Revision as of 15:30, 20 January 2024

Introduction

$\sin$ and $\cos$ are easy to define. I prefer the unit circle definition as it makes these proofs easier to understand. Next, we define some other functions:

$\tan = \frac{\sin}{\cos}$

$\cot = \frac{\cos}{\sin}$

$\sec = \frac{1}{\cos}$

$\csc = \frac{1}{\sin}$

Note: I've omitted $\theta$ because it's unnecessary and might clog things up a little.

With a bit of ingenuity, we can create the following diagram:

$[asy] import olympiad; markscalefactor = 1/96; real d = radians(40); unitsize(72); pair O = (0,0); draw(circle(O,1)); dot(O); label("O",O,dir(180+degrees(d)/2)); label("$\theta$",shift(dir(degrees(d)/2)/5)*O,dir(degrees(d)/2)); pair G = (0,1); label("G",G,N); pair A = (cos(d),0); label("A",A,S); pair B = (cos(d),sin(d)); label("B",B,dir(135+degrees(d))); pair C = (1,0); label("C",C,SE); pair D = (1,tan(d)); label("D",D,N); pair E = (1/tan(d),0); label("E",E,SE); pair F = (1/tan(d),1); label("F",F,N); pair G = (0,1); label("G",G,N); draw(D--O--C--D--B--A--E--F--G--O--F); draw(rightanglemark(G,O,C)); label("$\cos \theta$",O--A); label("$\sin \theta$",B--A); label("1",B--O); draw(shift(dir(270)/24)*brace(C,O)); label("$\cot \theta$",shift(dir(270)/4)*brace(E,O),S); draw(shift(dir(d+90)/24)*brace(O,D)); label("$\sec \theta$",shift(dir(degrees(d)+90)/24)*brace(O,D),dir(degrees(d)+90)); draw(shift(dir(270)/4)*brace(E,O)); label("1",shift(dir(270)/24)*brace(C,O),S); draw(shift(dir(270)/4)*O--shift(dir(270)/24)*O); draw(shift(dir(270)/4)*E--shift(dir(270)/24)*E); label("1",E--F,SE); label("$\tan \theta$",C--D); draw(shift(dir(degrees(d)+90)/4)*brace(O,F)); label("$\csc \theta$",shift(dir(degrees(d)+90)/4)*brace(O,F),dir(degrees(d)+90)); draw(shift(dir(degrees(d)+90)/4)*O--shift(dir(degrees(d)+90)/24)*O); draw(shift(dir(degrees(d)+90)/4)*F--shift(dir(degrees(d)+90)/24)*F); [/asy]$

We can note that the functions are correct by similar triangles.

Pythagorean identities

Pythagorean identities are easy and there's no algebra involved. In fact, the name Pythagorean is a giveaway of what we should do!

$\cos^2+\sin^2=1$

The proof here is very straightforward. We use the pythagorean theorem on $\triangle OAB$ giving us $OA^2+AB^2=OB^2$ or $\sin^2+\cos^2=1^2$ .

$\tan^2+1=\sec^2$

Same story here. Applying pythagorean to $\triangle OCD$ gives us $OC^2+CD^2=OD^2$ or $\tan^2+1^2=\sec^2$ .

$1+\cot^2=\csc^2$

Same. Pythagorean on $\triangle OEF$ gives $OE^2+EF^2=OF^2$ or $1^2+\cot^2=\csc^2$ .

@@ Line 32: / Line 32: @@
 label("B",B,dir(135+degrees(d)));
 pair C = (1,0);
-label("C",C,E);
+label("C",C,SE);
 pair D = (1,tan(d));
 label("D",D,N);
@@ Line 72: / Line 72: @@
 The proof here is very straightforward. We use the pythagorean theorem on <math>\triangle OAB</math> giving us <math>OA^2+AB^2=OB^2</math> or <math>\sin^2+\cos^2=1^2</math>.
-==<math>\tan^2+1=sec^2</math>==
+==<math>\tan^2+1=\sec^2</math>==
-Same story here. Applying pythagorean to <math>\triangle OCD</math>
+Same story here. Applying pythagorean to <math>\triangle OCD</math> gives us <math>OC^2+CD^2=OD^2</math> or <math>\tan^2+1^2=\sec^2</math>.
+==<math>1+\cot^2=\csc^2</math>==
+Same. Pythagorean on <math>\triangle OEF</math> gives <math>OE^2+EF^2=OF^2</math> or <math>1^2+\cot^2=\csc^2</math>.

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