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

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 $ZI = x$ $$sin(x) < x$$
Especially, when the angle $x$ is small, $sin(x)$ is getting closer and closer to $x$. We will use this fact to prove the Vieta formula.

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 $\pi$: $$\sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}} \cdot \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{\frac{1}{2}}}} \cdots = \frac{2}{\pi}$$