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MAS110 Core I: Problems for Week 1
1. Consider an isosceles right triangle ABC
with AB = BC = 1. (See 41 in Figure 1.) Then
√
∠BAC = ∠BCA = π/4 and AC = 2. Therefore sin π4 = cos π4 = √12 and tan π4 = 1.
Now consider an equilateral triangle ABC with side length 1. (See 42 in Figure √1.) Let D be the
√
1
mid-point of BC. Then ∠BAD = π6 , BD
=
and
AD
=
AB 2 − BD2 = 23 . We can now
2
√
√
use 4ABD to calculate sin π3 = cos π6 = 23 , cos π3 = sin π6 = 12 , tan π3 = 3 and tan π6 = √13 .
Figure 1: Graph (not to scale) of sin (in blue) and cos (in red)
sin0 (x) is the slope of the graph of y = sin(x). We observe that the slope sin0 (x)
• is positive for − π2 + 2nπ < π2 + 2nπ,
• is negative for π2 + 2nπ < x < 3π
+ 2nπ,
2
1
• is 0 when x = π2 + nπ,
• has maximum at 2nπ, and
• has minimum at (2n + 1)π,
where n is any integer. These agree with features of cos(x).
2. (a) Let ABC be an isosceles triangle with AB = AC = 1 and ∠A = 2θ. Let D be the midpoint
of BC. Then ∠BAD = θ and therefore the base BC = 2BD = 2 sin θ.
Figure 2: Diagram for Question 2(a)
(b) Join each vertex to the centre. This breaks up the polygon into n congruent isosceles triangles
with legs 1 and vertex angle 2π/n. By part (a) the base of each isosceles triangle has length
2 sin(π/n). Hence the perimeter of the polygon is 2n sin(π/n).
Figure 3: Inscribed regular polygons when n = 5 and n = 7
When n becomes big the perimeter of the polygon 2n sin(π/n) will approach the circumference of the circle 2π. So n sin(π/n) will tend to π when n is big.
2
3. The number of triangles removed at the n-th step is clearly the number of triangles before the
n-th step is performed. Now at each step, each triangle is replaced by 3 equilateral triangles of
half the length and quarter the area.
• Since the number of triangles grow by a factor of 3, there are 3n triangles after the n-th step
and the n-step removes 3n−1 triangles.
• The length of an equilateral triangle cut out at the n step is 21n a.
• Since the area decreases
by a factor of 34 at every step, the shape obtained just after the n-th
n
step has area 43 A.
4. We have
1
1
x2 − 2
− 2 =
.
2 x
2x2
√
√
0
2,
negative
when
0
<
x
<
2, and vanishes
Therefore
the
derivative
f
(x)
is
positive
when
x
>
√
√
at x = √2. Consequently, the function f (x)√is increasing when x ≥ 2 and decreasing when
0 < x ≤ 2. Also, it has a minimum at x = 2. So
√
√
1 √
2
f (x) ≥ f ( 2) =
2+ √
= 2
2
2
√
with equality precisely when x = 2.
f 0 (x) =
•
•
•
•
•
2
x1
2
x2
2
x3
2
x4
2
x5
= 1, so x2 =
=
=
=
=
3
2
= 1.5.
4
= 1.333 . . ., so x3 = 12 ( 32 + 34 ) = 17
= 1.416666 . . .
3
12
24
1 17
24
577
, so x4 = 2 ( 12 + 17 ) = 408 = 1.414215 . . ..
17
816
, so x5 = 12 ( 577
+ 816
) = 665857
= 1.41421356 . . ..
577
408
577
470832
665857
941664
, so x6 = 12 ( 470832
+ 941664
) = 886,731,088,897
= 1.414213562373095 . . ..
665857
665857
627,013,566,048
√
√
√
It seems that xn decreases to 2. Note that if x > 2 then x2 < 2. So if x and 2/x
√
agree in the first d decimal places, then x approximates
2 to at least d decimal places. Since
√
2
=
1.414213562373095
.
.
.,
x
approximates
2
to
at
least
15 decimal places. (I used the
6
x6
calculator on my phone. With a more powerful calculator you can check that they agree up to 23
decimal places.)
3
2
2
2
5. From Figure 4, we see that d2 = (r
√a . Since a is tiny compared to r, we can
√ + a) − r = 2ar +
2
2
ignore the a term in calculating 2ar + a . So d ≈ 2ar.
Figure 4: Exaggerated diagram for Question 5
To double d we will need to multiply a by 4. With r ≈ 6.37 × 106 m and a ≈ 1.6 m, we obtain
d ≈ 4.5 km.
From the diagram, we see that the ship is at the same distance on the other side when the mast
disappears. Thus it is at a distance 2d from the observer. To double this to a distance 4d, the
ship must be at a distance 3d from the horizon. This means that the height of the mast must
now be about 9a.
4