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Difference between revisions of "Proofs of trig identities"

(Created page with "=Introduction= <math>\sin</math> and <math>\cos</math> are easy to define. I prefer the unit circle definition as it makes these proofs easier to understand. Next, we define s...")
 
Line 10: Line 10:
  
 
<math>\csc = \frac{1}{\sin}</math>
 
<math>\csc = \frac{1}{\sin}</math>
 +
 +
Note: I've omitted <math>\theta</math> because it's unnecessary and might clog things up a little.
  
 
With a bit of ingenuity, we can create the following diagram:
 
With a bit of ingenuity, we can create the following diagram:
Line 45: Line 47:
 
label("1",B--O);
 
label("1",B--O);
 
draw(shift(dir(270)/24)*brace(C,O));
 
draw(shift(dir(270)/24)*brace(C,O));
−
label("$\cot \theta$",shift(dir(270)/24)*brace(C,O),S);
+
label("$\cot \theta$",shift(dir(270)/4)*brace(E,O),S);
 
draw(shift(dir(d+90)/24)*brace(O,D));
 
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));
 
label("$\csc \theta$",shift(dir(degrees(d)+90)/24)*brace(O,D),dir(degrees(d)+90));
 
draw(shift(dir(270)/4)*brace(E,O));
 
draw(shift(dir(270)/4)*brace(E,O));
−
label("1",shift(dir(270)/4)*brace(C,O),S);
+
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)*O--shift(dir(270)/24)*O);
 
draw(shift(dir(270)/4)*E--shift(dir(270)/24)*E);
 
draw(shift(dir(270)/4)*E--shift(dir(270)/24)*E);
−
label("1",E--F,SW);
+
label("1",E--F,SE);
−
label("$\tan \theta$",E--D);
+
label("$\tan \theta$",C--D);
 
draw(shift(dir(degrees(d)+90)/4)*brace(O,F));
 
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));
 
label("$\sec \theta$",shift(dir(degrees(d)+90)/4)*brace(O,F),dir(degrees(d)+90));
Line 59: Line 61:
 
draw(shift(dir(degrees(d)+90)/4)*F--shift(dir(degrees(d)+90)/24)*F);
 
draw(shift(dir(degrees(d)+90)/4)*F--shift(dir(degrees(d)+90)/24)*F);
 
</asy>
 
</asy>
 +
 +
We can note that the functions are correct by similar triangles.
 +
 +
=Pythagorean identities=
 +
 +
Pythagorean identities are easy and there's no algebra involved.
 +
 +
==<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>.

Revision as of 15:23, 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,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]

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

Pythagorean identities

Pythagorean identities are easy and there's no algebra involved.

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

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

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