 − 1 is divisible by 6.

7k+1 − 1 = 7 · 7k − 1 = 6 · 7k + 7k − 1 = 6 · 7k + (7k − 1)

By definition of divisibility, 6 · 7k is divisible by 6. Also, by the hypothesis(assumption), 7k − 1 is divisible by 6. Hence, their sum (which is equal to7k+1 − 1) is also divisible by 6.

Therefore, by the Principle of Math Induction, 7n − 1 is divisible by 6 for allpositive integers n. �

Note that 70 − 1 = 1− 1 = 0 = 6 · 0 is also divisible by 6. Hence, a strongerand more precise result in the preceding example is: 7n − 1 is divisible by 6 forevery nonnegative integer n. It does not make sense to substitute negative valuesof n since this will result in non-integer values for 7n − 1.

Example 2.3.5. Use mathematical induction to prove that, for every nonnega-tive integer n, n3 − n+ 3 is divisible by 3.

Solution. We again establish the two conditions in the Principle of MathematicalInduction.

105

Part 1. Note that claim of the statement is that it is true for every nonnegativeinteger n. This means that Part 1 should prove that the statement is true forn = 0.

03 − 0 + 3 = 3 = 3(1)

03 − 0 + 3 is divisible by 3.

Part 2. We assume that k3− k+3 is divisible by 3. By definition of divisibility,we can write k3 − k + 3 = 3a for some integer a.

To show: (k + 1)3 − (k + 1) + 3 is divisible by 3.

(k + 1)3 − (k + 1) + 3 = k3 + 3k2 + 2k + 3

= (k3 − k + 3) + 3k2 + 3k

= 3a+ 3k2 + 3k

= 3(a+ k2 + k)

Since a+k2+k is also an integer, by definition of divisibility, (k+1)3−(k+1)+3is divisible by 3.

Therefore, by the Principle of Math Induction, n3−n+3 is divisible by 3 forall positive integers n. �

Seatwork/Homework 2.3.1

Use mathematical induction to prove each divisibility statement for all nonnega-tive integers n.

(1) 72n − 3 · 5n + 2 is divisible by 12.

Answer:

Part 1

72(0) − 3 · 50 + 2 = 1− 3(1) + 2 = 0 = 12(0)

72(0) − 3 · 50 + 2 is divisible by 12

Part 2

Assume: 72k − 3 · 5k + 2 is divisible by 12

To show: 72(k+1) − 3 · 5(k+1) + 2 is divisible by 12

72(k+1) − 3 · 5(k+1) + 2

= 7272k − 3 · 5 · 5k + 2

106

= 49 · 72k − 15 · 5k + 2

= 72k + 48 · 72k − 3 · 5k − 12 · 5k + 2

=(72k − 3 · 5k + 2

)+ 48 · 72k − 12 · 5k

=(72k − 3 · 5k + 2

)+ 12

(4 · 72k − 5k

)

By the hypothesis, 72k − 3 · 5k + 2 is divisible by 12. The second term,12

(4 · 72k − 5k

), is divisible by 12 because 4 · 72k − 5k is an integer. Hence

their sum, which is equal to 72(k+1) − 3 · 5(k+1) + 2, is divisible by 12.

Therefore, by the Principle of Math Induction, 72n− 3 · 5n + 2 is divisible by12 for every nonnegative integer n.

(2) n3 + 3n2 + 2n is divisible by 3.

Answer:

Part 1

03 + 3 · 02 + 2(0) = 0 = 3(0)

Thus, 03 + 3 · 02 + 2(0) is divisible by 3.

Part 2

Assume: k3 + 3k2 + 2k is divisible by 3.

=⇒ k3 + 3k2 + 2k = 3a, a integer

To show: (k + 1)3 + 3(k + 1)2 + 2(k + 1) is divisible by 3.

(k + 1)3 + 3(k + 1)2 + 2(k + 1)

= k3 + 6k2 + 11k + 6

= (k3 + 3k2 + 2k) + 3k2 + 9k + 6

= 3a+ 3k2 + 9k + 6

= 3(a+ k2 + 3k + 2)

Since a+ k2+3k+2 is also an integer, by definition of divisibility, (k+1)3+3(k + 1)2 + 2(k + 1) is divisible by 3.

Therefore, by the Principle of Math Induction, n3 + 3n2 + 2n is divisible by3 for all positive integers n.

�2.3.3. Proving Inequalities

Finally, we now apply the Principle of Mathematical Induction in proving someinequalities involving integers.

Example 2.3.6. Use mathematical induction to prove that 2n > 2n for everyinteger n ≥ 3.

107

Solution. Just like the previous example, we establish the two conditions in thePrinciple of Mathematical Induction.

Part 1

23 = 8 > 6 = 2(3)

This confirms that 23 > 2(3).

Part 2

Assume: 2k > 2k, where k is an integer with k ≥ 3

To show: 2k+1 > 2(k + 1) = 2k + 2

We compare the components of the assumption and the inequality we need toprove. On the left-hand side, the expression is doubled. On the right-hand side,the expression is increased by 2. We choose which operation we want to apply toboth sides of the assumed inequality.

Alternative 1. We double both sides.

Since 2k > 2k, by the multiplication property of inequality, we have 2 · 2k >2 · 2k.

2k+1 > 2(2k) = 2k + 2k > 2k + 2 if k ≥ 3.

Hence, 2k+1 > 2(k + 1).

Alternative 2. We increase both sides by 2.

Since 2k > 2k, by the addition property of inequality, we have 2k+2 > 2k+2.

2(k + 1) = 2k + 2 < 2k + 2 < 2k + 2k if k ≥ 3.

The right-most expression above, 2k + 2k, is equal to 2(2k

)= 2k+1.

Hence, 2(k + 1) < 2k+1.

Therefore, by the Principle of Math Induction, 2n > 2n for every integern ≥ 3. �

We test the above inequality for integers less than 3.

20 = 1 > 0 = 2(0) True

21 = 2 = 2(1) False

22 = 4 = 2(2) False

The inequality is not always true for nonnegative integers less than 3. Thisillustrates the necessity of Part 1 of the proof to establish the result. However,the result above can be modified to: 2n ≥ 2n for all nonnegative integers n.

Before we discuss the next example, we review the factorial notation. Recall

108

that 0! = 1 and, for every positive integer n, n! = 1 · 2 · 3 · · ·n. The factorial alsosatisfies the property that (n+ 1)! = (n+ 1) · n!.

Example 2.3.7. Use mathematical induction to prove that 3n < (n + 2)! forevery positive integer n. Can you refine or improve the result?

Solution. We proceed with the usual two-part proof.

Part 1

31 = 3 < 6 = 3! = (1 + 2)! =⇒ 31 < (1 + 2)!

Thus, the desired inequality is true for n = 1.

Part 2

Assume: 3k < (k + 2)!

To show: 3k+1 < (k + 3)!

Given that 3k < (k + 2)!, we multiply both sides of the inequality by 3 andobtain

3(3k

)< 3 [(k + 2)!] .

This implies that

3(3k

)< 3 [(k + 2)!] < (k + 3) [(k + 2)!] , since k > 0,

and so3k+1 < (k + 3)!.

Therefore, by the Principle of Math Induction, we conclude that 3n < (n+2)!for every positive integer n.

The left-hand side of the inequality is defined for any integer n. The right-hand side makes sense only if n+ 2 ≥ 0, or n ≥ −2.

When n = −2: 3−2 =1

9< 1 = 0! = (−2 + 2)!

When n = −1: 3−1 =1

3< 1 = 1! = (−1 + 2)!

When n = 0: 30 = 1 < 2 = 2! = (0 + 2)!

Therefore, 3n < (n+ 2)! for any integer n ≥ −2. �

Seatwork/Homework 2.3.3

Use mathematical induction to prove that 2n+ 3 < 2n for n ≥ 4.

109

Answer:

Part 1

2(4) + 3 = 11 < 16 = 24

Thus, 2(4) + 3 <= 24.

Part 2

Assume: 2k + 3 < 2k, k ≥ 4

To show: 2(k + 1) + 3 < 2k+1

2(k + 1) + 3 = 2k + 5 = (2k + 3) + 2

< 2k + 2 < 2k + 2k = 2k+1

Therefore, by the Principle of Math Induction, 2n+ 3 < 2n for n ≥ 4.

Exercises 2.3

Prove the following statements by mathematical induction.

(1)n∑

i=1

(3i− 1) =3n2 + n

2

(2)1

1 · 2+

1

2 · 3+

1

3 · 4+ · · ·+

1

n(n+ 1)=

n

n+ 1

(3)n∑

i=1

2 · 3i−1 = 3n − 1

Hint:k+1∑

i=1

2·3i−1 =k∑

i=1

2·3i−1+2·3(k+1)−1 = 3k−1+2·3k = 3·3k−1 = 3k+1−1

(4)n∑

i=1

i3 =n2(n+ 1)2

4

(5) a1 + (a1 + d) + (a1 + 2d) + · · ·+ [a1 + (n− 1)d] =n [2a1 + (n− 1)d]

2

(6) 1 (1!) + 2 (2!) + · · ·+ n (n!) = (n+ 1)!− 1

Hint:k+1∑

i=1

i · i! =k∑

i=1

i · i! + (k + 1)(k + 1)! = (k + 1)!− 1 + (k + 1)(k + 1)! =

(k + 1)!(1 + k + 1)− 1 = (k + 2)!− 1

(7) 7n − 4n is divisible by 3

Hint: 7k+1 − 4k+1 = 7 · 7k − 4 · 4k = (3 + 4)7k − 4 · 4k = 3 · 7k + (7k − 4k)

110

(8) 10n + 3 · 4n+2 + 5 is divisible by 9

Hint: 10k+1+3·4k+3+5 = 10·10k+3·4·4k+2+5 = (9+1)10k+(9+3)4k+2+5 =9(10k + 4k+2) + 10k + 3 · 4k+2 + 5

(9) 11n+2 + 122n+1 is divisible by 133

Hint: 11k+3+122k+3 = 11·11k+2+122 ·122k+1 = 11·11k+2+(133+11)122k+1 =11(11k+2 + 122k+1) + 133 · 122k+1

(10) xn − yn is divisible by x− y for any positive integer n

Hint: xk+1 − yk+1 = x · xk − y · xk + y · xk − y · yk = (x− y)xk + y(xk − yk)

(11) xn + yn is divisible by x+ y for any odd positive integer n

Hint: xk+2+ yk+2 = x2xk + y2yk = x2xk +x2yk−x2yk + y2yk = x2(xk + yk)−yk(x− y)(x+ y)

(12) If 0 < a < 1, then 0 < an < 1 for any positive integer n

Hint: 0 < ak < 1 =⇒ 0 · a < ak · a < 1 · a =⇒ 0 < ak+1 < a < 1

(13) (1 + a)n > 1 + na for a > −1, a "= 0 and n an integer greater than 1

Hint: (1 + a)k+1 > (1 + ka)(1 + a) = 1 + (k + 1)a+ ka2 > 1 + (k + 1)a

(14) 2n > n2 for every integer n > 4

Hint: 2k+1 = 2 · 2k > 2k2 = k2 + k2 > k2 + 2k + 1 = (k + 1)2. The lastinequality follows from (k−1)2 > 2 for k > 4, which implies that k2 > 2k+1.

For k > 4, (k − 1)2 > 2

(15) 2n < n! for every integer n > 3

Hint: 2k+1 = 2 · 2k = 2k! < (k + 1)k! = (k + 1)!

4

Lesson 2.4. The Binomial Theorem

Time Frame: 4 one-hour sessions

Learning Outcomes of the Lesson

At the end of the lesson, the student is able to:

(1) illustrate Pascal’s Triangle in the expansion of (x + y)n for small positiveintegral values of n;

(2) prove the Binomial Theorem;

111

(3) determine any term in (x + y)n, where n is a positive integer, without ex-panding; and

(4) solve problems using mathematical induction and the Binomial Theorem.

Lesson Outline

(1) Expand (x+ y)n for small values of n using Pascal’s Triangle

(2) Review the definition of and formula for combination Teaching Notes

The concept ofcombination wasintroduced inGrade 10. Inparticular, theconcept wasdiscussed withcompetency codesfrom M10SP-IIIc-1to M10SP-IIId-e-1.

(3) State and prove the Binomial Theorem

(4) Compute all or specified terms of a binomial expansion

(5) Prove some combination identities using the Binomial Theorem

Introduction

In this lesson, we study two ways to expand (a + b)n, where n is a positiveinteger. The first, which uses Pascal’s Triangle, is applicable if n is not too big,and if we want to determine all the terms in the expansion. The second methodgives a general formula for the expansion of (a + b)n for any positive integer n.This formula is useful especially when n is large because it avoids the process ofgoing through all the coefficients for lower values of n obtained through Pascal’sTriangle. Teaching Notes

Calculations withbig numbers arerequired in manyof the examplesand exercises inthis section. Theuse of scientificcalculators isdesirable.

Moreover, if only a specific term is required, it can be computed directlyusing a simple formula. Lastly, the theorem can be used to derive and prove someuseful and interesting results about sums of combinations.

2.4.1. Pascal’s Triangle and the Concept of Combination

Consider the following powers of a+ b: Teaching Notes

You may ask thestudents to expandthese powers usinglong multiplication.

(a+ b)1 = a+ b

(a+ b)2 = a2 + 2ab+ b2

(a+ b)3 = a3 + 3a2b+ 3ab2 + b3

(a+ b)4 = a4 + 4a3b+ 6a2b2 + 4ab3 + b4

(a+ b)5 = a5 + 5a4b+ 10a3b2 + 10a2b3 + 5ab4 + b5

We list down the coefficients of each expansion in a triangular array as follows:

n = 1 : 1 1

n = 2 : 1 2 1

n = 3 : 1 3 3 1

n = 4 : 1 4 6 4 1

n = 5 : 1 5 10 10 5 1

112

The preceding triangular array of numbers is part of what is called the Pascal’sTriangle. Named after the French mathematician Blaise Pascal (1623-1662), someproperties of the Triangle are the following:

(1) Each row begins and ends with 1.

(2) Each row has n+ 1 numbers.

(3) The second and second to the last number of each row correspond to therow number.

(4) There is symmetry of the numbers in each row.

(5) The number of entries in a row is one more than the row number (or onemore than the number of entries in the preceding row).

(6) Every middle number after first row is the sum of the two numbers aboveit.

It is the last statement which is useful in constructing the succeeding rows of thetriangle.

Example 2.4.1. Use Pascal’s Triangle to expand the expression (2x− 3y)5.

Solution. We use the coefficients in the fifth row of the Pascal’s Triangle.

(2x− 3y)5 = (2x)5 + 5(2x)4(−3y) + 10(2x)3(−3y)2

+ 10(2x)2(−3y)3 + 5(2x)(−3y)4

+ (−3y)5

= 32x5 − 240x4y + 720x3y2 − 1080x2y3

+ 810xy4 − 243y5 �

Example 2.4.2. Use Pascal’s Triangle to expand (a+ b)8.

Solution. We start with the sixth row (or any row of the Pascal’s Triangle thatwe remember).

n = 6 : 1 6 15 20 15 6 1

n = 7 : 1 7 21 35 35 21 7 1

n = 8 : 1 8 28 56 70 56 28 8 1

Therefore, we get

(a+ b)8 = a8 + 8a7b+ 28a6b2 + 56a5b3

+ 70a4b4 + 56a3b5 + 28a2b6

+ 8ab7 + b8 �

113

We observe that, for each n, the expansion of (a + b)n starts with an

and the exponent of a in the succeeding terms decreases by 1, whilethe exponent of b increases by 1. This observation will be shown tobe true in general.

Let us review the concept of combination. Recall that C(n, k) or(nk

)counts

the number of ways of choosing k objects from a set of n objects. It is also usefulto know some properties of C(n, k):

(1) C(n, 0) = C(n, n) = 1,

(2) C(n, 1) = C(n, n− 1) = n, and

(3) C(n, k) = C(n, n− k).

These properties can explain some of the observations we made on the num-bers in the Pascal’s Triangle. Recall also the general formula for the number ofcombinations of n objects taken k at a time:

C(n, k) =

(n

k

)

=n!

k!(n− k)!,

where 0! = 1 and, for every positive integer n, n! = 1 · 2 · 3 · · ·n.

Example 2.4.3. Compute

(5

3

)

and

(8

5

)

.

Solution. (5

3

)

=5!

(5− 3)!3!=

5!

2!3!= 10

(8

5

)

=8!

(8− 5)!5!=

10!

3!5!= 56 �

You may observe that the value of(53

)and the fourth coefficient in the fifth

row of Pascal’s Triangle are the same. In the same manner,(85

)is equal to the

sixth coefficient in the expansion of (a+ b)8 (see Example 2.4.2). These observedequalities are not coincidental, and they are, in fact, the essence embodied in theBinomial Theorem, as you will see in the succeeding sessions.

Seatwork/Homework 2.4.1

1. Use Pascal’s Triangle to expand each expression.

(a) (x− 2y)4 Answer: x4 − 8x3y + 24x2y2 − 32xy3 + 16y4

114

(b) (2a− b2)3 Answer: 8a3 − 12a2b2 + 6ab4 − b6

(c) (a+ b)9

Answer: a9+9a8b+36a7b2+84a6b3+126a5b4+126a4b5+84a3b6+36a2b7+9ab8 + b9

2. Compute.

(a)

(5

2

)

Answer: 10

(b)

(9

7

)

Answer: 36

(c)

(12

10

)

Answer: 66

(d)

(20

5

)

Answer: 15504

3. Prove:

(n

2

)

=n(n− 1)

2.

Answer: (n

2

)

=n!

(n− 2)!2!=

n(n− 1)(n− 2)!

(n− 2)!2!=

n(n− 1)

2

2.4.2. The Binomial Theorem

As the power n gets larger, the more laborious it would be to use Pascal’s Triangle(and impractical to use long multiplication) to expand (a + b)n. For example,using Pascal’s Triangle, we need to compute row by row up to the thirtieth rowto know the coefficients of (a+ b)30. It is, therefore, delightful to know that it ispossible to compute the terms of a binomial expansion of degree n without goingthrough the expansion of all the powers less than n.

We now explain how the concept of combination is used in the expansion of(a+ b)n.

(a+ b)n = (a+ b)(a+ b)(a+ b) · · · (a+ b)︸ ︷︷ ︸

n factors

When the distributive law is applied, the expansion of (a + b)n consists ofterms of the form ambi, where 0 ≤ m, i ≤ n. This term is obtained by choosinga for m of the factors and b for the rest of the factors. Hence, m + i = n, orm = n − i. This means that the number of times the term an−ibi will appearin the expansion of (a + b)n equals the number of ways of choosing (n − i) or i

115

factors from the n factors, which is exactly C(n, i). Therefore, we have

(a+ b)n =n∑

i=0

(n

i

)

an−ibi.

To explain the reasoning above, consider the case n = 3.

(a+ b)3 = (a+ b)(a+ b)(a+ b)

= aaa+ aab+ aba+ abb+ baa+ bab+ bba+ bbb

= a3 + 3a2b+ 3ab2 + b3

That is, each term in the expansion is obtained by choosing either a or b in eachfactor. The term a3 is obtained when a is chosen each time, while a2b is obtainedwhen a is selected 2 times, or equivalently, b is selected exactly once.

We will give another proof of this result using mathematical induction. Butfirst, we need to prove a result about combinations.

Pascal’s Identity

If n and k are positive integers with k ≤ n, then

(n+ 1

k

)

=

(n

k

)

+

(n

k − 1

)

.

Proof. The result follows from the combination formula. Teaching Notes

The formula canalso be provedusing the fact that(

n

k

)

is the numberof ways to choose kfrom n distinctobjects. Suppose ais one of the nobjects. Then, inselecting k objects,either a is selectedor not. If a isincluded in the kobjects, then thereare

(

n

k−1

)

ways to

complete theselection of the kobjects; if a is notincluded, thenthere are

(

n

k

)

ways.

(n

k

)

+

(n

k − 1

)

=n!

k!(n− k)!+

n!

(k − 1)!(n− k + 1)!

=n!(n− k + 1) + n!(k)

k!(n− k + 1)!

=n!(n− k + 1 + k)

k!(n+ 1− k)!

=n!(n+ 1)

k!(n+ 1− k)!

=(n+ 1)!

k!(n+ 1− k)!

=

(n+ 1

k

)

�

Pascal’s identity explains the method of constructing Pascal’s Triangle, inwhich an entry is obtained by adding the two numbers above it. This identityis also an essential part of the second proof of the Binomial Theorem, which wenow state.

116

The Binomial Theorem

For any positive integer n,

(a+ b)n =n∑

i=0

(n

i

)

an−ibi.

Proof. We use mathematical induction.

Part 11∑

i=0

(1

i

)

a1−ibi =

(1

0

)

a1b0 +

(1

1

)

a0b1 = a+ b

Hence, the formula is true for n = 1.

Part 2. Assume that

(a+ b)k =k∑

i=0

(k

i

)

ak−ibi.

We want to show that

(a+ b)k+1 =k+1∑

i=0

(k + 1

i

)

ak+1−ibi.

(a+ b)k+1 = (a+ b)(a+ b)k

= (a+ b)k∑

i=0

(k

i

)

ak−ibi

= a

k∑

i=0

(k

i

)

ak−ibi + b

k∑

i=0

(k

i

)

ak−ibi

=k∑

i=0

(k

i

)

ak−i+1bi +k∑

i=0

(k

i

)

ak−ibi+1

=

(k

0

)

ak+1b0 +k∑

i=1

(k

i

)

ak+1−ibi +

(k

0

)

akb1 +

(k

1

)

ak−1b2

+

(k

2

)

ak−2b3 + · · ·+

(k

k − 1

)

a1bk +

(k

k

)

a0bk+1

= ak+1 +k∑

i=1

(k

i

)

ak+1−ibi +k∑

i=1

(k

i− 1

)

ak+1−ibi + bk+1

=

(k + 1

0

)

ak+1b0 +k∑

i=1

[(k

i

)

+

(k

i− 1

)]

ak+1−ibi +

(k + 1

k + 1

)

a0bk+1

117

=k+1∑

i=0

(k + 1

i

)

ak+1−ibi

The last expression above follows from Pascal’s Identity.

Therefore, by the Principle of Mathematical Induction,

(a+ b)n =n∑

i=1

(n

i

)

an−ibi

for any positive integer n. �

2.4.3. Terms of a Binomial Expansion

We now apply the Binomial Theorem in different examples.

Example 2.4.4. Use the Binomial Theorem to expand (x+ y)6.

Solution.

(x+ y)6 =6∑

k=0

(6

k

)

x6−kyk

=

(6

0

)

x6y0 +

(6

1

)

x5y1 +

(6

2

)

x4y2

+

(6

3

)

x3y3 +

(6

4

)

x2y4 +

(6

5

)

x1y5

+

(6

6

)

x0y6

= x6 + 6x5y + 15x4y2 + 20x3y3

+ 15x2y2 + 6xy5 + y6 �

Since the expansion of (a + b)n begins with k = 0 and ends with k = n, theexpansion has n + 1 terms. The first term in the expansion is

(n0

)an = an, the

second term is(n1

)an−1b = nan=1b, the second to the last term is

(n

n−1

)abn−1 =

nabn−1, and the last term is(nn

)bn = bn.

The kth term of the expansion is(

nk−1

)an−k+1bk−1. If n is even, there is a

middle term, which is the(n2+ 1

)th term. If n is odd, there are two middle

terms, the(n+12

)th and

(n+12

+ 1)th terms.

The general term is often represented by(nk

)an−kbk. Notice that, in any term,

the sum of the exponents of a and b is n. The combination(nk

)is the coefficient

of the term involving bk. This allows us to compute any particular term withoutneeding to expand (a+ b)n and without listing all the other terms.

118

Example 2.4.5. Find the fifth term in the expansion of(2x−√y

)20.Teaching Notes

To find a specificterm in the

expansion of(a+ b)n, it is

important to findthe value of k.

Solution. The fifth term in the expansion of a fifth power corresponds to k = 4.

(20

4

)

(2x)20−4 (−√y)4 = 4845(65536x16

)y2

= 317521920x16y2 �

Example 2.4.6. Find the middle term in the expansion of(x

2+ 3y

)6

.

Solution. Since there are seven terms in the expansion, the middle term is thefourth term (k = 3), which is

(6

3

)(x

2

)3

(3y)3 = 20

(x3

8

)(27y3

)=

135x3y3

2. �

Example 2.4.7. Find the term involving x (with exponent 1) in the expansion

of

(

x2 − 2y

x

)8

.

Solution. The general term in the expansion is

(8

k

)(x2

)8−k(

−2y

x

)k

=

(8

k

)

x16−2k·(−2)kyk

xk

=

(8

k

)

(−2)kx16−2k−kyk

=

(8

k

)

(−2)kx16−3kyk.

The term involves x if the exponent of x is 1, which means 16 − 3k = 1, ork = 5. Hence, the term is

(8

5

)

(−2)5xy5 = −1792xy5. �

Seatwork/Homework 2.4.3

1. Use the Binomial Theorem to expand (2a− b2)5.

Answer:

(2a− b2

)5=

(5

0

)

(2a)5 +

(5

1

)

(2a)4b2

119

+

(5

2

)

(2a)3(b2)2

+

(5

3

)

(2a)2(b2)3

+

(5

4

)

(2a)(b2)4

+

(5

5

)(b2)5

= 32a5 − 80a4b2 + 80a3b4 − 40a2b6

+ 10ab8 − b10

2. Find the two middle terms in the expansion of

(

x1/3 +2

y

)11

.

Answer: There are 12 terms in the expansion, so the two middle terms are the6th (corresponding to k = 5) and the 7th (corresponding to k = 6) terms.

(11

5

)(x1/3

)11−5(2

y

)5

= 462x2

(32

y5

)

=14784x2

y5

(11

6

)(x1/3

)11−6(2

y

)6

= 462x5/3

(64

y6

)

=29568x5/3

y6

3. Find the constant term in the expansion of

(x3

2+

3

x2

)10

.

Answer: The general term is

(10

k

)(x3

2

)10−k (3

x2

)k

=

(10

k

)(x30−3k

210−k

)(3k

x2k

)

=

(10

k

)3k

210−kx30−5k

The constant term contains x0, which means 30− 5k = 0, or k = 6.

(10

6

)36

24x0 =

76545

8

�2.4.4. Approximation and Combination Identities

We continue applying the Binomial Theorem.

�Example 2.4.8. (1) Approximate (0.8)8 by using the first three terms in theexpansion of (1− 0.2)8. Compare your answer with the calculator value.

(2) Use 5 terms in the binomial expansion to approximate (0.8)8. Is there animprovement in the approximation?

120

Solution.

(0.8)8 = (1− 0.2)8 =8∑

k=0

(8

k

)

(1)8−k(−0.2)k

=8∑

k=0

(8

k

)

(−0.2)k

(1)2∑

k=0

(8

k

)

(−0.2)k =(8

0

)

+

(8

1

)

(−0.2) +(8

2

)

(−0.2)2

= 1− 1.6 + 1.12 = 0.52

The calculator value is 0.16777216, so the error is 0.35222784.

(2)4∑

k=0

(8

k

)

(−0.2)k =(8

0

)

+

(8

1

)

(−0.2) +(8

2

)

(−0.2)2

+

(8

3

)

(−0.2)3 +(8

4

)

(−0.2)4

= 0.52− 0.448 + 0.112 = 0.184

The error is 0.01622784, which is an improvement on the previous estimate.�

Example 2.4.9. Use the Binomial Theorem to prove that, for any positive in-teger n,

n∑

k=0

(n

k

)

= 2n.

Solution. Set a = b = 1 in the expansion of (a+ b)n. Then

2n = (1 + 1)n =n∑

k=0

(n

k

)

(1)n−k(1)k =n∑

k=0

(n

k

)

. �

Example 2.4.10. Use the Binomial Theorem to prove that(100

0

)

+

(100

2

)

+

(100

4

)

+ · · ·+

(100

100

)

=

(100

1

)

+

(100

3

)

+

(100

5

)

+ · · ·+

(100

99

)

Solution. Let a = 1 and b = −1 in the expansion of (a+ b)100. Then

[1 + (−1)

]100=

100∑

k=0

(100

k

)

(1)100−k(−1)k.

121

0 =

(100

0

)

+

(100

1

)

(−1) +(100

2

)

(−1)2 +(100

3

)

(−1)3

+ · · ·+

(100

99

)

(−1)99 +(100

100

)

(−1)100

If k is even, then (−1)k = 1. If k is odd, then (−1)k = −1. Hence, we have

0 =

(100

0

)

−(100

1

)

+

(100

2

)

−(100

3

)

+ · · ·−(100

99

)

+

(100

100

)

Therefore, after transposing the negative terms to other side of the equation, weobtain

(100

0

)

+

(100

2

)

+

(100

4

)

+ · · ·+

(100

100

)

=

(100

1

)

+

(100

3

)

+

(100

5

)

+ · · ·+

(100

99

)

�

Seatwork/Homework 2.4.4

�1. Approximate (1.9)10 using the first three terms in the expansion of(2− 0.1)10, and find its error compared to the calculator value.

Answer:

(1.9)10 = (2− 0.1)10 ≈2∑

k=0

(10

k

)

210−k(−0.1)k

= 210 − 10 · 29 · 0.1 + 45 · 28 · 0.12

= 627.2

Calculator value = 613.1066258

Error from the calculator value = 14.09337422

2. Prove that, for any positive integer n,

n∑

k=0

(n

k

)

3k = 4n.

Answer: 4n = (1 + 3)n =n∑

k=0

(n

k

)

1n−k3k =n∑

k=0

(n

k

)

3k

122

Exercises 2.4

1. Use the Binomial Theorem to expand each expression.

(a) (x− 2)5 Answer: x5 − 10x4 + 40x3 − 80x2 + 80x− 32

(b)

(

x+1

y

)7

Answer: x7 +7x6

y+

21x5

y2+

35x4

y3+

35x3

y4+

21x2

y5+

7x

y6+

1

y7

(c)

(

3− 1

2

)4

Answer: 81− 3(27)

(1

2

)

+ 3(9)

(1

4

)

− 1

8=

377

8

2. Without expanding completely, compute the indicated term(s) in the expan-sion of the given expression.

(a)

(

x3 +1

2x

)15

, first 3 terms Answer: x45 +15x41

2+

105x37

4

(b) (4− 3x)6, last 3 terms Answer: 19440x4 − 5832x5 + 729x6

(c)

(

x+3

2

)12

, 9th term Answer:3247695

256x4

(d)

(√x− 1

y

)25

, 6th term Answer: −53130x10

y5

(e)

(1

2p+

1

q

)18

, middle term Answer:12155p9

128q9

(f)

(2

a+

a2

3

)11

, two middle terms Answer:9856

81a4 +

4928

243a7

(g)(√

y + x)9, term involving y3 Answer: 84x3y3

(h)

(1

x3− 2x

)16

, constant term Answer:366080

729

(i) (xy − 2y−2)21, term that does not contain y Answer: −14883840x14

(j)

(√x

y2− y

x

)18

, term in which the exponents of x and y are equal

Answer:43758

x6y6

�3. Approximate (1.1)10 by using the first 4 terms in the expansion of(1 + 0.1)10. Compare your answer with the calculator result.

Answer: 2.57, with an error of 0.0237424601 from the calculator value of2.59374246

123

4. Use the Binomial Theorem to prove that

n∑

k=0

(n

k

)

2k = 3n.

Hint: Expand (1 + 2)n.

5. Use the Binomial Theorem to prove that

50∑

k=0

(50

k

)

(−2)k = 1.

Hint: Expand (1− 2)50.

4

124

Unit 3

Trigonometry

https://commons.wikimedia.org/wiki/File%3AUnderground River.jpg

By Giovanni G. Navata (Own work)[CC BY-SA 3.0 (http://creativecommons.org/licenses/by-sa/3.0)],

via Wikimedia Commons

Named as one of the New Seven Wonders of Nature in 2012, the PuertoPrincesa Subterranean River National Park is world-famous for its limestonekarst mountain landscape with an underground river. The Park was also listedas UNESCO World Heritage Site in 1999. The underground river stretches about8.2 km long, making it one of the world’s longest rivers of its kind.

125

Lesson 3.1. Angles in a Unit Circle

Time Frame: 3 one-hour sessions

Learning Outcomes of the Lesson

At the end of the lesson, the student is able to:

(1) illustrate the unit circle and the relationship between the linear and angularmeasures of arcs in a unit circle.

(2) convert degree measure to radian measure, and vice versa.

(3) illustrate angles in standard position and coterminal angles.

Lesson Outline

(1) Linear and angular measure of arcs

(2) Conversion of degree to radian, and vice versa

(3) Arc length and area of the sector

(4) Angle in standard position and coterminal angles

Introduction

There are many problems involving angles in several fields like engineering,medical imaging, electronics, astronomy, geography and many more. Survey-ors, pilots, landscapers, designers, soldiers, and people in many other professionsheavily use angles and trigonometry to accomplish a variety of practical tasks.In this lesson, we will deal with the basics of angle measures together with arclength and sectors.

3.1.1. Angle Measure

An angle is formed by rotating a ray about its endpoint. Teaching Notes

Angles intrigonometry differfrom angles inEuclideangeometry in thesense of motion.An angle ingeometry is definedas a union of rays(that is, static)and has measurebetween 0◦ and180◦. An angle intrigonometry is arotation of a ray,and, therefore, hasno limit. It haspositive andnegative directionsand measures.

In the figure shownbelow, the initial side of ∠AOB is OA, while its terminal side is OB. An angleis said to be positive if the ray rotates in a counterclockwise direction, and theangle is negative if it rotates in a clockwise direction.

126

An angle is in standard position if it is drawn in the xy-plane with its vertexat the origin and its initial side on the positive x-axis. The angles α, β, and θ inthe following figure are angles in standard position.

To measure angles, we use degrees, minutes, seconds, and radians.

A central angle of a circle measures one degree, written 1◦, if it inter-cepts 1

360of the circumference of the circle. One minute, written 1′, is

160

of 1◦, while one second, written 1′′, is 160

of 1′.

For example, in degrees, minutes, and seconds,

10◦30′18′′ = 10◦(

30 +18

60

)′

= 10◦30.3′

=

(

10 +30.3

60

)◦

= 10.505◦

and

79.251◦ = 79◦(0.251× 60)′

= 79◦15.06′

= 79◦15′(0.06× 60)′′

= 79◦15′3.6′′.

Recall that the unit circle is the circle with center at the origin and radius 1unit.

127

A central angle of the unit circle that intercepts an arc of the circlewith length 1 unit is said to have a measure of one radian, written 1rad. See Figure 3.1.

Figure 3.1

In trigonometry, as it was studied in Grade 9, the degree measure is often used.On the other hand, in some fields of mathematics like calculus, radian measure ofangles is preferred. Radian measure allows us to treat the trigonometric functionsas functions with the set of real numbers as domains, rather than angles.

Example 3.1.1. In the following figure, identify the terminal side of an angle instandard position with given measure.

(1) degree measure: 135◦, −135◦, −90◦, 405◦

(2) radian measure: π

4rad, −3π

4rad, 3π

2rad, −π

2rad

128

Solution. (1) 135◦:−→OC; −135◦: −−→OD; −90◦: −−→OE; and 405◦:

−−→OB

(2) radian measure: π

4rad:

−−→OB; −3π

4rad:

−−→OD; 3π

2rad:

−−→OE; and −π

2rad:

−−→OE �

Since a unit circle has circumference 2π, a central angle that measures 360◦

has measure equivalent to 2π radians. Thus, we obtain the following conversionrules.

Converting degree to radian, and vice versa

1. To convert a degree measure to radian, multiply it by π

180.

2. To convert a radian measure to degree, multiply it by 180π.

Figure 3.2 shows some special angles in standard position with the indicatedterminal sides. The degree and radian measures are also given.

Figure 3.2

129

Example 3.1.2. Express 75◦ and 240◦ in radians.

Solution.

75( π

180

)

=5π

12=⇒ 75◦ =

5π

12rad

240( π

180

)

=4π

3=⇒ 240◦ =

4π

3rad �

Example 3.1.3. Express π

8rad and 11π

6rad in degrees.

Solution.π

8

(180

π

)

= 22.5 =⇒ π

8rad = 22.5◦

11π

6

(180

π

)

= 330 =⇒ 11π

6rad = 330◦ �

Seatwork/Homework 3.1.1

1. Convert the following degree measures to radian measure.

(a) 60◦ Answer: π

3rad

(b) 90◦ Answer: π

2rad

(c) 150◦ Answer: 5π6rad

2. Convert the following radian measures to degree measure.

(a) π

9rad Answer: 20◦

(b) 3π4rad Answer: 135◦

3.1.2. Coterminal Angles

Two angles in standard position that have a common terminal side are calledcoterminal angles. Observe that the degree measures of coterminal angles differby multiples of 360◦.

Two angles are coterminal if and only if their degree measures differby 360k, where k ∈ Z.

Similarly, two angles are coterminal if and only if their radian mea-sures differ by 2πk, where k ∈ Z.

130

As a quick illustration, to find one coterminal angle with an angle that mea-sures 410◦, just subtract 360◦, resulting in 50◦. See Figure 3.3.

Figure 3.3

Example 3.1.4. Find the angle coterminal with −380◦ that has measure

(1) between 0◦ and 360◦, and

(2) between −360◦ and 0◦.

Solution. A negative angle moves in a clockwise direction, and the angle −380◦lies in Quadrant IV.

(1) −380◦ + 2 · 360◦ = 340◦

(2) −380◦ + 360◦ = −20◦ �

Seatwork/Homework 3.1.2

1. Find the angle between 0◦ and 360◦ (if in degrees) or between 0 rad and 2π rad(if in radians) that is coterminal with the given angle.

(a) 736◦ Answer: 16◦

(b) −28◦48′65′′ Answer: 331◦10′55′′

(c) 13π2

rad Answer: π

2rad

�(d) 10 rad Answer: 3.72 rad

2. Find the angle between −360◦ and 0◦ (if in degrees) or between −2π rad and0 rad (if in radians) that is coterminal with the given angle.

131

(a) 142◦ Answer: −218◦

(b) −400◦1′23′′ Answer: −40◦1′23′′

(c) π

6rad Answer: −11π

6rad

�(d) −20 rad Answer: −1.15 rad

3.1.3. Arc Length and Area of a Sector

In a circle, a central angle whose radian measure is θ subtends an arc that is thefraction θ

2πof the circumference of the circle. Teaching Notes

Review how arcswere measured inGrade 10. Whatunit of measurewas used? For twocircles withdifferent radii, doequal centralangles interceptarcs of the samemeasure?Conclude thatprevious notion ofarc measure is notthe same as length.Arcs are nowmeasured in termsof length andmeasure changeswith the radius ofthe circle.

Thus, in a circle of radius r (seeFigure 3.4), the length s of an arc that subtends the angle θ is

s =θ

2π× circumference of circle =

θ

2π(2πr) = rθ.

Figure 3.4

In a circle of radius r, the length s of an arc intercepted by a centralangle with measure θ radians is given by

s = rθ.

Example 3.1.5. Find the length of an arc of a circle with radius 10 m thatsubtends a central angle of 30◦.

Solution. Since the given central angle is in degrees, we have to convert it intoradian measure. Then apply the formula for an arc length.

30( π

180

)

=π

6rad

s = 10(π

6

)

=5π

3m �

132

Example 3.1.6. A central angle θ in a circle of radius 4 m is subtended by anarc of length 6 m. Find the measure of θ in radians.

Solution.

θ =s

r=

6

4=

3

2rad �

A sector of a circle is the portion of the interior of a circle bounded by theinitial and terminal sides of a central angle and its intercepted arc. It is like a“slice of pizza.” Note that an angle with measure 2π radians will define a sectorthat corresponds to the whole “pizza.” Therefore, if a central angle of a sectorhas measure θ radians, then the sector makes up the fraction θ

2πof a complete

circle. See Figure 3.5. Since the area of a complete circle with radius r is πr2, wehave

Area of a sector =θ

2π(πr2) =

1

2θr2.

Figure 3.5

In a circle of radius r, the area A of a sector with a central anglemeasuring θ radians is

A =1

2r2θ.

Example 3.1.7. Find the area of a sector of a circle with central angle 60◦ ifthe radius of the circle is 3 m.

Solution. First, we have to convert 60◦ into radians. Then apply the formula forcomputing the area of a sector.

60( π

180

)

=π

3rad

A =1

2(32)

π

3=

3π

2m2

�

133

Example 3.1.8. A sprinkler on a golf course fairway is set to spray water overa distance of 70 feet and rotates through an angle of 120◦. Find the area of thefairway watered by the sprinkler.

Solution.

120( π

180

)

=2π

3rad

A =1

2(702)

2π

3=

4900π

3≈ 5131 ft2 �

Seatwork/Homework 3.1.3

1. In a circle of radius 7 feet, find the length of the arc that subtends a centralangle of 5 radians. Answer: 35 ft

2. A central angle θ in a circle of radius 20 m is subtended by an arc of length15π m. Find the measure of θ in degrees. Answer: 135◦

3. Find the area of a sector of a circle with central angle that measures 75◦ if theradius of the circle is 6 m. Answer: 7.5 m2

Exercises 3.1

1. Give the degree/radian measure of the following special angles.

134

2. Convert each degree measure to radians. Leave answers in terms of π.

(a) 330◦ Answer: 11π6

rad

(b) 480◦ Answer: 8π3rad

(c) 15◦ Answer: π

12rad

(d) 105◦ Answer: 7π12

rad

(e) 265◦ Answer: 53π36

rad

(f) −120◦ Answer: −2π3rad

(g) −315◦ Answer: −7π4rad

3. Convert each radian measure to degree-minute-second measure (approximateif necessary).

(a) 5π6rad Answer: 150◦

(b) 8π3rad Answer: 480◦

(c) 15π4

rad Answer: 675◦

(d) −π

6rad Answer: −30◦

(e) −7π20

rad Answer: −63◦�(f) 20 rad Answer: 1145◦54′56.12′′

�(g) −35 rad Answer: −2005◦21′8.22′′�(h) −5 rad Answer: −286◦28′44.03′′

4. Find the angle between 0◦ and 360◦ (if in degrees) or between 0 rad and 2π rad(if in radians) that is coterminal with the given angle.

(a) 685◦ Answer: 325◦

(b) 451◦ Answer: 91◦

(c) −1400◦ Answer: 40◦

(d) 960◦45′34′′ Answer: 240◦45′34′′

(e) −728◦15′43′′ Answer: 352◦15′43′′

(f) 29π6

rad Answer: 5π6rad

(g) −3π2rad Answer: π

2rad

�(h) 16 rad Answer: 3.43 rad

�(i) −20 rad Answer: 5.13 rad

5. Find the angle between −360◦ and 0◦ (if in degrees) or between −2π rad and0 rad (if in radians) that is coterminal with the given angle.

135

(a) 685◦ Answer: −35◦

(b) 451◦ Answer: −269◦

(c) −1400◦ Answer: −320◦

(d) 960◦45′34′′ Answer: −120◦45′34′′

(e) −728◦15′43′′ Answer: −8◦15′43′′

(f) 29π6

rad Answer: −7π6rad

(g) −3π2rad Answer: −3π

2rad

�(h) 16 rad Answer: ≈ −2.850 rad

�(i) −20 rad Answer: ≈ −1.150 rad

6. Find the length of an arc of a circle with radius 21 m that subtends a centralangle of 15◦. Answer: 7π

4m

7. A central angle θ in a circle of radius 9 m is subtended by an arc of length 12m. Find the measure of θ in radians. Answer: 4

3rad

8. Find the radius of a circle in which a central angle of π

6rad determines a sector

of area 64 m2. Answer: 16 m

9. If the radius of a circle is doubled, how is the length of the arc intercepted bya fixed central angle changed? Answer: The length is doubled.

10. Radian measure simplifies many formulas, such as the formula for arc length,s = rθ. Give the corresponding formula when θ is measured in degrees insteadof radians. Answer: s = πrθ

180

�11. As shown below, find the radius of the pulley if a rotation of 51.6◦ raises theweight by 11.4 cm. Answer: 12.7 cm

�12. How many inches will the weight rise if the pulley whose radius is 9.27 inchesis rotated through an angle of 71◦50′? Answer: 11.6 in

136

�13. Continuing with the previous item, through what angle (to the nearest minute)must the pulley be rotated to raise the weight 6 in? Answer: 37◦5′

�14. Given a circle of radius 3 in, find the measure (in radians) of the central angleof a sector of area 16 in2. Answer: 3.6 rad

�15. An automatic lawn sprinkler sprays up to a distance of 20 feet while rotating30◦. What is the area of the sector the sprinkler covers? Answer: 104.72 ft2

�16. A jeepney has a windshield wiper on the driver’s side that has total arm andblade 10 inches long and rotates back and forth through an angle of 95◦. Theshaded region in the figure is the portion of the windshield cleaned by the7-inch wiper blade. What is the area of the region cleaned? Answer: 75.4 in2

17. If the radius of a circle is doubled and the central angle of a sector is unchanged,how is the area of the sector changed? Answer: The area is quadrupled.

18. Give the corresponding formula for the area of a sector when the angle ismeasured in degrees. Answer: A = πr2θ

360

�19. A frequent problem in surveying city lots and rural lands adjacent to curvesof highways and railways is that of finding the area when one or more of theboundary lines is the arc of a circle. Approximate the total area of the lotshown in the figure. Answer: 1909.0 m2

137

20. Two gears of radii 2.5 cm and 4.8 cm are adjusted so that the smaller geardrives the larger one, as shown. If the smaller gear rotates counterclockwisethrough 225◦, through how many degrees will the larger gear rotate?

Answer: 117◦

4

Lesson 3.2. Circular Functions

Time Frame: 2 one-hour sessions

Learning Outcomes of the Lesson

At the end of the lesson, the student is able to:

(1) illustrate the different circular functions; and

(2) use reference angles to find exact values of circular functions.

Lesson Outline

(1) Circular functions

(2) Reference angles

Introduction

We define the six trigonometric function Teaching Notes

The teacher cangive a review oftrigonometricratios as discussedin Grade 9.

in such a way that the domain ofeach function is the set of angles in standard position. The angles are measuredeither in degrees or radians. In this lesson, we will modify these trigonometricfunctions so that the domain will be real numbers rather than set of angles.

138

3.2.1. Circular Functions on Real Numbers

Recall that the sine and cosine functions (and four others: tangent, cosecant,secant, and cotangent) of angles measuring between 0◦ and 90◦ were defined inthe last quarter of Grade 9 as ratios of sides of a right triangle. It can be verifiedthat these definitions are special cases of the following definition.

Let θ be an angle in standard position and P (θ) = P (x, y) the pointon its terminal side on the unit circle. Define

sin θ = y csc θ =1

y, y "= 0

cos θ = x sec θ =1

x, x "= 0

tan θ =y

x, x "= 0 cot θ =

x

y, y "= 0

Example 3.2.1. Find the values of cos 135◦, tan 135◦, sin(−60◦), and sec(−60◦).

Solution. Refer to Figure 3.6(a).

(a) (b)

Figure 3.6

From propertiesTeaching Notes

A 45◦-45◦ righttriangle is isosceles.

Moreover, theopposite side of the

30◦-angle in a30◦-60◦ right

triangle is half thelength of itshypotenuse.

of 45◦-45◦ and 30◦-60◦ right triangles (with hypotenuse 1unit), we obtain the lengths of the legs as in Figure 3.6(b). Thus, the coordinatesof A and B are

A =

(

−√2

2,

√2

2

)

and B =

(

1

2,−√3

2

)

.

139

Therefore, we get

cos 135◦ = −√2

2, tan 135◦ = −1,

sin(−60◦) = −√3

2, and sec(−60◦) = 2. �

From the last example, we may then also say that

cos(π

4rad

)

=

√2

2, sin

(

−π3rad

)

= −√3

2,

and so on.

From the above definitions, we define the same six functions on real numbers.These functions are called trigonometric functions.

Let s be any real number. Suppose θ is the angle in standard positionwith measure s rad. Then we define

sin s = sin θ csc s = csc θ

cos s = cos θ sec s = sec θ

tan s = tan θ cot s = cot θ

From the last example, we then have

cos(π

4

)

= cos(π

4rad

)

= cos 45◦ =

√2

2

and

sin(

−π3

)

= sin(

−π3rad

)

= sin(−60◦) = −√3

2.

In the same way, we have

tan 0 = tan(0 rad) = tan 0◦ = 0.

Example 3.2.2. Find the exact values of sin 3π2, cos 3π

2, and tan 3π

2.

Solution. Let P(3π2

)be the point on the unit circle and on the terminal side of

the angle in the standard position with measure 3π2rad. Then P

(3π2

)= (0,−1),

and so

sin3π

2= −1, cos

3π

2= 0,

but tan 3π2is undefined. �

140

Example 3.2.3. Suppose s is a real number such that sin s = −34and cos s > 0.

Find cos s.

Solution. We may consider s as the angle with measure s rad. Let P (s) = (x, y)be the point on the unit circle and on the terminal side of angle s.

Since P (s) is on the unit circle, we know that x2 + y2 = 1. Since sin s = y =−3

4, we get

x2 = 1− y2 = 1−(

−3

4

)2

=7

16=⇒ x = ±

√7

4.

Since cos s = x > 0, we have cos s =√74. �

Let P (x1, y1) and Q(x, y) be points on the terminal side of an angle θ instandard position, where P is on the unit circle and Q on the circle of radius r(not necessarily 1) with center also at the origin, as shown above. Observe thatwe can use similar triangles to obtain

cos θ = x1 =x1

1=

x

rand sin θ = y1 =

y11

=y

r.

We may then further generalize the definitions of the six circular functions.

141

Let θ be an angle in standard position, Q(x, y) any point on the ter-minal side of θ, and r =

√

x2 + y2 > 0. Then

sin θ =y

rcsc θ =

r

y, y "= 0

cos θ =x

rsec θ =

r

x, x "= 0

tan θ =y

x, x "= 0 cot θ =

x

y, y "= 0

We then have a second solution for Example 3.2.3 as follows. With sin s = −34

and sin s = yr, we may choose y = −3 and r = 4 (which is always positive). In

this case, we can solve for x, which is positive since cos s = x4is given to be

positive.

4 =√

x2 + (−3)2 =⇒ x =√7 =⇒ cos s =

√7

4

Seatwork/Homework 3.2.1

1. Given θ, find the exact values of the six circular functions.

(a) θ = 30◦

Answer: sin 30◦ = 12, cos 30◦ =

√32, tan 30◦ =

√33, csc 30◦ = 2, sec 30◦ =

2√3

3, cot 30◦ =

√3

(b) θ = 3π4

Answer: sin 3π4=

√22, cos 3π

4= −

√22, tan 3π

4= −1, csc 3π

4=√2, sec 3π

4=

−√2, cot 3π

4= −1

(c) θ = −150◦

Answer: sin(−150◦) = −12, cos(−150◦) = −

√32, tan(−150◦) =

√33, csc(−150◦) =

−2, sec(−150◦) = −2√3

3, cot(−150◦) =

√3

(d) θ = −4π3

Answer: sin(−4π3) =

√32, cos(−4π

3) = −1

2, tan(−4π

3) = −

√3, csc(−4π

3) =

2√3

3, sec(−4π

3) = −2, cot(−4π

3) = −

√33

2. Given a value of one circular function and sign of another function (or thequadrant where the angle lies), find the value of the indicated function.

(a) sin θ = 12, θ in QI; cos θ Answer:

√32

(b) cos θ = 35, θ in QIV; csc θ Answer: 5

4

(c) sin θ = −37, sec θ < 0; tan θ Answer: 3

√10

20

(d) cot θ = −29, cos θ > 0; csc θ Answer: −

√859

142

3.2.2. Reference Angle

We observe that if θ1 and θ2 are coterminal angles, the values of the six circularor trigonometric functions at θ1 agree with the values at θ2. Therefore, in findingthe value of a circular function at a number θ, we can always reduce θ to a numberbetween 0 and 2π. For example, sin 14π

3= sin

(14π3− 4π

)= sin 2π

3. Also, observe

from Figure 3.7 that sin 2π3= sin π

3.

Figure 3.7

In general, if θ1, θ2, θ3, and θ4 are as shown in Figure 3.8 with P (θ1) =(x1, y1), then each of the x-coordinates of P (θ2), P (θ3), and P (θ4) is ±x1, whilethe y-coordinate is ±y1. The correct sign is determined by the location of theangle. Therefore, together with the correct sign, the value of a particular circularfunction at an angle θ can be determined by its value at an angle θ1 with radianmeasure between 0 and π

2. The angle θ1 is called the reference angle of θ.

Figure 3.8

143

The signs of the coordinates of P (θ) depends on the quadrant or axis whereit terminates. It is important to know the sign of each circular function in eachquadrant. See Figure 3.9. It is not necessary to memorize the table, since thesign of each function for each quadrant is easily determined from its definition.We note that the signs of cosecant, secant, and cotangent are the same as sine,cosine, and tangent, respectively.

Figure 3.9

Using the fact that the unit circle is symmetric with respect to the x-axis, they-axis, and the origin, we can identify the coordinates of all the points using thecoordinates of corresponding points in the Quadrant I, as shown in Figure 3.10for the special angles.

Figure 3.10

144