Math 120
"Explanation Exam" Prep #1
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SOLUTION:
The answers are in bold, below:
1) Given Sin
θ = 135 , where θ is in the second quadrant,
find all the other trig functions using definitions:
1. First, we draw out a triangle, and fill in the sides we know:
(It's good, but not mandatory to draw it like this, on the x & y axes, so stuff is clearer later on)
2. We want to use the Pythagorean Theorem, so we write out the formula, and explain it:
a 2  b2  c 2
a is the line segment between A & C (length is 13)
b is the line segment between B & C (length is 5)
c is the line segment between A & B (length is not yet known)
3. Substitute values:
a 2  52  132
4. Square the two numbers:
a 2  25  169
5. Simplify (Subtraction):
a 2  169  25
6. Simplify (Subtraction):
a 2  144
7. Square root of each side:
8. Simplify (Do the radical)
a 2  144
a  12
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Math 120
"Explanation Exam" Prep #1
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9. Fill in the missing side:
(It's ok to put the missing side onto the same triangle as you drew in step 1)
10. At this point, you need to either:
a. Write out (in English) a quick explanation about why its actually a negative 12
("Because the angle is in the second quadrant, and we need to go in the
negative X direction in order to get to point C, the 12 is negative"), or else
b. Draw the triangle on an XY coordinate system (clearly labeling the axes), and
list 12 as being negative.
11. Optionally, write out the definitions of the Trig functions.
12. Since the answer is provided for you, you're done!
5
13
13
Csc θ =
5
Sin θ =
Answers:
12
Cos θ = 
13
13
Sec θ = 
12
5
12
12
Cot θ = 
5
Tan θ = 
5
, where θ is in the second quadrant,
13
with find all the other trig functions using trig identities:
2) Given Sin θ =
There are 5 different functions you need to find. You can do them in any order, and you
don't need to do them exactly like this (i.e., there are multiple, valid, ways you can approach
these problems)
<You're given Sin, so there's no explanation needed >
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Math 120
"Explanation Exam" Prep #1
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CSC:
1
 Csc
Sin
1
 Csc
2. Substitute the known value in: 5
13
1. Reciprocal Identity:
13
5  Csc
3. Multiply numerator and denominator by the same value: 5 3

13 5
13
1
5  Csc
4. Simplify (denominator goes to 1):
1
1
Simplify (denominator of 1 can be ignored)
(This step may be combined with the previous one):
13
 Csc
5
COS:
1. Pythagorean Identity: Cos2θ + Sin2θ = 1
2. Substitute the known value in:
3. Do the exponent: Cos2θ +
Cos2θ
5
+ 
 13 
2
=1
25
=1
169
4. Simplify (Subtraction): Cos2θ = 1 -
25
169
5. Simplify (Common denominator, for subtraction): Cos2θ =
6. Simplify (Subtraction): Cos2θ =
7. Square root of each side:
144
169
cos 2  
8. Simplify (Do the radical) cos  
169 25

169 169
144
169
144
169
12
9. Simplify (Do the radical) cos  
13
10. Explain (Via picture or English sentence) that being in Quadrant II means that the
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Math 120
"Explanation Exam" Prep #1
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cosine is negative
SEC:
1
 Sec
Cos
1
2. Substitute the known value in:  12  Sec
13
1. Reciprocal Identity:
13
12  Sec
3. Multiply numerator and denominator by the same value: 12
3
 
13
12
13
1 
12  Sec
4. Simplify (denominator goes to 1):
1
1 
Simplify (denominator of 1 can be ignored)
(This step may be combined with the previous one):
TAN:
1. Quotient Identity: Tan 

13
 Sec
12
Sin 
Cos
5
13
2. (We already know Sinθ, Cosθ, so plug those in: Tan  12

13
5
13

13 12
3. Multiply numerator and denominator by the same value: Tan  12
13
 
13 12
4. Simplify: Denominator cancels to 1, and is ignored, cross-cancel on top: Tan  
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5
12
Math 120
"Explanation Exam" Prep #1
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COT:
1. Reciprocal Identity:
1
 Tan
Cot
2. Multiply each side by Cotθ, and by
1
1
 Cot
, in order to get:
Tan
Tan
1
3. Substitute the known value in:  5  Cot
12
5
12  Cot
4. Multiply numerator and denominator by the same value:
5
2
 
12
5
5
1 
12  Cot
5. Simplify (denominator goes to 1):
1
1 
Simplify (denominator of 1 can be ignored)
(This step may be combined with the previous one):
5
Sin θ =
13
13
Csc θ =
5
Answers:
12
Cos θ = 
13
13
Sec θ = 
12
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
5
 Cot
12
5
12
12
Cot θ = 
5
Tan θ = 
Math 120
"Explanation Exam" Prep #1
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3)
You're sitting in class one day, and you overhear a classmate, Bob, explaining a problem to someone else. What you hear is the following:
"Tangent of 30 degrees? Without using a calculator? No problem. You just draw out your handy-dandy 30-60-90 triangle like this, and notice that
the side adjacent to 30° is
3 , the opposite is 1, and the hypotenuse is 2.
but where's the fun in that? Let's use the fact that tan =
hypotenuse
adjacent
hypotenuse
cos
, and that Sin  
sin
opposite
2
2
(which is
We could just use the definition of tan, namely that tan =
). Plug those into the equation for tan, get
3
3
2
1
(which is
, which simplifies to
1
3
2
),
1
adjacent
opposite
,
and that cos θ =
- Voila!" List all the mistakes that Bob
has made:
(These may be listed in any order)
Opposite
1. Tangent is
Adjacent
Opposite
2. Sine is
Hypotenuse
3. Cosine is
4. Tan is
Adjacent
Hypotenuse
Sin
Cos
Explain why Bob still got the right answer:
Basically, Bob's mistakes w/ Sin & Cos 'canceled out' the mistakes with Tan:
cos
Some math might illustrate this better: According to Bob, tan =
. Plugging in what Bob
sin
2
thinks is cos & sin, we get tan 
by ½, and get
1
3
3
. Ignore the 1 in the denominator, and multiply top & bottom
2
1
. This just happens to be the same values we get if we plug in the correct values
1
3
1
sin
of sin ( ) and cos (
) into the correct identity for tan (
, or 2 ), which then simplifies to
2
cos
2
3
2
1 2

2
3 , or 1 .
3
3 2

2
3
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