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There's not much trigonometry in this proof, just the fact that sine has period 2pi is zero at pi. Any function that has these two properties could directly rep
by bonoboTP 10y ago
There's not much trigonometry in this proof, just the fact that sine has period 2pi is zero at pi. Any function that has these two properties could directly replace sine in this proof.
And actually any periodic function that is zero somewhere could be used in a tiny bit modified version of this proof.
- yequalsx 10y agoThat is true but in Trigonometry Sine the function one studies and so using this result in a trigonometry course is nice.
- JadeNB 10y ago> And actually any periodic function that is zero somewhere could be used in a tiny bit modified version of this proof. I think that, for the modification to be tiny, you need something like: there is an interval on which the function is positive, and on the boundary of which it vanishes. (Some sign condition is necessary; taking the constant function at 0 doesn't work! Of course, this particular condition is automatically satisfied for any continuous, periodic function that vanishes somewhere, but is not identically 0.)