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CUmath
QUIZ 6 – detailed solutions
1) Convert 5 / 3 radians to degrees.
This is a classic ratio proportion question. Since  radians is equivalent to 180 (degrees).
We can convert a radian measure of an angle to a degree measure by simply replacing  by
180. Doing so we obtain
5 / 3 5(180) / 3 = 900 / 3 = 300
so 5 / 3 radians is equivalent to 300 degrees.
2) Convert 80 degrees to radians.
This is a classic ratio proportion question. Since 180 degrees is equivalent to radians we can
convert a degree measure of an angle to radian measure by writing and solving a proportion. Do
so we get
Degrees
80 / 180
=
Radians
x / 
Note, above we have let x represent the radian measure of an angle that measures 80 degrees.
Cross multiplying
80 = 180 x
Dividing both sides by 180
80 / 180 = x
Reducing
4 / 9 = x
So 80 degrees is equivalent to 4 / 9 radians.
3) Given the diagram below, find cos 

4
90
5
Trigonometry. SOHCAHTOA (a classic nomonic memory gimmick for recollecting the 3 basic
trigonometry functions) in this problem we are asked to find the cosine of the angle marked .
Since, cosine of an angle is the ratio of the leg opposite the angle to the hypotenuse, we begin by
employing the Pythagorean theorem to find the length of the hypotenuse. We let x represent the
length of the hypotenuse
4 2+ 5 2= x 2
16 + 25 = x 2
41 = x 2
sqrt(41) = x
[Note: we reject the negative value obtained when we perform the square root since distance can
not be negative.]
Next, with the leg adjacent to given as 4 and the hypotenuse known to be sqrt(41) we assert
that cos = 4 / sqrt(41).
4) Given the figure below, solve for x
4
x
90
/4
Two approaches exist to solve this problem we will use the standard trigonometric technique
(SOHCAHTOA). Given the acute angle /4 radians (45 degrees) and the Hypotenuse we need
to find the leg Opposite the given angle. This calls for the use of the sine ratio. Doing so
sin( /4) = x/4
since sin( /4) is well known (or should be) to be sqrt{2}/2. We substitute.
sqrt{2}/2 = x/4
Multiplying both sides by 4 we obtain
4 sqrt{2} / 2 = x
Simplifying: 2 sqrt{2} = x is our final answer.
5) Evaluate sin 2 /3
Since  radians is equivalent to 180 degrees we can convert radian measure to degree measure
by substituting 180 for and then simplifying doing so.
sin(2 /3) = sin ( 2(180)/3 )
= sin(120 o )
Now, 120 o is a second quadrant angle and since the sine function is positive in the second
quadrant and since 120 o is a reference angle of 60 o we know
sin(120 o ) = sin(60 o )
Finally sin 60 o is well known to be sqrt{3}/2 – our final answer.
6) Evaluate cos(-5 /6)
We begin by noting that cos(- 5 /6) is equivalent to cos(-150 o ). A –150 o angle will have
trigonometric values referenced by a 30 o angle in the third quadrant. Since we are looking
for the cosine of the angle and cosine is negative in the third quadrant. We know;
cos(- 5 /6) = -cos(30 o )
Next, cos 30 o is well kown to be sqrt(3)/2.
So, cos( - 5 /6) = -sqrt(3)/2.
7) Evaluate tan 3 /4.
We begin by noting that tan(3 /4) is equivalent to tan(135 o ). A 135 o angle will have a
trigonometric value referenced by a 45 o angle in the second quadrant. Since we are looking
for the tangent of the angle and tangent is negative in the second quadrant. We know,
tan(3 pi /4) = -tan(45 o )
Next, tan(45 o ) is well known to be 1. So tan(3 /4) = -1.
8) Evaluate sec(17 /4).
We begin by noting that 17 /4 is equivalent to sec(765 o ). Next 765 o is equivalent to a
reference angle of 45 o in the first quadrant. Since, all trigonometric functions are positive in
the first quadrant, we know;
sec(17 /4) = sec(45 o )
Next,
sec  = 1/cos 
so
sec(17 /4) = 1/cos(45 o ).
Now, it is well known that cos 45 o is sqrt{2}/2, so
sec(17 /4) = 1/ (sqrt{2}/2).
Simplifying
1 / ( sqrt{2}/2 ) we get
sec(17 /4) = 2 / sqrt{2}
or
sec(17 /4) = sqrt{2}.
9) Evaluate cot(11 /6).
We begin by noting 11 /6 is equivalent to 330 o . Now 330 o is equivalent to a reference
angle of 30 o in the fourth quadrant. Since cot values are negative in the fourth quadrant, we
know;
cot(11 /6) = -cot(30 o ).
Next, since cotangent is the reciprocal of tangent
cot(11 /6) = - 1/ tan 30 o .
Now tan(30 o ) is well known to be 1/ sqrt{3}. So
cot(11 /6) = - 1/(1/sqrt(3)).
And if we rationalize this expression
cot(11 /6) = - sqrt{3}.
10) Evaluate csc(-3 ).
We begin by noting -3  is equivalent to -540 o . Now, -540 o is equivalent to a reference
angle of 180 o . Also, csc is positive in the second quadrant. So
csc(-3 ) = csc(180 o ).
Next, since cosecant is the reciprocal of sine w know
csc(-3 ) = 1/sin 180 .
Now, it is well known that sin 180 =0. So
csc(-3 ) = 1/0.
which is clearly undefined. We conclude
csc(-3 ) is undefined.
B. Smith
CUMATH
Clarkson University
Potsdam NY 13699