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Semester Review
FOR
Trigonometry
Zohra Noor Hafeez
Chapter 1&2
Chapter 2-2: Solving Systems of Equations in Three
Variables
Example
1. Solve the System of Equations
X – 2y + z = 15
2x +3y -3z = 1
4x +10y – 5z = -3
-2(x-2y+z) = -2(15) →
Get rid of the x by multiplying the first
equation by -2.
-2x + 4y – 2z = -30
-2x +4y – 2z = -30
2x + 3y – 3z = 1
Add the result to the second equation and
add the like terms.
7y – 5z = -29
-4(x – 2y + z) = -4(15) →
-4x + 8y -4z = -60
-4x +8y – 4z = -60
4x +10y – 5z = -3
18y – 9z = -63
Once again, get rid of x with the first and
third equation by multiplying -4 to the first
equation.
Add together the first and third equations.
Add together the two results to solve for one
variable, in this case y.
-9[7y – 5z = -29] → -63y + 45z = 261
5[18y – 9z = -63] → 90y - 45z = -315
27y
= -54
Y=2
7(2) -5z = -29 → 14 – 5z = -26
Substitute y into one of the 2nd
derived equations to solve for z
X – 2(2) + (3) = 15
Z=3
Plug in all the known variables into
the original equation to find x.
X=8
Common Mistake: Don’t get all the variables mixed up by choosing two variables to eliminate first;
you’re cancelling out one variable each time.
Chapter 2-6: Solving Systems of Linear Inequalities
2. Find the maximum and minimum values of f(x,y) = x – y +2 for the polygonal convex set
determined by the system of inequalities.
X +4y ≤ 12
3x – 2y ≥ -6
x + y ≥ -2
3x – y ≤ 10
1. Write the inequalities in slope-intercept form
2. Graph the inequalities and find the coordinates of the vertices
3. Evaluate the function by plugging in the vertices coordinates into the original
equation
4. The highest number evaluated will be the maximum and the lowest number
evaluated will be the minimum.
Common Mistake : don’t forget the dotted lines or solid lines from the inequalities.
Chapter 3
Chapter 3-2: Families of Graphs
Example
1. Describe how the graphs of f(x) and g(x) are related
F(x) = x
G(x) = x + 6
The original graph would be a horizontal line through the origin (0,0). The second graph would
move 6 units up on the y-axis because it is the “c” and the “c” shifts the graph vertically.
Common Mistake: Students often mistake the “c” shifting the graph horizontally but that is the variable
in the parenthesis with the “x.”
Chapter 3-7: The Nature of Graphs
2.
Determine the slant asymptote for f(x) =
𝟐𝒙𝟐 − 𝟑𝒙+𝟏
.
𝒙−𝟐
a. Use long division to rewrite the function
3
X – 2 / 2x2 – 3x + 1 → f(x) = 2x + 1 + 𝑥−2
3
b. As x approaches ∞, 𝑥−2 approaches 0. Y = 2x + 1 is the slant asymptote, verified by
graphing the function. The graph was not pasting correctly so it cannot be shown.
Slant asymptote- the oblique line for a function if the graph approaches x→∞ or as
x→-∞.
Common Mistake : When using long division, it often occurs that the variables are not listed in
chronological order for the exponents, which causes inaccurate information.
Chapter 4
Chapter 4-3 The Remainder and Factor Theorems:
Synthetic Division – a shortcut for dividing a polynomial by a binomial.







Step 1: arrange the terms of the polynomial in descending powers of x. Insert zeros for
any missing powers of x. Then write the coefficients
Step 2: Write the constant r of the divisior.
Step 3: bring down the first coefficient
Step 4: multiply the first coefficient by r. Then write the product under the next
coefficient. Add.
Step 5: Multiply the sum by r. Then write the product under the next coefficients. Add.
Step 6: Repeat Step 3 for all coefficients in the dividend
Step 7: The final sum represents the remainder.
Example 1
Divide x3 – x2 + 2 by x + 1 using synthetic division.
−1⌋
1
1
-1
0
2
-1
2
-2
-2
2
0
The quotient is x2 – 2x + 2 w/ a remainder of 0.
Common Mistake: Remember when dividing to bring down the first term because the position serves
as a placeholder.
Chapter 4-7 Radical Equations and Inequalities
Example 2:
Solve √𝑥 + 10 = 5 – √3 − 𝑥.
√𝑥 + 10 = 5 – √3 − 𝑥.
Square each side.
X + 10 = 25 - 10√3 − 𝑥 + 3 – x
Isolate the radical
2x – 18 = -10√3 − 𝑥
Square each side
4x2 – 72x + 324 = 100(3-x)
4x2 – 72x + 324 = 300 – 100x
4x2 + 28x + 24 = 0
X2 + 7x + 6 = 0
(x+6)(x+1) = 0
X = -6
x = -1
Common Mistake : When squaring radicals that are polynomials, remember to FOIL the entire
polynomial instead of getting rid of the radical and squaring the number.
Chapter 5
Chapter 5-6: The law of Sines
The law of sines can be used to solve triangles that are not right triangles.
The law of Sines:
𝒂
𝒔𝒊𝒏 𝑨
𝒃
𝒄
= 𝒔𝒊𝒏 𝑩 = 𝒔𝒊𝒏 𝑪
Example 1:
Solve ΔABC if A = 33˚, B = 105˚, and b = 37.9˚
a
b
sin A
𝒂
𝒔𝒊𝒏 𝟑𝟑˚
a=
c
= sin B
𝟑𝟕.𝟗
= sin B
𝒃𝒄
= 𝒔𝒊𝒏 𝟏𝟎𝟓˚
37.9 sin 33
sin 105
b
sin C
𝒔𝒊𝒏 𝟒𝟐
a = 21. 37
c=
𝟑𝟕.𝟗
= 𝒔𝒊𝒏 𝟏𝟎𝟓
37.9 sin 42
sin 105
c = 26.25
Common Mistake: When the case for the Law of sines can be ambiguous, remember to check all the
possible solutions.
Chapter 5-8: The law of Cosines
The law of cosines:
A2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2bc cos C
Example 2: Solve the triangle
A = 120˚, b = 9, c = 5
A2 = b2 + c2 – 2bc cos A
a2 = 151
a2 = 92 + 52 – 2(9)(5) cos A
a = 12.29
𝑎
𝑏
=
sin 𝐴 sin 𝐵
12.3
sin 120
=
9
Sin B =
sin 𝐵
9 sin 120
12.3
B = 39.3
C = 180 – (39.3 + 120)
C = 20.7
a= 12.29 B= 39.9 C= 20.7
Common Mistake: Don’t mix up the variables for the sides with the variables for the angles and
don’t forget to use law of sines as well.
Chapter 6
Chapter 6-4 Amplitude and Period of Sine and Cosine
Functions
Y = A sin 𝜽
The absolute value of A is the amplitude of the functions y = A sin θ and y = A cos θ
The period is obtained by simply dividing 2𝜋 by B.
Example: State the amplitude and period or the function y = ½ sin θ. Then graph the function
Amplitude = ½
T = 2 𝜋/ B =
2𝜋
1
=2𝜋
Graph:
Common Mistake: when listing the increments, don’t forget that the graph starts with the
horizontal shift and ends by adding the period. Also, don’t forget to add the vertical shifts when
they’re there.
Chapter 6-8: Trigonometric Inverses and their Graphs
Example 2: find the value:
Arcsin(−
√2
2
Sin θ = (−
)
You’re trying to find the angle that (−
√2
2
)
√2
2
) forms.
Look at the Unit Circle to find the angle of (−
√2
2
)
𝜋
Θ=−4
Common mistake: sometimes I put the angle degree for the coterminal instead of the actual
answer. All of the answers are in the first quadrant
Chapter 7
Chapter 7-1 Basic Trigonometric Identities:
Sin2 θ + cos2 θ = 1
tan2 θ + 1 = sec2 θ
1 + cot2 θ = csc2 θ
Example 1: Use the given information to find the trigonometric value
If sec θ = 3/2, find cos θ
cos θ = 1/ sec θ
1
cos θ = 2/3 = 3/2.
Common Mistake: remember to use the bread and butter equations to derive other ones to
make the problem easier instead of plugging the givens into tangent squared plus 1.
Chapter 7-3 Sum and Difference Identities
Cos (𝛼 ± 𝛽) = cos 𝛼 cos 𝛽 ∓ sin 𝛼sin 𝛽
Sin (𝛼 ± 𝛽) = sin 𝛼 sin 𝛽 ± cos 𝛼cos 𝛽
𝑡𝑎𝑛𝛼±𝑡𝑎𝑛𝛽
Tan (𝛼 ± 𝛽) = 1∓𝑡𝑎𝑛𝛼𝑡𝑎𝑛𝛽
Example 2: use the sum or difference identity for tangent to find the exact value of tan 285˚
Tan 285˚ = tan(240 = 45)
𝑡𝑎𝑛240±𝑡𝑎𝑛45
= 1∓𝑡𝑎𝑛240𝑡𝑎𝑛45
=
√3+ 1
1−(√3)(1)
= -2 - √3
Common Mistake: find the common angles to use for the 𝛼 𝑎𝑛𝑑 𝛽.
Chapter 7-1 Basic Trigonometric Identities:
Example 2
Use the half angle identity to find the exact value of each function
7𝜋
Sin 12
7𝜋
Sin 12 = sin
= √1 − cos
=√1 − (−
2
= √2 +
7𝜋
6
2
7𝜋
6
/2
√3
)/2
2
√3
2
Common mistake: forgetting to divide the given by two and using that cosine.