Category Archives: Trigonometry

De Moivre’s Theorem application

Question:

If f_{r}(\alpha)=(\cos{\frac{\alpha}{r^{2}}}+i\sin{\frac{\alpha}{r^{2}}}) \times (\cos{\frac{2\alpha}{r^{2}}}+i\sin{\frac{2\alpha}{r^{2}}}) \ldots (\cos{\frac{\alpha}{r}}+i\sin{\frac{\alpha}{r}}), then

\lim_{n \rightarrow \infty}f_{n}{\pi} equals

(a) -1

(b) 1

(c) -i

(d) i

Solution.

Using De Moivre’s theorem,

f_{r}{\alpha}=e^{i\frac{\alpha}{r^{2}}}e^{i\frac{2\alpha}{r^{2}}}\ldots e^{i\frac{\alpha}{r}}

which in turn equals e^{(i \frac{\alpha}{r^{2}})(1+2+\ldots+r)}=e^{(i\frac{\alpha}{r^{2}})(\frac{r(r+1)}{2})}=e^{i(\frac{\alpha}{2})(1+\frac{1}{r})}

Hence, \lim_{n \rightarrow \infty}f_{n}(\pi)=\lim_{n \rightarrow \infty}e^{(i)(\frac{\pi}{2})(1+\frac{1}{n})}=e^{i(\frac{\pi}{2})}=\cos{\frac{\pi}{2}}+i\sin{\frac{\pi}{2}}=i.

More complex stuff to be continued in next blog (pun intended) 🙂

Nalin Pithwa

More trigonometry practice

Problem.

Let ABC be a triangle and h_{a} be the altitude through A. Prove that

(b+c)^{2} \geq a^{2}+4(h_{a})^{2}.

(As usual, a, b, c denote the sides BC, CA, AB respectively.)

Proof.

The given inequality is equivalent to (b+c)^{2}-a^{2} \geq 4(h_{a})^{2}=\frac{16\triangle^{2}}{a^{2}}.

where \triangle is the area of triangle ABC. Using the identity

16\triangle^{2}=[(b+c)^{2}-a^{2}][a^{2}-(b-c)^{2}], we see that the inequality to be proved is

a^{2}-(b-c)^{2} \leq a^{2} (here we use a<b+c), which is true. Observe that equality holds iff b=c. QED.

More later,

Nalin Pithwa