18.01 PROBLEM SET 1 – SOLUTIONS

2.

(15 + \sqrt{220})^{5} =15^{5} + 5 · 15^{4} · \sqrt{220} + 10 · 15^{3} · (\sqrt{220})^{2} + 10 · 15^{2} · (\sqrt{220})^{3}+
+ 5 · 15 · (\sqrt{220})^{4} + (\sqrt{220})^{5}
(15 − \sqrt{220})^{5} =15^{5} − 5 · 15^{4} · \sqrt{220} + 10 · 15^{3} · (\sqrt{220})^{2} − 10 · 15^{2} · (\sqrt{220})^{3}+
+ 5 · 15 · (\sqrt{220})^{4} − (\sqrt{220})^{5},

hence

(15 + \sqrt{220})^{5} + (15 − \sqrt{220})^{5} = 2 · 15^{5} + 2 · 10 · 15^{3} · (220) + 2 · 5 · 15 · (220)^{2}

is indeed an integer, divisible by 10. (15 − \sqrt{220})^{5} < 1, so the last digit preceding the decimal point in the decimal expression of (15 + \sqrt{220})^{5} is 9.

(3 points)

3. Using similar triangles

\frac{t}{a} = \frac{sin θ}{cos θ} = tan θ.

But a = 1, so t = tan θ. Similarly,

\frac{s}{z} = \frac{t}{sin θ} = \frac{tan θ}{sin θ} = sec θ.

Again b = 1, so s = sec θ.

The similar statements for cot θ and csc θ are shown on the picture to the right. The proof of these statements are basically the same as the ones for tan θ and sec θ.

(3 points)

4. The area of the OAB triangle is \frac{1}{2} sin θ.

The area of the OAB circle sector is \frac{1}{2}θ.

The area of the OAT triangle is \frac{1}{2} tan θ.

The statement now follows from the fact that the OAB triangle is contained in the OAB circle sector, which is contained in the OAT triangle.

(3 points)

Typeset by AMS-TEX 2 18.01 PROBLEM SET 1 – SOLUTIONS

5.

sin θ < θ < tan θ

implies

1 < \frac{θ}{sin θ} < \frac{1}{cos θ} .

Since lim_{θ→0} cos θ = cos 0 = 1, this implies, that lim_{θ→0} \frac{sin θ}{θ} = 1.

(3 points)

6. A possible solution: f(x) = \{ \begin{matrix}1 & if x is a rational number \\ 0 & if x is not a rational number \end{matrix} (3 points)