Revision as of 20:58, 20 January 2024

Shortcut:

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:

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

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

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

We can also solve for other expressions:

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

$\cos^2=\frac{\cos(2)+1}{2}$

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. I hope this helps you. $\sin^2(\alpha+\beta)=(\sin\alpha\cos\beta+\cos\alpha\sin\beta)(\sin\alpha\cos\beta+\cos\alpha\sin\beta)=\sin^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta+2\sin\alpha\sin\beta\cos\alpha\cos\beta$

and

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

There's something we can cancel.

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

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

If $f()=\sin\cos$ , then it simplifies to

$(f(\alpha)-f(\beta))^2$

Notice $f\left(\frac{k\pi}{2}\right)=0$ . If we let $\beta=0$ :

$\cos(2\alpha+k\pi)=\sin^2\alpha\cos^2\alpha$

Halved angles

Starting with the identities from the double section:

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

$\cos^2=\frac{1+\cos(2)}{2}$

We take the square root to obtain:

$\sin=\pm\sqrt{\frac{1-\cos(2)}{2}}$

$\cos=\pm\sqrt{\frac{1+\cos(2)}{2}}$

For tangent:

$\tan=\frac{\sin}{\cos}=\frac{\pm\sqrt{\frac{1-\cos(2)}{2}}}{\pm\sqrt{\frac{1+\cos(2)}{2}}}=\pm\sqrt{\frac{1-\cos(2)}{1+\cos(2)}}$

There are two nice variations to know.

$\pm\sqrt{\frac{1-\cos(2)}{1+\cos(2)}}\times\sqrt{\frac{1-\cos(2)}{1-\cos(2)}}=\pm\frac{1-\cos(2)}{\sin(2)}$

$\pm\sqrt{\frac{1-\cos(2)}{1+\cos(2)}}\times\sqrt{\frac{1+\cos(2)}{1+\cos(2)}}=\pm\frac{\sin(2)}{1+\cos(2)}$

Triple angles and more

Triple sums

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

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

$\tan(\alpha+\beta+\gamma)=\tan(\alpha+(\beta+\gamma))$ $=\frac{\tan\alpha+\tan(\beta+\gamma)}{1-\tan\alpha\tan(\beta+\gamma)}$ $=\frac{\tan\alpha+\frac{\tan\beta+\tan\gamma}{1-\tan\beta\tan\gamma}}{1-\tan\alpha\frac{\tan\beta+\tan\gamma}{1-\tan\beta\tan\gamma}}$ $=\frac{\frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\beta\tan\gamma}}{\frac{1-\tan\beta\tan\gamma-\tan\alpha\tan\beta-\tan\alpha\tan\gamma}{1-\tan\beta\tan\gamma}}$ $=\frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\beta\tan\gamma-\tan\alpha\tan\gamma-\tan\alpha\tan\beta}$

Triple angles

$\sin 3\theta=3\cos^2\theta\sin\theta-\sin^3\theta=3\sin\theta-3(1-\cos^2\theta)\sin\theta-\sin^3\theta=3\sin\theta-4\sin^3\theta$

$\cos 3\theta=\cos^3\theta-3\cos\theta\sin^2\theta=\cos^3\theta+3(1-\sin^2\theta)\cos\theta-3\cos\theta=4\cos^3\theta-3\cos\theta$

$\tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$

Third angles

Let $\sin\theta = x$ and $\sin 3\theta = y$ . We get this depressed cubic:

$0=3x-4x^3-y$

First, divide both sides by -4 and rearrange: $x^3-\frac{3}{4}x+y=0$ . The discriminant $\Delta = \frac{y^2}{4}-\frac{1}{64}=\frac{16y^2-1}{64}$

Then, $u=\frac{-4y\pm\sqrt{16y^2-1}}{8}$

The solutions are $\sqrt[3]{-4y+\sqrt{16y^2-1}}$ , $\frac{\sqrt[3]{-4y+\sqrt{16y^2-1}}+\sqrt[3]{4y-\sqrt{16y^2-1}}}{2}$ , and $\sqrt[3]{-4y-\sqrt{16y^2-1}}$ .

A tiny adjustment gives us the cosine third-angle formulas:

$\sqrt[3]{4y+\sqrt{16y^2-1}}$ , $\frac{\sqrt[3]{4y+\sqrt{16y^2-1}}+\sqrt[3]{4y-\sqrt{16y^2-1}}}{2}$ , and $\sqrt[3]{4y-\sqrt{16y^2-1}}$ .

For tangent:

$\frac{\sqrt[3]{-4y+\sqrt{16y^2-1}}}{\sqrt[3]{4y+\sqrt{16y^2-1}}}$ , $\frac{\sqrt[3]{-4y+\sqrt{16y^2-1}}+\sqrt[3]{-4y-\sqrt{16y^2-1}}}{\sqrt[3]{4y+\sqrt{16y^2-1}}+\sqrt[3]{4y-\sqrt{16y^2-1}}}$ , and $\frac{\sqrt[3]{-4y-\sqrt{16y^2-1}}}{\sqrt[3]{4y-\sqrt{16y^2-1}}}$

All identities

Definition

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

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

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

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

Symmetric

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

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

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

Pythagorean

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

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

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

Sum

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

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

$\tan(\alpha + \beta)=\frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

Double

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

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

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

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

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

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

Sum $\iff$ Product

$\theta =\alpha +\beta$

$\phi =\alpha -\beta$

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

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

$2\sin\alpha\cos\beta\iff\sin\theta+\sin\phi$

$2\sin\alpha\sin\beta\iff\cos\phi-\cos\theta$

$2\cos\alpha\cos\beta\iff\cos\theta+\cos\phi$

Product

$\cos(2\alpha+k\pi)=\sin^2\alpha\cos^2\alpha$

Halves

$\sin\theta=\pm\sqrt{\frac{1-\cos(2\theta)}{2}}$

$\cos\theta=\pm\sqrt{\frac{1+\cos(2\theta)}{2}}$

$\tan\theta=\pm\sqrt{\frac{1-\cos(2\theta)}{1+\cos(2\theta)}}$

$\tan\theta=\pm\frac{1-\cos(2\theta)}{\sin 2\theta}$

$\tan\theta=\pm\frac{\sin 2\theta}{1+\cos(2\theta)}$

3 Sums

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

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

$\tan(\alpha+\beta+\gamma)=\frac{\tan\alpha+\tan\beta+\tan\gamma-\tan\alpha\tan\beta\tan\gamma}{1-\tan\beta\tan\gamma-\tan\alpha\tan\gamma-\tan\alpha\tan\beta}$

Triple

$\sin 3\theta=3\sin\theta-4\sin^3\theta$

$\cos 3\theta=4\cos^3\theta-3\cos\theta$

$\tan 3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$

Thirds

$\sin\theta=\sqrt[3]{-4\sin 3\theta+\sqrt{16\sin^2(3\theta)-1}},\frac{\sqrt[3]{-4\sin 3\theta+\sqrt{16\sin^2(3\theta)-1}}+\sqrt[3]{-4\sin 3\theta-\sqrt{16\sin^2(3\theta)-1}}}{2}, \\ \text{or }\sqrt[3]{4\sin 3\theta-\sqrt{16\sin^2 3\theta-1}}$

$\cos\theta=\sqrt[3]{4\cos 3\theta+\sqrt{16\cos^2(3\theta)-1}},\frac{\sqrt[3]{4\cos 3\theta+\sqrt{16\cos^2(3\theta)-1}}+\sqrt[3]{4\cos 3\theta-\sqrt{16\cos^2(3\theta)-1}}}{2}, \\ \text{or }\sqrt[3]{4\cos 3\theta-\sqrt{16\cos^2 3\theta-1}}$

$\tan\theta=\frac{\sqrt[3]{-4\sin 3\theta+\sqrt{16\sin^2(3\theta)-1}}}{\sqrt[3]{4\cos 3\theta+\sqrt{16\cos^2(3\theta)-1}}},\frac{\sqrt[3]{-4\sin 3\theta+\sqrt{16\sin^2(3\theta)-1}}+\sqrt[3]{-4\sin 3\theta-\sqrt{16\sin^2(3\theta)-1}}}{\sqrt[3]{4\cos 3\theta+\sqrt{16\cos^2(3\theta)-1}}+\sqrt[3]{4\cos 3\theta-\sqrt{16\cos^2(3\theta)-1}}}, \\ \text{or }\frac{\sqrt[3]{-4\sin 3\theta-\sqrt{16\sin(3\theta)^2-1}}}{\sqrt[3]{4\cos 3\theta-\sqrt{16\cos^2(3\theta)-1}}}$

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-{{shortcut|[[Trig identities]]}}
+{{shortcut|[[Trig identity proof]]}}
 =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.

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Difference between revisions of "Proofs of trig identities"

Revision as of 20:58, 20 January 2024

Contents

Introduction

Symmetric identities

Pythagorean identities

Conclusion

Angle addition and subtraction

Double angle formulas

Variations

Sum to Product to Sum

Chapter 1

Chapter 2

Chapter 3

Bonus: Product identity

Halved angles

Triple angles and more

Triple sums

Triple angles

Third angles

All identities

Definition

Symmetric

Pythagorean

Sum

Double

Sum Product

Product

Halves

3 Sums

Triple

Thirds

See also

Sum $\iff$ Product