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

Radian

In our normal life, when we talk about angles, we often use the degree. For example, a right angle is equal to 90 degrees, an angle of an equilateral triangle is 60 degrees, a straight angle is $180^{o}$. However, for all trigonometric functions in mathematics, such as sin(x), cos(x), etc., the angle $x$ is always understood as in radian.

So what is the radian?

To understand angle measurement in radians, we draw a unit circle. Unit circle is a circle with radius equal to 1. We know that by definition, $\pi$ is the length of the semicircle.

An angle's measurement in radians is equal to the length of the corresponding arc of the unit circle.

In radians, $x$ is the length of the arc |

Let us look at an example of a right angle. A right angle is subtended by a quarter of the circle, and the length of a quarter of the circle is $\frac{\pi}{2}$, so a right angle is equal to $\frac{\pi}{2}$ in radians.

A straight angle ($180^{o}$) occupies half of the circle, and the length of a half of the circle is $\pi$, so a straight angle is equal to $\pi$ in radians.

We can easily remember the translation from degrees to radians by the following correspondence

straight angle 180 degrees $\to$ semicircle $\to ~~ \pi$Common angles are

$$180^{o} ~~\to ~~ \pi$$ $$360^{o} ~~\to ~~ 2\pi$$ $$90^{o} ~~\to ~~ \frac{\pi}{2}$$ $$45^{o} ~~\to ~~ \frac{\pi}{4}$$ $$60^{o} ~~\to ~~ \frac{\pi}{3}$$ $$30^{o} ~~\to ~~ \frac{\pi}{6}$$

Let us stop here for now. We will return to our series on mathematical identities in our next post. Hope to see you there.

*Homework.*

In this homework section, we will prove the Vieta formula of the number $\pi$ that we mention in the previous post $$ \frac{2}{\pi} = \sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}} \cdots $$ In the following figure, we can see that $ZA = sin(x)$ is a line segment which is shorter than the circle arc $ZI = x$ $$ sin(x) < x $$

1. Use the trigonometric cos formula for double angle $$cos(2x) = 2 cos^2(x) - 1$$ to show that $$ cos \frac{\pi}{4} = \sqrt{\frac{1}{2}} $$ $$ cos \frac{\pi}{8} = \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}} $$ $$ cos \frac{\pi}{16} = \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}} $$ Thus, $$ \sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}} = cos \frac{\pi}{4} \cdot cos \frac{\pi}{8} \cdot cos \frac{\pi}{16} $$

2. Use the trigonometric sin formula for double angle $$sin(2x) = 2 sin(x) ~ cos(x)$$ to show that $$ cos \frac{\pi}{4} \cdot cos \frac{\pi}{8} \cdot cos \frac{\pi}{16} = \frac{\frac{1}{8}}{sin \frac{\pi}{16} } = \frac{2}{\pi} \cdot \frac{\frac{\pi}{16}}{sin \frac{\pi}{16} } $$

3. As we have discussed above, since the angle $\frac{\pi}{16}$ is very small, $$ sin \frac{\pi}{16} \approx \frac{\pi}{16}$$ and $$ cos \frac{\pi}{4} \cdot cos \frac{\pi}{8} \cdot cos \frac{\pi}{16} \approx \frac{2}{\pi} $$

4. In general, prove that $$ cos \frac{\pi}{4} \cdot cos \frac{\pi}{8} \cdots cos \frac{\pi}{2^n} = \frac{2}{\pi} \cdot \frac{\frac{\pi}{2^n}}{sin \frac{\pi}{2^n} } $$ and $$ \lim_{n \to \infty} cos \frac{\pi}{4} \cdot cos \frac{\pi}{8} \cdots cos \frac{\pi}{2^n} = \frac{2}{\pi} $$ This is the Vieta formula of the number $\pi$: $$\sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}} \cdots = \frac{2}{\pi}$$

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