Pythagorean Identities
The Pythagorean Theorem
This is one of the older and more famous theorems in mathematics. It states that given a right triangle, the sum of the square of the legs equals the square of the hypotenuse. When I say “square” I mean squared in the algebra sense, and a square in the geometric sense.
There are literally hundreds of proofs of the theorem (this book lists 371 of them and Cut the Knot also has many beautiful proofs). I will provide what I think is the simplest proof.
We can see that in both diagrams, all we’ve done is change the positions of the red triangles. Thus, the blue regions have equal areas. In the left diagram, the blue region has area $c^2$. In the right diagram, the blue region has $a^2 + b^2$. Thus, we’ve proven the Pythagorean Theorem.
You can also prove it algebraically from these diagrams. Notice that the area of the blue region equals the area of the entire square minus the area of the red. Written algebraically,
\[\begin{align} &c^2 = (a + b)^2 - 4 \cdot \frac{1}{2} ab \\[10pt] &c^2 = a^2 + 2ab + b^2 - 2ab \\[10pt] &c^2 = a^2 + b^2 \end{align}\]Pythagoras Identities Derivations
Recall the canonical diagram for trigonometry
We see that we have a right triangle, so let’s use Pythagoras!
\[\sin^2 \theta + \cos^2 \theta = 1\]While simple, this is probably the most commonly used identity in trigonometry and higher mathematics. From this identity, we can derive a few others
\[\begin{align} &\sin^2 \theta + \cos^2 \theta = 1 \\[10pt] &\frac{1}{\cos^2 \theta} [ \sin^2 \theta + \cos^2 \theta ] = \frac{1}{\cos^2 \theta} [1] \\[10pt] &\tan^2 \theta + 1 = \sec^2 \theta \end{align}\]Likewise
\[\begin{align} &\sin^2 \theta + \cos^2 \theta = 1 \\[10pt] &\frac{1}{\sin^2 \theta} [ \sin^2 \theta + \cos^2 \theta ] = \frac{1}{\sin^2 \theta} [1] \\[10pt] &1 + \cot^2 \theta = \csc^2 \theta \end{align}\]