Author Topic: Non-transcendental trig values (Read 1148 times) Tweet Share

kamil9876 · « **on:** February 26, 2009, 11:05:58 pm »

Upon noticing that values such as $\ cos(\pi /8)$ are algebraic(non-transcendental(can be written using natural numbers, fractions and roots)) because of compound angle formulas/double angle formulas I also realised that $\ cos(\frac{\pi}{2^k})$ can also be solved by using the double angle formula repitively and hence obtaining an algebraic answer since it would just involve manipulating surds. A formula for triple angle, quadruple angle etc. can also be derived using compound angle formulas. Does this implu that values of the sort $\ cos(\frac{\pi}{k^n})$ can be found and are hence algebriac since the formulas are polynomial in nature?. I'm wondering if a combination of all these formulas can be used to find and hence prove that all values of the form $cos(q\pi)$ are algebraic, where q is a rational number.

An interesting fact: refer to diagram attatched; 666 degrees is of the form $q\pi$ and $\phi /2$ (half of golden ratio) is obviously algebraic.

/0 · « **Reply #1 on:** February 27, 2009, 03:30:09 pm »

$(\mbox{cis}\theta)^\frac{p}{q}=(\cos\theta+i\sin\theta)^\frac{p}{q}$

$\Rightarrow \mbox{cis}\left(\frac{p}{q}\theta\right) = (\cos\theta+i\sin\theta)^\frac{p}{q}$

$\cos\left(\frac{p}{q}\theta\right) = Re\left[(\cos\theta+i\sin\theta)^\frac{p}{q}\right]$

i.e. $\cos\left(\frac{p}{q}\theta\right)$ can be expressed as a finite combination of powers of other algebraic terms so long as $\frac{p}{q}$ is rational.

kamil9876 · « **Reply #2 on:** February 27, 2009, 05:08:07 pm »

thank's bro, very clever stuff

Wonder if there is analogous proof which doesn't involve comlpex numbers. But very nice stuff

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Author Topic: Non-transcendental trig values (Read 1148 times) Tweet Share

kamil9876

Non-transcendental trig values

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