Radian

On this $\pi$ day occasion, we will learn about the concept of radian.

Gauss' trigonometric identities for heptadecagon

Today, we write down Gauss' magical trigonometric identities for regular heptadecagon.

Trigonometric multiple-angle formulas

In our previous post, we show a compass-and-straightedge construction of a regular pentagon based on the following trigonometric formula $$\cos{\frac{\pi}{5}} = \frac{1 + \sqrt{5}}{4}.$$
We derive this formula of $\cos{\frac{\pi}{5}}$ by observing that $\cos{\frac{2 \pi}{5}} = -\cos{\frac{3 \pi}{5}}$ and then applying the trigonometric formulas for double angle and triple angle:
$$\cos{2 x} = 2 \cos^2{x} - 1,$$ $$\cos{3 x} = 4 \cos^3{x} - 3 \cos{x}$$ to set up a cubic equation for $\cos{\frac{\pi}{5}}$.

It seems a good occasion now for us to learn about trigonometric multiple-angle formulas. In this post, we will show how to derive formulas for $\sin{nx}$, $\cos{nx}$, $\tan{nx}$ and $\cot{nx}$ using de Moivre's identity of the complex numbers.

Construction of a regular pentagon

Today we will look at a compass-and-straightedge construction of a regular pentagon based on the following trigonometric formula $$\cos{\frac{\pi}{5}} = \frac{1 + \sqrt{5}}{4}.$$

Sequence - Part 9

This is the last post of our series; here is the link to "Sequence - Part 1" if you haven't read it. Today we will do some more exercises on sequence. We will prove some interesting identities. For the Pell sequence $$P_0=0, ~~P_1 = 1, ~~P_n = 2 P_{n-1} + P_{n-2},$$ and the companion Pell sequence $$H_0=1, ~~H_1 = 1, ~~H_n = 2 H_{n-1} + H_{n-2},$$ we will show that $$H_n^2 - 2 P_n^2 = (-1)^n.$$
For the Fibonacci sequence $$F_0 = 0, ~~F_1 = 1, ~~F_n = F_{n-1} + F_{n-2},$$ we will prove the following identity $$\frac{F_{2013(n+1)} - F_{2013 (n−1)}}{F_{2013 n}} = \frac{F_{2013(n^{2013}+1)} - F_{2013 (n^{2013}−1)}}{F_{2013 n^{2013}}}.$$

Sequence - Part 8

In the last post, we learn how to determine a trigonometric formula for a sequence in the case when the characteristic equation has complex roots. Today we will solve more exercises for this case.

Sequence - Part 7

This is the 7th part of our series on sequences. Today we will learn how to solve a linear recurrence equation in the case when the characteristic equation has complex roots. In this case, we can express the complex roots of the characteristic equation in trigonometric form and then use de Moivre's identity to obtain a trigonometric formula for the sequence.

de Moivre's formula

In previous post we have learned about complex numbers. Today, we will learn about the trigonometric form of a complex number and the famous de Moivre's formula.

Complex number

Today we will learn about the complex numbers. The crucial point about complex numbers is that we accept a very special number that we will denote it by $i$. This number $i$ is very special because it satisfies the following identity $$i^2 = -1.$$

So a complex number will have the form $$a + ib$$ where $a$ and $b$ are two real numbers. When $b=0$ then $a + ib = a$ is just a normal real number, and when $a=0$ then $a + ib = ib$ is called a pure imaginary number. Here are some examples of complex numbers: $$1+ i, ~~ 2 - 3i, ~~ -\sqrt{3} + 4i, ~~5i - 4, ~~6, ~~i, ~~-3i, ~~4 + 2i, \dots$$

Mathematical induction III

Today, we will solve some more problems using mathematical induction.

Problem 7. Observe that $$\cos 2 \alpha = 2 \cos^2 \alpha - 1$$

Prove that we can write $\cos n\alpha$ as a polynomial of $\cos \alpha$.