Exam Rehearsal for Identities and Counting

Hey people! It's JessiccaI_198 as scribe today. Today we had two exam rehearsals; one for identities in the morning and the other one was for counting in the afternoon. We talked a bit about the arc sine and about functions, but I didn't really get many notes on that. But here are the questions and solutions for the Exam Rehearsal Identities:

1) The expression (sinx + cosx)^2 - 1 is equivalent to:

a) sin2x

b) cos2x

c) tan^2x

d) 0

The answer is a) sin2x.

Solution:

sin^2x + 2sinxcosx + cos^2x - 1

2sinxcosx

sin2x

2) If sinA = 3/5 and sinB = 2/3, and A and B are acute angles, what is the value of cos(A - B)?

a) -2/3

b) (4√5 - 6)/15

c) (4√5 + 2)/5

d) (4√5 + 6)/15

The answer is B

Solution:

cos(A - B) = cosAcosB + sinAsinB

= (4/5)(√5/3) + (4/5)(√5/3)

= (4√5 +6)/15

3) Find all values of x in the interval 0 < x < cos2x =" -3sinx">

The answer is x = 7π/6, 11π/6

Solution:

cos2x = -3sinx - 1

1 - 2sin^2x = -3sinx - 1

0 = 2sin^2x - 3sinx - 1

= (2sinx + 1) (sinx - 2)

sinx = -1/2 sinx = 2

2 is rejected because in the unit circle, no value can be bigger than 1.

x = 7π/6, 11π/6

4) If sinΘ = √5/3, what is the value of cos2Θ?

The answer is -1/9

Solution:

cos2Θ = 1 - 2sin^2Θ

= 1 - 2(5/9)

= 1 - 10/9

= 9/9 - 10/9

= -1/9

5) Prove the identity: sinΘ/(1-cosΘ) + sinΘ/(1+cosΘ) = 2cscΘ

Here are the problems from this afternoons class on counting:

1) The number of different arrangements of 3 boys and 4 girls in a row, if the girls must stand together, is represented by:

a) 4!x4!

b) 3!x4!

c) 4!x4!x2!

d) 3!x4!x2!

The answer is a.

2) The students in a music department have practised 6 contemporary and 5 traditional choruses. For their concert, they will choose a program in which they present 4 of the contemporary and 3 of the traditional choruses. How many different programs can be presented, if the order of the choruses does not matter?



a) 25


b) 35


c) 150


d) 330

The answer is c.

Solution:

6C4 . 5C3

= 150

3) All telephone numbers are preceded by a 3-digit area code. In the original Bell Telephone System of assigning area codes, the first digit could be any number from 2 to 9, the second digit was either 0 or 1, and the third digit could be any number except 0. In this system, the number of different are codes possible was:

The answer is 144 different ways.

Solution:

8C1 . 2C1 . 9C1 = 144 different ways

8C1 - there are eight numbers in the first digit - 2, 3, 4, 5, 6, 7, 8, and 9.

2C1 - two number in the second digit - 0 and 1

9C1 - all the numbers, excluding 0







4) A paperboy who delivers papers on his bike can travel only on the trails represented in the diagram. The number of different trails that the paperboy can take to get from house A to house B without backtracking is:






Solution:
There are 60 different trails






5. a) How many groups of 3 chairs can be chosen from 7 chairs if the chairs are all different colours?



b) How many different ways can 7 chairs be arranges in a row if 2 of the chair are blue, 3 are yellow, 1 is red and 1 is green? (Assume that all of the chairs are identical except for colour.)



c) How many different ways can the chairs in (b) above be arranged in a circle?

5. Answers:
a) 35
b) 420
c) 60

Solution:
5a) 7C3 = 35
- since order doesn't matter. 7 is for the total number of chairs, and 3 is for the number of chairs to be chosen.

b) 7! / 3!2! = 420
- 7! represents the total number of arrangements, and 3! 2! represent the non-distinguishable objects.

c) 6! / 3!2! = 60
- one of the chairs is a reference point, and the rest are seated in a circle. 3!2! are the non-distinguishable objects.