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Aryth · #2 February 13th, 2008, 09:13 AM

Quote:

Originally Posted by ThePerfectHacker

1)Prove that $\sin 1^{\circ}$ and $\cos 1^{\circ}$ are irrational.

Suppose we let:

$f = cos(x),g = sin(x)$

Then:

$e^{xi} = cos(x) + i*sin(x) = f + ig$

Therefore:

$e^{5xi} = cos(5x) + i*sin(5x) = (f + ig)^5$

$(f + ig)^5 = f^5 + 5f^4gi - 10f^3g^2 - 10f^2g^3i + 5fg^4 + g^5$

Now we take the real parts:

$cos(5x) = f^5 - 10f^3g^2 + 5fg^4$

Using $g^2 = 1 - f^2$ throughout:

$cos(5x) = 16f^5 - 20f^3 + 5f$

We know that:

$cos(45-30) = cos(45)cos(30) + sin(45)sin(30) = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

This is clearly irrational, therefore:

$cos(15) = 16cos(3) - 20cos(3) + 5cos(3) = \frac{\sqrt{3} + 1}{2\sqrt{2}}$

If $cos(3)$ was rational, $16cos(3) - 20cos(3) + 5cos(3)$ would also be rational, which is a contradiction.

Finally:

Using $cos(3x) = 4cos^3\!(x) - 3cos(x)$

We get:

$cos(3) = 4cos^3\!(1) - 3cos(1)$

If $cos(3)$ was rational, then $4cos^3\!(1) - 3cos(1)$ would also be rational, and if $cos(1)$ was rational, then the statement above would also be rational, we have thus arrived at a contradiction:

$cos(1)$ is irrational.

$Q.E.D.$

I have to go to class someone else do $sin(1)$ or I'll do it later.

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