Difference between revisions of "Proofs of trig identities"
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=Pythagorean identities= | =Pythagorean identities= | ||
| − | Pythagorean identities are easy and there's no algebra involved. | + | Pythagorean identities are easy and there's no algebra involved. In fact, the name Pythagorean is a giveaway of what we should do! |
==<math>\cos^2+\sin^2=1</math>== | ==<math>\cos^2+\sin^2=1</math>== | ||
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>. | 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>. | ||
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| + | ==<math>\cos^2+\sin^2=1</math>== | ||
Revision as of 15:26, 20 January 2024
Contents
Introduction
and 
are easy to define. I prefer the unit circle definition as it makes these proofs easier to understand.
Next, we define some other functions:
Note: I've omitted 
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,E); 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("$\csc \theta$",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("$\sec \theta$",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]](http://latex.artofproblemsolving.com/1/0/8/1089cc23fab235b2d3fb44680116557b8dfacd16.png)
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!
The proof here is very straightforward. We use the pythagorean theorem on 
giving us 
or 
.
