×

# Can you prove ?!

Prove that $$cos(1^{o})$$ is irrational

Note by A K
3 years, 2 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

The following is the standard approach.

Using the method of proof by contradiction, assume that $$\cos(1^{\circ})$$ is rational.

Then $$\cos(2^{\circ}) = 2*\cos^{2}(1^{\circ}) - 1$$ must also be rational.

Now $$\cos(k^{\circ} + 1^{\circ}) + \cos(k^{\circ} - 1^{\circ}) = 2\cos(k^{\circ})\cos(1^{\circ})$$.

Using this equation, now that we have, by assumption, $$\cos(1^{\circ})$$ and hence $$\cos(2^{\circ})$$ as rational, (in addition to the fact that $$\cos(0^{\circ}) = 1$$ is rational), we see that in turn $$\cos(n^{\circ})$$ is rational for each successive integer. But $$\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$$ is clearly irrational, thus implying that the original assumption, i.e., that $$\cos(1^{\circ})$$ is rational, is in fact false.

- 3 years, 2 months ago

HAVE U USED INDUCTION

- 3 years, 2 months ago

Yes, I have essentially used strong induction on the equation

$$\cos(k^{\circ} + 1^{\circ}) = 2\cos(k^{\circ})\cos(1^{\circ}) - \cos(k^{\circ} - 1^{\circ})$$.

- 3 years, 2 months ago

×