Talk:Pythagoras's Theorem

There seems to be a bit of circular logic in the algebric proof: The Equivalence of Definitions of Sine and Cosine uses the fact that the Sum of Squares of Sine and Cosine property is satisfied by the geometric definition. The geometric proof of this property uses (unsurprisingly) The Pythagoras's Theorem.--Nngnna (talk) 16:57, 23 July 2019 (EDT)


 * The inadequacy of Equivalence of Definitions of Sine and Cosine has already been raised on the talk page for that page. But in any case, this should not be a problem. There are many proofs of Pythagoras's Theorem here, only one of which uses the definitions of sine and cosine. So we can prove Pythagoras's Theorem using e.g. Proof 1, then from that we have Sum of Squares of Sine and Cosine, and from there Equivalence of Definitions of Sine and Cosine follows. From there we can show the algebraic proof. --prime mover (talk) 17:42, 23 July 2019 (EDT)
 * You are correct, we are not at danger of not proving pythagoras's Theorem. But if we need to use a different proof it means the Algebric one is useless.--Nngnna (talk) 00:40, 24 July 2019 (EDT)


 * It's a proof. It stands alone. It's valid. --prime mover (talk) 01:21, 24 July 2019 (EDT)