April 1, 2011
18.01 Problem Set 9 solutions
Part II: 15 points
1a) Find the average value of the function 1 = cos0 (x) on the interval 0 ≤ x ≤ 2π.
By definition this is
2π
Z
1
2π
0
2π
1
1
1dx =
x
=
[2π − 0] = 1.
2π
2π
0
b) Find the average value of the function cos2 (x) on the interval 0 ≤ x ≤ 2π. (You can just
quote an answer from the last problem set if you prefer.) By definition this is
1
2π
Z
2π
cos2 (x)dx
0
The systematic method to deal with even powers of sine and cosine is to use the doubling formulas
cos2 (x) =
1
(1 + cos(2x)),
2
sin2 (x) =
1
(1 − cos(2x)).
2
Doing this, and changing the variable to u = 2x (with dx = 12 du) gives
=
1
4π
Z
2π
(1 + cos(2x))dx =
0
1 1
2 4π
Z
4π
(1 + cos(u))du.
0
The integral is easy to compute; or you can observe that this is half of the average value of 1 (giving
1/2) plus half of the average value of cos over two periods (giving zero). So the answer is 1/2.
c) Find the average value of the function cos4 (x) on the interval 0 ≤ x ≤ 2π.
Working as in (b) gives
1
2π
Z
2π
cos4 (x)dx =
0
1 1
4 2π
Z
2π
(1 + cos(2x))2 dx =
0
1 1
4 4π
Z
4π
(1 + cos(u))2 du
0
Expanding the integrand gives 1 + 2 cos(u) + cos2 (u). We can integrate the first two terms immediately,
and then integrate the third by another use of the doubling formula; or we can observe that what is
written is
1
((average value of 1) + 2 · (average value of cos) + (average value of cos2 )).
4
The middle term is zero, since we are averaging over two periods of cosine. The end terms were
computed in parts (a) and (b) as 1 and 1/2; so the answer is 3/8.
d) Find the average value of the function cos6 (x) on the interval 0 ≤ x ≤ 2π.
Working as in (b) gives
1
2π
Z
2π
0
1 1
cos (x)dx =
8 2π
6
Z
0
2π
1 1
(1 + cos(2x)) dx =
8 4π
3
Now the integrand is
1 + 3 cos(u) + 3 cos2 (u) + cos3 (u).
Z
0
4π
(1 + cos(u))3 du
If we continue our program of trying just to “see” what the answer must be, we see that need to know
the average value of cos3 (u) from 0 to 2π. In fact the average value of any odd power of cosine (or
sine) from 0 to 2π is zero. (The reason is that the interval [0, π/2] where cos2m+1 is positive precisely
cancels the interval [π/2, π] where cos2m+1 is negative.) So we get
1
3
1
((average value of 1) + 3 · (average value of cos2 )) = · (1 + ) = 5/16.
8
8
2
e) Explain why the answers for (a)–(d) are decreasing. Because cos takes values between −1
and 1, cos2 takes values between 0 and 1. For every value of x, it follows that the powers of cos2 (x) are
decreasing:
1 ≥ cos2 (x) ≥ cos4 (x) ≥ cos6 (x) ≥ cos8 (x) ≥ · · ·
Taking integrals respects these inequalities, so the average values must decrease.
It turns out that the average values decrease in a very simple way:
(average value of cos2n+2 (x)) =
2n + 1
· (average value of cos2n (x)).
2n + 2
This formula is not so easy to see from the computational method I described above, but it should
become obvious at least by the end of 18.03.
2) You can use the formulas
cos(ax) cos(bx) =
1
(cos((a + b)x) + cos((a − b)x))
2
cos(ax) sin(bx) =
1
(sin((a + b)x) + sin((a − b)x))
2
sin(ax) sin(bx) =
1
(− cos((a + b)x) + cos((a − b)x))
2
to write things like sinM (x) cosN (x) as sums of terms like cos(nx) and sin(nx) (with n
smaller than M + N ).
a) Use this idea to find a formula of the form
cos3 (x) = A cos(3x) + B cos(2x) + C cos(x) + D.
Begin by using the first formula to write
cos2 (x) =
1
(cos(2x) + 1).
2
Multiply by cos(x) to get
cos3 (x) =
1
(cos(2x) cos(x) + cos(x)).
2
Use the first identity to rewrite
cos(2x) cos(x) =
1
(cos(3x) + cos(x)).
2
Plugging this in gives
cos3 (x) =
1
1
(cos(3x) + cos(x) + 2 cos(x)) = (cos(3x) + 3 cos(x)).
4
4
So A = 1/4, B = 0, C = 3/4, and D = 0.
b) Use the formula in (a) to calculate
Z
Z
cos3 (x)dx =
1
4
Z
cos3 (x)dx.
1
(cos(3x) + 3 cos(x))dx = (sin(3x)/3 + sin(x)) .
4
c) Use this idea to say as much as you can about the trigonometric identity
cosn (x) = an cos(nx) + an−1 cos((n − 1)x) + an−2 cos((n − 2)x) + · · · + a1 cos(x) + a0 .
(Best answer is a formula for every coefficient an . But if you can say something like, “the
last term a0 is 11 when n is odd,” or “every fifth term is zero,” that’s good too.) I’ll do
two more by hand to help look for patterns. Multiplying the identity in (b) by cos(x) gives
cos4 (x) =
1
(cos(3x) cos(x) + 3 cos2 (x)).
4
The first identity in the problem gives
cos(3x) cos(x) =
1
(cos(4x) + cos(2x)),
2
cos2 (x) =
1
(cos(2x) + 1).
2
Plugging these in gives
cos4 (x) =
1
1
(cos(4x) + cos(2x) + 3 cos(2x) + 3) = (cos(4x) + 4 cos(2x) + 3).
8
8
In exactly the same way, you can find
cos5 (x) =
1
(cos(5x) + 5 cos(3x) + 10 cos(x)).
16
So here is a list of the first six identities that we’re looking for:
cos0 (x)
=
1
cos (x)
=
2
cos (x)
=
3
cos (x)
=
4
cos (x)
=
5
=
cos (x)
1
1/2 (
1
1(
1
2(
1
4(
1
8(
1
16 (
1/2 )
cos(x)
cos(2x)
cos(3x)
cos(4x)
cos(5x)
+
+
+
+
1
3 cos(x)
4 cos(2x)
5 cos(3x)
+
)
+
)
3
10 cos(x)
)
)
)
I’ve written the one for cos0 in a peculiar way, to keep the “powers of two” pattern in the coefficients.
The most obvious things about the list are these: the formula for cosn (x) involves only cos((n − 2m)x)
(for 0 ≤ m ≤ n/2); there is a coefficient of 2−(n−1) in front; and all the other entries (except for the
funny one for cos) ) are positive integers. If you stare at the integers for a while, you may notice a
pattern like Pascal’s triangle: each entry appears to be the sum of the two entries above it (and to
either side). For example, the 5 in 5 cos(3x) in the last row is the sum of the 4 in 4 cos(2x) and the 1 in
cos(4x) in the row above. This works perfectly except on the cos(x) column: the coefficient of cos(x) is
the sum of the coefficient of cos(2x) and twice the constant term in the row above. (For example, the
10 in 10 cos(x) is the 4 from 4 cos(2x) plus twice the 3.)
If you think about the way I constructed the formulas in this table, each one from the one before
using the first identity in the problem, you will see that the construction satisfies these addition rules.
Therefore the entries in the formula really do behave like Pascal’s triangle. In fact the entries are
exactly the entries in the left-hand side of Pascal’s triangle, except that the entries in the middle must
be divided by two. Here’s a picture showing Pascal’s triangle on the left and the entries from our table
of identities on the right:
1
1
1
1
1
2
3
4
1/2
1
1
−→
1
3
6
1
1
4
1
1
1
1
3
4
3
Here is a formal statement of the identity.
Theorem. For every non-negative integer n, there is an identity
cosn (x) = an cos(nx) + an−2 cos((n − 2)x) + an−4 cos((n − 4)x) + · · ·
The coefficient an−2m is equal to
1
2n−1
(binomial coefficient n choose m) =
n!
2n−1 m!(n
− m)!
for 0 ≤ m < n/2. If n is even, the coefficient of the constant term is half this number
a0 =
n!
.
2n (n/2)!(n/2)!