Re: Trigonometry

On Sun, 16 Dec 2007 03:46:37 -0800 (PST), Deep <deepkdeb@xxxxxxxxx>
wrote:

On Dec 16, 2:09 am, quasi <qu...@xxxxxxxx> wrote:
On Sat, 15 Dec 2007 15:39:11 -0800 (PST), Deep <deepk...@xxxxxxxxx>
wrote:



Consider the following situations under the given conditions.

Sin D = V^(1/2) (1) Sin kD = F^(k/
2) (2)

Cos D = U^(1/2) (3) Cos kD = E^(k/
2) (4)

Conditions: V, F, U, E are rational but not perfect squares. V, F, U,
E < 1; 0 < D < pi/2;
odd k > 3

Assertion: (1), (2), (3) and (4) can simultaneously be satisfied

For any particular k, (1), (2), (3), (4) (together with your specified
conditions) probably can't be simultaneously satisfied.

Proving it for k=5, is doable, I think, but not elementary -- it
reduces to the question of whether or not a certain elliptic curve has
a rational point.

For a given k > 5, if the claim is true, finding a proof is unlikely.
The associated diophantine equation is just too difficult to analyze.

As far as the general case (with k left unknown), it may not even be
true, but if it is, forget it -- you'll never prove it.

only if V = F and U = E

The exceptional conditions you specify above are silly. It's not
possible to have V = F and U = E. That's easy to prove.

quasi
***
Kindly prove your assertion that V = F and U = E are not possible.
***

Suppose V = F and U = E, subject to your stated conditions.

By hypothesis, 0 < D < Pi/2.

Hence

sin(D) = V^(1/2) => 0 < V < 1

cos(D) = U^(1/2) => 0 < U < 1

Then

sin^2(D) + cos^2(D) = 1 => V + U = 1

sin^2(k*D) + cos^2(k*D) = 1 => F^k + E^k = 1 => V^k + U^k = 1

But V^k < V and U^k < U, so V^k + U^k < V + U, contradiction.

quasi
.

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