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Section 1: Some Trigonometry Review Exercises
1
Some Supplemental Exercises
1. Some Trigonometry Review Exercises
Exercise 1. How many radians is 145 degrees? How many degrees is 7π/6 radians?
Exercise 2. Does it make sense to talk about 180 radians? How about π degrees?
Exercise 3. Criticize the statement: π = 180.
Exercise 4. Why do you think the ancient Greeks used degrees instead of radians?
Why do we primarily use radians instead of degrees?
Exercise 5. Using our association of real numbers to points on the unit circle, find the
point associated with the real number t = 5π/6. Then use this point to find the value
of all of the standard trigonometric functions at 5π/6.
Exercise 6. Repeat the last exercise, but use 4π/3.
Exercise 7. Draw a copy of the unit circle, and label the points associated with the
real numbers t = 0, t = π/6, t = π/4, t = π/3, and t = π/2. Then label all of the points
on the unit circle which one would obtain by reflecting these points about the y axis,
about the x axis, and about the origin. Finally, find the value of t between 0 and 2π
which would be associated with the points you found above.
Exercise 8. What would the sign of the various trigonometric functions be for t = 4?
In other words, is tan(4) (for example) positive or negative? (You should be able to do
this problem without using a calculator, and without worrying about the value of tan(4),
etc..
√
Exercise 9. Repeat the last exercise, using t = 7.
Exercise 10. Find the point on the unit circle associated with t = 2π/3, and then find
at least two other values of t which are also associated with this point. Then find all
values of t associated with this point.
Exercise 11. Solve the equation 2 sin(x) + 1 = 0 for x.
Exercise 12. Solve 2 sin(2x) + 1 = 0 for x.
Exercise 13. Where does sin(x) − cos(x) = 0?
Exercise 14. What is the domain of the function f (x) = sec(2x)?
Exercise 15. Draw a sketch of f (x) = sec(2x) by thinking about the domain, where
the function is positive and where it is negative, what is happening near its asymptotes,
etc.
Section 1: Some Trigonometry Review Exercises
2
Exercise 16. Without looking back at the text, show how to get the trigonometric
identity for sin(2x) from the addition identity for sin(x + y). Do the same for the
identity involving cos(2x).
Exercise 17. Does sin2 (4) + cos2 (6) = 1? Use this to help explain what is wrong with
writing sin2 + cos2 = 1. How would you explain to a friend that it is a “sin” to write sin,
when you mean to write sin(x).
Exercise 18. What is the period of the secant function? Show that your answer is
correct.
Exercise 19. Use the fact that 7π/12 = π/3 + π/4 to find cos(7π/12) without using a
calculator. Then find the value of the other trigonometric functions at 7π/12.
Exercise 20. If tan(z) = 2 and you are given that z is between π and 2π, find the
values of the other trigonometric functions at z.
Exercise 21. Consider f (x) = sin(cos(x)) and g(x) = sin(x) cos(x). Are these the same
or different? Evaluate f (1) and g(1), using a calculator if necessary.
Exercise 22. Given that a circle of radius r has area π · r2 , what would be the area of
a slice of the circle (think pizza) whose angle is h? Hint: think about the fact that if
h = 2π, you get the whole pizza which has area πr2 , while if h = π/2, you would have
area (1/4) · (πr2 ).
Solutions to Exercises
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Solutions to Exercises
Exercise 1. Recall that one degree is π/180 radians. Thus 145 degrees is 145 · π/180
radians, which can be written as 29π/36 radians. Also recall that one radian is 180/π
degrees. Thus 7π/6 radians is 7π/6 · 180/π degrees, which can also be written as 210
degrees.
Exercise 1
Exercise 2. Sure, why not?
Exercise 2
Exercise 3. What is wrong with this statement is that it isn’t true. π is a real number
that is bigger than 3 but less than 4. The real number 180 is much different.
Exercise 3
Exercise 4. Most historians think that the Greeks chose 360 because there are about
that many days in a year, and because 360 can easily be divided by many small numbers
(2, 3, 4, 5, 6, 8, 10, 12, . . . ). Note that the number 360 is somewhat arbitrary, however.
There is nothing intrinsic in a circle which says that it is a good idea to divide it into
360 parts. On the other hand, radians are more natural in the sense that 2π does have
something to do with a circle, namely it is the circumference of the unit circle. Thus the
relationship between the length of an arc cut off by a central angle and the measure in
radians of the central angle is given by this decision to define radians as we have defined
Exercise 4
them.
√
Exercise 5. The point associated with 5π/6 would be (− 3/2, 1/2). Thus we have
√
√
√
cos(5π/6) = − 3/2, sin(5π/6) = 1/2, tan(5π/6) = −1/ 3, cot(5π/6) = − 3, sec(5π/6) =
√
−2/ 3, csc(5π/6) = 2.
Exercise 5
√
Exercise 6. The point associated with 4π/3 is (−1/2, − 3/2). Thus we have sin(4π/3) =
√
√
√
− 3/2, cos(4π/3) = −1/2, tan(4π/3) = 3, cot(4π/3) = 1/ 3, sec(4π/3) = −2, and
√
csc(4π/3) = −2/ 3.
Exercise 6
Exercise 7. You should have all of the following labelled on your circle:
Solutions to Exercises
4
t
(cos(t), sin(t))
0
(1, 0)
√
( 3/2, 1/2)
√
√
( 2/2, 2/2)
√
(1/2, 3/2)
π/6
π/4
π/3
π/2
2π/3
3π/4
5π/6
π
7π/6
5π/4
4π/3
3π/2
5π/3
7π/4
11π/6
(0, 1)
√
(−1/2, 3/2)
√
√
(− 2/2, 2/2)
√
(− 3/2, 1/2)
(−1, 0)
√
(− 3/2, −1/2)
√
√
(− 2/2, − 2/2)
√
(−1/2, − 3/2)
(0, −1)
√
(1/2, − 3/2)
√
√
( 2/2, − 2/2)
√
( 3/2, −1/2)
Exercise 7
Exercise 8. Since π ≈ 3.1 and 3π/2 ≈ 4.7, we note that the point associated with
the real number 4 lies in the 3rd quadrant. Thus, both sin(4) and cos(4) would be
negative, as would sec(4) and csc(4), since these have the same sign as cos(4) and sin(4),
respectively. However, tan(4) and cot(4) would be positive.
Exercise 8
√
Exercise 9. Note that 7 is about 2.6, so the point on the unit circle associated with
√
7 is between π/2 and π. Thus the point on the circle associated with this real number
√
√
is in the 2nd quadrant, so sin( 7) > 0 while cos( 7) < 0. Thus tangent and cotangent
at this point are negative, while the cosecant is positive and the secant is negative.
Exercise 9
√
Exercise 10. The point associated with 2π/3 is (−1/2, 3/2). Other values of t associated with this point include t = 8π/3, t = 14π/3, etc. The set of all such points can
be written {x : x = 2π/3 + 2kπ, k an integer}.
Exercise 10
Exercise 11. This equation would be valid wherever sin(x) = −1/2. Using our chart
and our memory, we recall that this occurs at the point directly across from the point
Solutions to Exercises
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associated with x = π/6. Thus one solution of this equation is π + π/6 = 7π/6. Another
solution would be at the other point on the unit circle which intersects the line y = −1/2,
namely the point associated with 11π/6. Thus the set of all solutions is {x : x =
7π/6 + 2kπ} ∪ {x : x = 11π/6 + 2lπ}, where k and l are any integers.
Exercise 11
Exercise 12. This is similar to the last problem, but we are looking for points where
sin(2x) = −1/2. This would occur where 2x = 7π/6 + 2kπ or 2x = 11π/6 + 2lπ. Thus
the solution set is {x : x = 7π/12 + kπ k an integer} ∪ {x : x = 11π/12 + lπ l an integer}.
Exercise 12
Exercise 13. This equation is satisfied wherever sin(x) = cos(x), which we recall is (for
example) where x = π/4. Also the point directly across the unit circle will have this
property. Thus the solution set is {x : x = π/4 + kπ k an integer}.
Exercise 13
Exercise 14. The function f (x) = sec(2x) = 1/ cos(2x) is defined wherever cos(2x) 6= 0.
But cos(2x) = 0 exactly where 2x = π/2 + kπ, which is where x = π/4 + kπ/2. Thus the
domain of sec(2x) is {x : x 6= π/4 + kπ/2. Some examples of numbers not in the domain
include π/4, 3π/4, 5π/4,. . . .
Exercise 14
Exercise 15. The function has asymptotes at the points that aren’t in the domain. The
function is always either greater than one or less than −1.
Exercise 15
Exercise 16. sin(2x) = sin(x + x) = sin(x) cos(x) + sin(x) cos(x) = 2 sin(x) cos(x).
Likewise, cos(2x) = cos(x + x) = cos(x) cos(x) − sin(x) sin(x) = cos2 (x) − sin2 (x).
Exercise 16
Exercise 17. No, sin2 (4) + cos2 (6) 6= 1. The statement sin2 + cos2 = 1 doesn’t make
sense for the same reason that sin doesn’t really make sense. For example, we know
that sin(π) is the real number 0, and that sin(x) is a nice function that we have been
discussing, but plain old sin doesn’t really have any meaning when used in an equation.
The point of this problem is to help convince you to always use arguments when writing
trig functions.
Exercise 17
1
1
Exercise 18. The period is 2π, because sec(x + 2π) = cos(x+2π)
= cos(x)
= sec(x). And
if sec(x + m) = sec(x) for any m < 2π, this would imply that the period of the cosine
function was less than 2π.
Exercise 18
√
√
√
√
Exercise 19. cos(7π/12) = ( 2 − 6)/4. sin(7π/12) = ( 6 + 2)/4. tan(7π/12) =
√
√
√
√
√
√
√
√
( 6 + 2)/( 2 − 6). sec(7π/12) = (4/( 2 − 6). csc(7π/12) = 4/( 6 + 2).
√
√
√
√
cot(7π/12) = ( 2 − 6)/( 6 + 2).
Exercise 19
Exercise 20. Since sin(z)/ cos(z) = 2, we know that sin(z) = 2 cos(z). Also, we know
that sin2 (z) + cos2 (z) = 1. Thus (by substituting). we have 4 cos2 (z) + cos2 (z) = 1, so
p
cos2 (z) = 1/5, or | cos(z)| = 1/5. Now since z is between π and 2π, the point associated
with z on the unit circle must either lie in the 3rd or 4th quadrant. But tangent is only
positive in the 3rd quadrant. So we must have that both sin(z) and cos(z) are negative.
Solutions to Exercises
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p
p
√
Thus tan(z) = 2. cot(z) = 1/2. sin(z) = −2 1/5. cos(z) = − 1/5. sec(z) = − 5.
√
Exercise 20
csc(z) = (−1/2) 5.
Exercise 21. These two functions are different. f (x) is the composition of sin(x) and
cos(x), while g(x) is the product of these two functions. f (1) = sin(cos(1)) ≈ .5144,
while sin(1) · cos(1) ≈ .4546.
Exercise 21
Exercise 22. The area of a slice of pizza is (h/2)r2 . Note that this formula gives the
correct result for h = 2π (the whole pizza) and h = π/2 (1/4 of the pizza.)
Exercise 22