Math Help Forum - View Single Post

Aryth · #3 February 13th, 2008, 12:27 PM

For $sin(1) = irrational$

Using the math above and taking the imaginary parts:

$isin(5x) = 5f^4gi - 10f^2g^3i$

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

$sin(5x) = 5g^9 - 10g^7 + 15g^5 - 5g$

We know that:

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

That is clearly irrational.

$sin(15) = 5sin^9\!(3) - 10sin^7\!(3) + 15sin^5\!(3) - 5sin(3) = \frac{\sqrt{3} + 1}{2}$

sin(3) is irrational since sin(15) is irrational.

Finally:

$sin(3x) = 3sin(x) - 4sin^3\!(x)$

Therefore:

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

Therefore, since sin(3) is irrational, sin(1) is also irrational.

$Q.E.D.$

Using this as logic:

$sin(nx) \ is \ irrational \Rightarrow sin(x) \ is \ also \ irrational$

Same with cos(nx)

The Following 2 Users Say Thank You to Aryth For This Useful Post:
colby2152, qspeechc
$Donate to MHF$