Revision as of 18:48, 22 August 2024

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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$",shift(dir(degrees(d)+90)/4)*brace(O,F),dir(degrees(d)+90)); draw(shift(dir(degrees(d)+90)/4)*O--shift(dir(degrees(d)+90)/24)*O); draw(shift(dir(degrees(d)+90)/4)*F--shift(dir(degrees(d)+90)/24)*F); [/asy]$

It has three main triangles: cos-sin-1, 1-tan-sec, and cot-1-csc.

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

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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.

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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.

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Angle addition and subtraction

$\sin(\alpha + \beta)$

When does sin appear? When does sin appear? In the first triangle, of course. Let's make a diagram!

$[asy] unitsize(216); real d = 1/cos(radians(35)); real d1 = d * cos(radians(55)); real d2 = d * sin(radians(55)); pair O = (0,0); pair A = (cos(radians(20)),0); pair B = (cos(radians(20)),sin(radians(20))); pair C = (cos(radians(20)),d2); pair D = (d1,d2); draw(O--A--B--O--D--B--O--D--C--B); dot(O); dot(B); dot(A,red); dot(C,green); dot(D,blue); label("O",O,SW); label("$\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.

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, let's use the tan-1-sec triangle.

$[asy] unitsize(216); real d = 1/cos(radians(35)); real d1 = d * cos(radians(55)); real d2 = d * sin(radians(55)); pair O = (0,0); pair A = (cos(radians(20)),0); pair B = (cos(radians(20)),sin(radians(20))); pair C = (cos(radians(20)),d2); pair D = (d1,d2); draw(O--A--B--O--D--B--O--D--C--B); dot(O); dot(B); dot(A,red); dot(C,green); dot(D,blue); label("O",O,SW); label("$\alpha$",shift(dir(10)/5)*O); label("$\beta$",shift(dir(37.5)/5)*O); label("A",A,SE,red); label("B",B,E); label("C",C,NE,green); label("D",D,dir(122.5),blue); label("1",O--A,S); label("$\tan \alpha$",A--B,E); label("$\sec \alpha$",O--B,dir(302.5)); label("$\tan \beta$",B--C,E); label(scale(0.75)*"$\tan \alpha \tan \beta$",C--D,N); label(scale(0.75)*"$\sec \alpha \tan \beta$",B--D,dir(200)); label(scale(0.75)*"$\sec \alpha \sec \beta$",D--O,dir(325)); [/asy]$

Wait, is that just the same diagram? No! the labels have changed!

Note: I did some algebra when noting that sin * sec = tan and cos * sec = 1

Looking at the diagram, the height of the new triangle is $\tan \alpha + \tan \beta$ , but the width is only $1-\tan\alpha\tan\beta$ , so we arrive at $\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$

$\sec(\alpha + \beta)$

I dunno why most people never use this, but it's right up there in the diagram. $\frac{\sec\alpha\sec\beta}{1-\tan\alpha\tan\beta}$

$\csc(\alpha + \beta)$

Hey, if you don't need this, stop reading.

We need the third triangle here. I'm going to do something weird, you'll see why when I complete the diagram (I= afly. This page was all made by afly) $[asy] unitsize(216); real d = 1/cos(radians(35)); real d1 = d * cos(radians(55)); real d2 = d * sin(radians(55)); pair O = (0,0); pair A = (cos(radians(20)),0); pair B = (cos(radians(20)),sin(radians(20))); pair C = (cos(radians(20)),d2); pair D = (d1,d2); draw(O--A--B--O--D--B--O--D--C--B); dot(O); dot(B); dot(A,red); dot(C,green); dot(D,blue); label("O",O,SW); label("$\alpha$",shift(dir(10)/5)*O); label("$\beta$",shift(dir(37.5)/5)*O); label("A",A,SE,red); label("B",B,E); label("C",C,NE,green); label("D",D,dir(122.5),blue); label(scale(0.75)*"$\cot \alpha \cot \beta$",O--A,S); label("$\cot \beta$",A--B,E); label(scale(0.75)*"$\csc \alpha \cot \beta$",O--B,dir(302.5)); label("$\cot \alpha$",B--C,E); label("1",C--D,N); label(scale(0.75)*"$\csc \alpha$",B--D,dir(200)); label(scale(0.75)*"$\csc \alpha \csc \beta$",D--O,dir(325)); [/asy]$

Can you look at the diagram? what do you have? $\csc(\alpha+\beta)=\frac{\csc\alpha\csc\beta}{\cot\alpha\cot\beta-1}$

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:

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

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

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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 $\alpha,\beta\iff\theta,\phi$ )

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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 $\beta=0$ :

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

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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.

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

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

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

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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

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

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

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

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

$\csc(\alpha + \beta)=\frac{\csc \alpha \csc \beta}{\cot \alpha + \cot \beta}$

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

Memory aids: sin is different, cos is same, tan is sum over one minus product, cot is product minus one over sum, sec is like tan but with product of sec on top, csc is like cot but with product of csc on top.

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

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

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

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

@@ Line 233: / Line 233: @@
 label(scale(0.75)*"$\csc \alpha \csc \beta$",D--O,dir(325));
 </asy>
+Can you look at the diagram? what do you have? <math>\csc(\alpha+\beta)=\frac{\csc\alpha\csc\beta}{\cot\alpha\cot\beta-1}</math>
 =Double angle formulas=

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Difference between revisions of "Proofs of trig identities"

Revision as of 18:48, 22 August 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

Sum

Double

Sum Product

Product

Halves

3 Sums

Triple

Thirds

See also

Sum $\iff$ Product