Feedback - Random trig proofs

This blog post was once used for a thing. It is now entirely repurposed to serve as a trig proof only. Enjoy.

Let ΔxyR be a right triangle in the x y plane where x and y each sit on the respective x and y axes.
Thus, by pythagorean theorem, x² + y² = R²
Converting to polar, (x = Rcos(θ), y = Rsin(θ))
(Rcos(θ))² + (Rsin(θ))² = R²cos(θ)² + R²sin(θ))² = R²
Dividing through by R² we get
cos(θ)² + sin(θ)² = 1
Q.E.D.

Edit: Now consolidating the proof from a later blog into this one. Here's a Pythagorean theorem proof:

Let there be two squares of dimensions a + b. The first square is made of four identical right triangles with side lengths a and b, a square with side length a, and a square with side length b. The second square is made of four identical right triangles with side lengths a and b and a square of side length c whose sides also comprise the hypotenuse of each of the four right triangles.

¹

Square one may be represented as A₁ = (a+b)² = 4*(½*a*b)+a²+b²
and square two may be represented as A₂ = (a+b)² = 4*(½*a*b)+c²
Because A₁ = A₂, we may set 4*(½*a*b)+a²+b²=4*(½*a*b)+c²
And when we subtract 4*(½*a*b) from both sides, we get a²+b²=c²

Thus, a²+b²=c²
Q.E.D.

RQK Note: The law of cosines a²+b²-2abcos(θ)=c²may be proven in a similar fashion. It should be noted that Pythagorean theorem is exactly when cos(90°) = 0

¹ Angie Head, University of Georgia

Comments

[Mac349]

You assume that R is nonzero. You failed to account for triangles that have side lengths of 0.

[RQK]

4257741 Ah but you see, ΔxyR with side length 0 is just a point, but even then the proof does not depend on side length, merely that the legs of the triangle may be expressed in terms of R, the hypotenuse.

[Mellow]

4257761 This proof more than validates the existance of two blogs! The Trigonometry Godtm is pleased!

[SIGAWESOME]

You and your fancy maths!
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