Trigonometric Proof of Pythagorean Theorem

This proof is similar to Bhaskara's second proof, but at the end we also derive an important trigonometric identity. In the diagram, A, B, C are the lengths of the sides of a right triangle. The lower case, a, represents the angle opposite of side A.

From the definitions of the common trigonometric functions of sin(a) and cos(a), sin(a) = A/C and cos(a) = B/C. These two equations can be rewritten as A = C sin(a) and B = C cos(a).

On the hypotenuse side of the triangle, C can be broken into two sub-lengths, A’ and B’, which represent the projection of sides A and B, respectively, onto side C.

From the similar triangles that are formed, sin(a) = A’/A and cos(b) = B’/B. These two equations can be rewritten as A’ = A sin(a) and B’ = B cos(a).

Substituting for sin(a) and cos(b), A’= A (A/C) and B’ = B (B/C) or A’ = A²/C and B’ = B²/C.

Given that side C = A’ + B’, we get C = A²/C and B’ = B²/C. Multiplying both sides by C, we now derive the Pythagorean Theorem:

C² = A² + B².

The above proof of the Pythagorean Theorem used the trigonometric functions to derive the terms of A² and B². Alternatively, we can keep the sin(a) and cos(a) functions to show another important identity. Returning to the terms for A’ and B’, we can also express these two lengths as A’ = A sin(a) and B’ = B cos(a). Substituting for A and B in these two equations, A’ = C sin(a) sin(a) and B’ = C cos(a) cos(a). Trigonometric convention allows these terms to be stated as A’ = C sin²(a) and B’ = C cos²(a).

Substituting these last two equations into the relationship of side C being the sum of A’ and B’,

C = A’ + B’

C = C sin²(a) + C cos²(a)

Next divide both sides of the equation by C, we now have an important trigonometric identity:

1 = sin²(a) + cos²(a)