YATPOPT, Yet Another Trigonometric Proof of the Pythagorean Theorem

John Molokach

Let $\alpha$ be an acute angle of a right triangle with legs $a$ and $b,$ and hypotenuse $c,$ such that $\cos\alpha = a/c$ and $\sin\alpha = b/c.$ Then

$\cos\alpha=\cos(2\alpha-\alpha)=\cos 2\alpha\cos\alpha-\sin 2\alpha\sin\alpha,$

implying

$\displaystyle\cos 2\alpha=1-\frac{\sin 2\alpha\sin\alpha}{\cos\alpha}.$

It is easily shown without using the Pythagorean theorem that $\sin 2\alpha = 2\sin\alpha\cos\alpha$ by taking an isosceles triangle with base $2a$ and legs $c,$ and drawing an altitude from a base angle to a leg. Therefore,

$\displaystyle\cos 2\alpha = 1-\frac{\sin 2\alpha\sin\alpha}{\cos\alpha}=1-2\sin^{2}\alpha.$

But also

$\cos 2\alpha =\cos^{2}\alpha-\sin^{2}\alpha.$

Now equating the two expressions for $\cos 2\alpha,$ we can write

$1-2\sin^{2}\alpha=\cos^{2}\alpha-\sin^{2}\alpha,$

which is $\cos^{2}\alpha+\sin^{2}\alpha=1.$

|Pythagorean Theorem| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2017 Alexander Bogomolny

 62680654

Search by google: