Other Trigonometric Fundamental Formulae

From the angle sum identities for sine and cosine, we get the following: $$ \begin{eqnarray*} \sin(2\alpha) &=& 2\sin(\alpha)\cos(\alpha) \\ \cos(2\alpha) &=& \cos^2(\alpha) - \sin^2(\alpha) \end{eqnarray*} $$ and $$ \begin{eqnarray*} \sin(x) &=& 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) \\ \cos(x) &=& \cos^2\left(\frac{x}{2}\right) - \sin^2\left(\frac{x}{2}\right) \end{eqnarray*} $$ Then, from the above the following holds: \[ \sin^2(x) + \cos^2(x) = \left( \cos^2\left(\frac{x}{2}\right) + \sin^2\left(\frac{x}{2}\right) \right)^2 \] Repeating this process we have the following: $$ \begin{eqnarray*} \sin^2(x) + \cos^2(x) &=& \left( \cos^2\left(\frac{x}{2}\right) + \sin^2\left(\frac{x}{2}\right) \right)^2 \\ &=& \left( \cos^2\left(\frac{x}{4}\right) + \sin^2\left(\frac{x}{4}\right) \right)^4 \\ &\vdots& \\ &=& \left( \cos^2\left(\frac{x}{2^2}\right) + \sin^2\left(\frac{x}{2^n}\right) \right)^{2^n} \end{eqnarray*} $$ It is not difficult to prove the above using mathematical induction. Because all results on the right hand side are equal for every \( n \), what is the result when \( n \) approaches infinity? We shall prove the following using L'Hôpital's Rule: \[ \sin^2(x) + \cos^2(x) = \lim_{n\rightarrow\infty} \left( \cos^2\left(\frac{x}{2^2}\right) + \sin^2\left(\frac{x}{2^n}\right) \right)^{2^n} = 1\] We will also prove the same result with an elementary way without using L'Hôpital's Rule. Therefore, the Pythagorean Identity is established without using the Pythagorean Theorem and the Pythagorean Identiy

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