MAT 1332, Assignment #2 Solutions
1. If you simply integrate, the answer is zero. So that obviously isn’t right. It’s best to sketch the two graphs.
[5]
See Figure 1.
0
1
0.5
0
0.5
1
B
C
A
sin(2x)
cos(2x)
5
π
/8
π
/8
π
/2
π
/4
3
π
/4
π
Figure 1: Area between sin(2
x
) and cos(2
x
).
Clearly the area between them isn’t zero.
In fact, it’s area A, plus area B, plus area C on the graph.
And before we can find them, we need to find the points of intersection.
That is, the values of
x
where
sin(2
x
) = cos(2
x
). If you look at Figure 1 again, you may be able to subtly discern these. But let’s figure
them out:
sin(2
x
)
=
cos(2
x
)
tan(2
x
)
=
1
(dividing both sides by cos(2
x
))
2
x
=
...

7
π
4
,

3
π
4
,
π
4
,
5
π
4
,
9
π
4
...
x
=
...

7
π
8
,

3
π
8
,
π
8
,
5
π
8
,
9
π
8
...
Actually, since 0
≤
x
≤
π
, we only need values that fall into this range. So the points of intersection within
this range are
x
=
π
8
and
x
=
5
π
8
.
Now we’re ready to integrate. The area is the sum of areas A, B and C, so we have
Area
=
A
+
B
+
C
=
Z
π/
8
0
(cos(2
x
)

sin(2
x
))
dx
+
Z
5
π/
8
π/
8
(sin(2
x
)

cos(2
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 Spring '07
 MUNTEANU
 Derivative, Sin, Cos

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