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Difference between revisions of "Proofs of trig identities"

m (→‎Double angle formulas)
m (→‎Sum to Product to Sum)
Line 189: Line 189:
 
<math>-\cos(\alpha -\beta)=-\cos\alpha\cos\beta-\sin\alpha\sin\beta</math>
 
<math>-\cos(\alpha -\beta)=-\cos\alpha\cos\beta-\sin\alpha\sin\beta</math>
  
−
So <math>\cos(\alpha+\beta)-cos(\alpha-\beta)=-2\sin\alpha\sin\beta</math>
+
So <math>\cos(\alpha+\beta)-\cos(\alpha-\beta)=-2\sin\alpha\sin\beta</math>
  
 
Or, <math>2\sin\alpha\sin\beta\iff\cos\phi-\cos\theta</math> (same conversions of <math>\alpha,\beta\iff\theta,\phi</math>)
 
Or, <math>2\sin\alpha\sin\beta\iff\cos\phi-\cos\theta</math> (same conversions of <math>\alpha,\beta\iff\theta,\phi</math>)
Line 198: Line 198:
 
<math>\cos(\alpha -\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta</math>
 
<math>\cos(\alpha -\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta</math>
  
−
So <math>\cos(\alpha+\beta)+cos(\alpha-\beta)=2\cos\alpha\cos\beta</math>
+
So <math>\cos(\alpha+\beta)+\cos(\alpha-\beta)=2\cos\alpha\cos\beta</math>
  
 
Or, <math>2\cos\alpha\cos\beta\iff\cos\theta+\cos\phi</math> (same conversions of <math>\alpha,\beta\iff\theta,\phi</math>)
 
Or, <math>2\cos\alpha\cos\beta\iff\cos\theta+\cos\phi</math> (same conversions of <math>\alpha,\beta\iff\theta,\phi</math>)

Revision as of 17:41, 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.

Symmetric identities

If we draw a few copies of the triangle, we get:

$\sin()=\cos(90-)=-\cos(90+)=\sin(180-)=-\sin(180+)=\cos(270-)=-\cos(270+)=-\sin(-)$

$\cos()=\sin(90-)=\sin(90+)=-\cos(180-)=-\cos(180+)=\sin(270-)=-\sin(270+)=\cos(-)$

$\tan()=\cot(90-)=-\cot(90+)=-\tan(180-)=\tan(180+)=\cot(270-)=-\cot(270+)=-\tan(-)$

The other three can be derived by taking the reciprocals of these three.

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

Conclusion

Even though with the first one and the definitions, we can make the rest from algebra, having a geometric meaning is nice when we want to know what it actually means.

Angle addition and subtraction

[asy] unitsize(216); real d = 1/cos(radians(35)); real d1 = d * cos(radians(55)); real d2 = d * sin(radians(55)); pair O = (0,0); pair A = (cos(radians(20)),0); pair B = (cos(radians(20)),sin(radians(20))); pair C = (cos(radians(20)),d2); pair D = (d1,d2); draw(O--A--B--O--D--B--O--D--C--B); dot(O); dot(B); dot(A,red); dot(C,green); dot(D,blue); label("O",O,SW); label("$\alpha$",shift(dir(10)/5)*O); label("$\beta$",shift(dir(37.5)/5)*O); label("A",A,SE,red); label("B",B,E); label("C",C,NE,green); label("D",D,dir(122.5),blue); label("$\cos \alpha$",O--A,S); label("$\sin \alpha$",A--B,E); label("1",O--B,dir(302.5)); label("$\frac{\cos \alpha \sin \beta}{\cos \beta}$",B--C,E); label("$\frac{\sin \alpha \sin \beta}{\cos \beta}$",C--D,N); label("$\frac{\sin \beta}{\cos \beta}$",B--D,dir(200)); label("$\frac{1}{\cos \beta}$",D--O,dir(325)); [/asy]

where $\triangle OAB \sim \triangle BCD$

The diagram illustrates the identities nicely.

$\sin(\alpha + \beta)$

The diagram shows the height of point $D$ is $\sin(\alpha)+\frac{\cos \alpha \sin \beta}{\cos \beta}$. However, the length of $OD$ is $\frac{1}{\cos\beta}$. To compensate, we must divide by $\frac{1}{\cos\beta}$ to make it the sine. After some *easy* algebra, we arrive at $\sin(\alpha+\beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha$.

$\cos(\alpha + \beta)$

The diagram says that it is $\cos(\alpha)-\frac{\sin \alpha \sin \beta}{\cos \beta}$, but we need to divide by $\frac{1}{\cos\beta}$ again. We arrive at $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$.

$\tan(\alpha + \beta)$

This time we can't get it from our diagram. We need to go back to the original definition of tangent. This is the summary of my algebra: $\tan (\alpha + \beta ) = \frac{\sin (\alpha + \beta )}{\cos (\alpha + \beta )} = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta - \sin \alpha \sin \beta} = \frac{\frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta}}{\frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\cos \alpha \cos \beta}} = \frac{\frac{\sin \alpha}{\cos \alpha} + \frac{\sin \beta}{\cos \beta}}{1 - \frac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta}} = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

Double angle formulas

This is a breeze. Just sub in for sum:

$\sin(2)=2\sin\cos$

$\cos(2)=\cos^2-\sin^2$

$\tan(2)=\frac{2\tan}{1-\tan^2}$

Variations

Since $\sin^2+\cos^2=1$, we can edit the double angle cosine formula a bit. Here are the three most helpful variants:

$\cos(2)=2\cos^2-1$

$\cos(2)=\cos^2-\sin^2$

$\cos(2)=1-2\sin^2$

Sum to Product to Sum

Our angle addition formulas look nasty. Let's try to cancel something.

Chapter 1

Let's start with the formula $\sin(\alpha + \beta)=\sin\alpha\cos\beta+\sin\beta\cos\alpha$. Both terms are symmetrical, let's try to cancel out the first one.

$\sin()=-\sin(-)$ and $\cos() = \cos(-)$ might give us an idea. $\sin(\beta -\alpha)=-\sin\alpha\cos\beta+\sin\beta\cos\alpha$

Therefore, $\sin(\beta +\alpha)+\sin(\beta -\alpha)=\sin\alpha\cos\beta+\sin\beta\cos\alpha-\sin\alpha\cos\beta+\sin\beta\cos\alpha=2\sin\beta\cos\alpha$.

To put this in a nicer form, $2\sin\alpha\cos\beta\iff\sin\theta+\sin\phi$ where:

$\theta =\alpha +\beta$

$\phi =\alpha -\beta$

$\alpha =\frac{\theta +\phi}{2}$

$\beta =\frac{\theta -\phi}{2}$

Chapter 2

We did all we could. Now let's try doing something to the other formula: $\cos(\alpha + \beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$. Let's try to cancel the first one first. $\sin$ is a lot easier to cancel than $\cos$ so we have to subtract.

$-\cos(\alpha -\beta)=-\cos\alpha\cos\beta-\sin\alpha\sin\beta$

So $\cos(\alpha+\beta)-\cos(\alpha-\beta)=-2\sin\alpha\sin\beta$

Or, $2\sin\alpha\sin\beta\iff\cos\phi-\cos\theta$ (same conversions of $\alpha,\beta\iff\theta,\phi$)

Chapter 3

Next: $\cos(\alpha + \beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$. Now let's cancel the second one.

$\cos(\alpha -\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta$

So $\cos(\alpha+\beta)+\cos(\alpha-\beta)=2\cos\alpha\cos\beta$

Or, $2\cos\alpha\cos\beta\iff\cos\theta+\cos\phi$ (same conversions of $\alpha,\beta\iff\theta,\phi$)

Bonus: Product identity

This is a special identity most people don't know about. I hope this helps you.

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