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Given a right triangle, and its two legs $a$ and $b$, I know of two ways to find the hypotenuse $c$:

Equation 1: $$c=\sqrt{a^2+b^2}$$ Equation 2: $$c=\frac{b}{\sin{(\tan^{-1}{(\frac{b}{a})})}}$$ The first equation is just the pythagorean theorem. The second equation represents finding one of the angles from the legs, and using the definition of $\sin{\theta}=\frac{opp}{hyp}$ to find the hypotenuse. I also know that the expansions of $\sin{x}$ and $\tan^{-1}{x}$ are: $$\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}\ldots$$ $$\tan^{-1}{x}=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}\ldots$$ My question is, can one show that the second equation "reduces" to the first equation based on the series expansions of $\sin$ and $\tan^{-1}$ (assuming positive values of $a$ and $b$)? Is this possible?

FuzzyCat444
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  • Equation 2 is a consequence of the definitions of the sine and arc tangent functions (if you define them by using ratios of sides of right triangles), and does not actually involve the Pythagorean Theorem. – David K Feb 17 '18 at 22:57
  • @DavidK But they can be defined as series too. – FuzzyCat444 Feb 17 '18 at 23:10

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Are you willing to use the fact that $$ \sin(\tan^{-1} z) = \frac{z}{\sqrt{z^2+1}} ? $$ I can't tell from your question if you would be interested in reducing the second equation to the first by a means other than series. If you only care about series, I can delete this answer.

Matthew Conroy
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  • If I am not mistaken, that equation is a direct consequence of the pythagorean theorem. I think an answer involving series would be more satisfying, but this answer is definitely intuitive. – FuzzyCat444 Feb 17 '18 at 21:41