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Comments
The following is the standard approach.
Using the method of proof by contradiction, assume that cos(1∘) is rational.
Then cos(2∘)=2∗cos2(1∘)−1 must also be rational.
Now cos(k∘+1∘)+cos(k∘−1∘)=2cos(k∘)cos(1∘).
Using this equation, now that we have, by assumption, cos(1∘) and hence cos(2∘) as rational, (in addition to the fact that cos(0∘)=1 is rational), we see that in turn cos(n∘) is rational for each successive integer. But cos(30∘)=23 is clearly irrational, thus implying that the original assumption, i.e., that cos(1∘) is rational, is in fact false.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"\(...\)or\[...\]to ensure proper formatting.2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
The following is the standard approach.
Using the method of proof by contradiction, assume that cos(1∘) is rational.
Then cos(2∘)=2∗cos2(1∘)−1 must also be rational.
Now cos(k∘+1∘)+cos(k∘−1∘)=2cos(k∘)cos(1∘).
Using this equation, now that we have, by assumption, cos(1∘) and hence cos(2∘) as rational, (in addition to the fact that cos(0∘)=1 is rational), we see that in turn cos(n∘) is rational for each successive integer. But cos(30∘)=23 is clearly irrational, thus implying that the original assumption, i.e., that cos(1∘) is rational, is in fact false.
HAVE U USED INDUCTION
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Yes, I have essentially used strong induction on the equation
cos(k∘+1∘)=2cos(k∘)cos(1∘)−cos(k∘−1∘).