Trigonometry

This chapter provides notes on trigonometry.

Angles

If two rays originate from the same point, they form an angle.

The ray lying along the 𝑥-axis is called the initial side. The ray that "sweeps out" is called the terminal side. The distance between these two rays is an angle. When terminal side sweeps out counter-clockwise, we have a positive angle measure. And when it sweeps out clockwise, we have a negative angle measure:

If we sweep the terminal side all the way around to the initial side, we get 360∘:{360^\circ:}

If we swept a fourth of the way, we get 90∘,{90^\circ,} given that:

3604=90 \dfrac{360}{4} = 90

And if we swept half-way, we'd get 180∘{180^\circ} (this is where the notion of a line having 180∘{180^\circ} comes from):

3602=180 \dfrac{360}{2} = 180

At this point, we should probably ask: Why 360? No real reason, really. At least not mathematically. We use 360 because the Mesopotamians — a cradle of both human civilization and mathematics — employed a sexigesimal system (a system of base 60). The idea of a circle comprising 360{360} degrees stuck long after the Mesopotamians disappeared because later civilizations found the the unit incredibly helpful. 360{360} is considered a highly composite integer. That is, an integer with many divisors:

{1234568910121518202430364045607290120180360} \begin{Bmatrix} 1 & 2 & 3 & 4 & 5 \\ 6 & 8 & 9 & 10 & 12 \\ 15 & 18 & 20 & 24 & 30 \\ 36 & 40 & 45 & 60 & 72 \\ 90 & 120 & 180 & 360 \end{Bmatrix}

And with more divisors, splitting the pie "fairly" is much easier. We appreciate that ease far more when we realize that arithmetic was done with Roman numerals up until the 15th century. It's not so easy to give someone XVth of something (the division algorithm didn't exist yet). But, if we can pretend that something is a pie — 360 — it does become easy: There are XV of XXIV total, so just give them a XXIV.

The problem with degrees: They truly have no meaning. It's an entirely arbitrary choice. We could have just as easily gotten stuck with 100{100} or 720.{720.} Had the Mayans taken the mantle, we might be using some multiple of 20 today.

Because 360 is an entirely arbitrary choice, degrees have no meaningful, reason-grounded connections to other branches of mathematics. And because those connections aren't there, degrees lead to nothing but pain when we want to actually talk about angles. That is, treating them as genuine mathematical objects, worthy of study, rather than just some convenience in life. To give a brief glimpse of this suffering, here's a famous equation, using degrees:

sin⁡x∘=π180x∘+π35832000+(x∘)33!+π5188956800000−…=∑i=0∞(−1)iπ2i+11802i+1x∘2i+1(2i+1)! \sin x^\circ = \dfrac{\pi}{180}x^\circ + \dfrac{\pi^3}{5832000} + \dfrac{(x^\circ)^3}{3!} + \dfrac{\pi^5}{188956800000} - \ldots = \sum_{i=0}^{\infty}(-1)^i \dfrac{\pi^{2i+1}}{180^{2i+1}} \dfrac{x^{\circ^{2i+1}}}{(2i+1)!}

the same equation, in a far better unit called radians:

sin⁡x=x−x33!+x55!−…=∑i=0∞(−1)ix2i+1(2i+1)! \sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \ldots = \sum_{i=0}^{\infty} (-1)^i \dfrac{x^{2i+1}}{(2i+1)!}

Trigonometric Ratios

In the triangle below, △abc{\triangle abc} and △(a+e)d(c+f){\triangle (a+e)d(c+f)} are, proportionally, exactly the same. This is because of the property of similarity.

triangle

The concepts of sine, cosine, tangent, are simply ratios of the sides of the triangles.

Sine

A helpful mnemonic for sine: "Snakes often hide." (sine, opposite, hypotenuse).

sine. Let θ{\theta} be an interior angle of a right triangle. Then the sine of θ,{\theta,} denoted sin⁡θ,{\sin \theta,} is the ratio of the length of the side opposite the angle to the length of the hypotenuse.

sin⁡θ=oppositehypotenuse=ah \sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{a}{h}
sine

Cosine

Cosine: "Centipedes adorn hell." (cosine, adjacent, hypotenuse).

cosine. Let θ{\theta} be an interior angle of a right triangle. Then the cosine of θ,{\theta,} denoted cos⁡θ,{\cos \theta,} is the ratio of the length of the side adjacent to the angle to the length of the hypotenuse.

cos⁡θ=adjacenthypotenuse=bh \cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{b}{h}
cosine

Tangent

Tangent: "Termites offend ants" (tangent, opposite, adjacent).

tangent. Let θ{\theta} be an interior angle of a right triangle. Then the tangent of θ,{\theta,} denoted tan⁡θ,{\tan \theta,} is the ratio of the length of the side opposite to the angle to the length of the side adjacent.

tan⁡θ=oppositeadjacent=ab \tan \theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{a}{b}
tangent

Cotangent

Cotangent: "Cottages are ominous." (cotangent, adjacent, opposite).

cotangent. Let θ{\theta} be an interior angle of a right triangle. Then the cotangent of θ,{\theta,} denoted cot⁡θ,{\cot \theta,} is the ratio of the length of the side adjacent to the angle to the length of the side opposite.

cot⁡θ=adjacentopposite=ba \cot \theta = \dfrac{\text{adjacent}}{\text{opposite}} = \dfrac{b}{a}
cotangent

Secant

Secant: "Secure houses always" (secant, hypotenus, adjacent).

secant. Let θ{\theta} be an interior angle of a right triangle. Then the secant of θ,{\theta,} denoted sec⁡θ,{\sec \theta,} is the ratio of the length of the hypotenuse to the length of the side adjacent to θ.{\theta.}

sec⁡θ=hypotenuseadjacent=hb \sec \theta = \dfrac{\text{hypotenuse}}{\text{adjacent}} = \dfrac{h}{b}
secant

Cosecant

Cosecant: "Coruscant helped Obiwan" (cosecant, hypotenuse, opposite).

cosecant. Let θ{\theta} be an interior angle of a right triangle. Then the cosecant of θ,{\theta,} denoted csc⁡θ,{\csc \theta,} is the ratio of the length of hypotenuse to the angle to the length of the side opposite to θ.{\theta.}

csc⁡θ=hypotenuseopposite=ha \csc \theta = \dfrac{\text{hypotenuse}}{\text{opposite}} = \dfrac{h}{a}
cosecant

For the rest of these materials, we will use the following variables:

O=the side opposite to θA=the side adjacent to θH=the hypotenuse \eqs{ \Oo &= \text{the side opposite to $\theta$} \\ A &= \text{the side adjacent to $\theta$} \\ H &= \text{the hypotenuse} \\ }

Ratio Identities

Recall that the divison identity from real number arithmetic tells us that:

ab=acbc=ac⋅cb \dfrac{a}{b} = \dfrac{ \dfrac{a}{c} } { \dfrac{b}{c} } = \dfrac{a}{\cancel{c}} \cdot \dfrac{\cancel{c}}{b}

where a,b,c∈R{a,b,c \in \R} and c≠0.{c \neq 0.} We can thus rewrite many of the preceding definitions to alternative forms. These forms are called ratio identities.

Tangent-sine-cosine Relation (TSCR)

Because tangent is defined as:

tan⁡θ=OA \tan \theta = \dfrac{\Oo}{A}

the division identity allows us to write the right-hand side as:

tan⁡θ=OA=OHAH \tan \theta = \dfrac{\Oo}{A} = \dfrac{\dfrac{\Oo}{H}}{\dfrac{A}{H}}

Notice that the numerator and denominator are the definitions of sine and cosine:

sin⁡θ=OHcos⁡θ=AH \sin \theta = \dfrac{\Oo}{H} \\[1em] \cos \theta = \dfrac{A}{H}

Thus, we have the relationship:

tan⁡θ=sin⁡θcos⁡θ \tan \theta = \dfrac{\sin \theta}{\cos \theta}

Cotangent-tangent Relation (CTR)

Cotangent is often characterized as the inverse of tangent. Why? Recall that the cotangent is defined as:

cot⁡θ=AO \cot \theta = \dfrac{A}{\Oo}

Using the division identity, we can rewrite right-hand side as:

cot⁡θ=AAOA=1OA \cot \theta = \dfrac{\dfrac{A}{A}}{\dfrac{\Oo}{A}} = \dfrac{1}{\dfrac{\Oo}{A}}

That denominator is simply tangent:

cot⁡θ=1tan⁡θ \cot \theta = \dfrac{1}{\tan \theta}

Cotangent-Cosine-Sine Relation (CCSR)

From the relationship of tangent to sine and cosine, we can also define cotangent as:

cot⁡θ=1tan⁡θ=1sin⁡θcos⁡θ \cot \theta = \dfrac{1}{\tan \theta} = \dfrac{1}{\dfrac{\sin \theta}{\cos \theta}}

This reduces to:

cot⁡θ=cos⁡θsin⁡θ \cot \theta = \dfrac{\cos \theta}{\sin \theta}

Secant-cosine Relation (SCR)

The definition of secant tells us that:

sec⁡θ=HA \sec \theta = \dfrac{H}{A}

Using the division identity, we get:

sec⁡θ=HHAH=HH⋅HA \sec \theta = \dfrac{\dfrac{H}{H}}{\dfrac{A}{H}} = \dfrac{H}{\cancel{H}} \cdot \dfrac{\cancel{H}}{A}

The denominator in the rewritten expression is simply cosine:

sec⁡θ=1cos⁡θ \sec \theta = \dfrac{1}{\cos \theta}

Cosecant-sine Relation (CSR)

Cosecant is defined as:

csc⁡θ=HO \csc \theta = \dfrac{H}{\Oo}

Rewriting the right-hand side with the division identity, we get:

csc⁡θ=HHOH=HH⋅HO \csc \theta = \dfrac{\dfrac{H}{H}}{\dfrac{\Oo}{H}} = \dfrac{H}{\cancel{H}} \cdot \dfrac{\cancel{H}}{\Oo}

That denominator can be expressed as sine:

csc⁡θ=1sin⁡θ \csc \theta = \dfrac{1}{\sin \theta}

Tangent-cosecant-secant Relation (TCSR)

From CSR, we know that:

csc⁡θ=1sin⁡θ \csc \theta = \dfrac{1}{\sin \theta}

and from SCR we know that:

sec⁡θ=1cos⁡θ \sec \theta = \dfrac{1}{\cos \theta}

We can therefore infer that:

sin⁡θ=1csc⁡θ \sin \theta = \dfrac{1}{\csc \theta}

and:

cos⁡θ=1sec⁡θ \cos \theta = \dfrac{1}{\sec \theta}

(The inverse of the inverse gives you the verse). Then, from TSCR, we know that:

tan⁡θ=sin⁡θcos⁡θ \tan \theta = \dfrac{\sin \theta}{\cos \theta}

Using the alternative definitions of inverse definitions of sine and cosine, we have:

tan⁡θ=sin⁡θcos⁡θ=1csc⁡θ1sec⁡θ=1csc⁡θ⋅sec⁡θ1 \tan \theta = \dfrac{\sin \theta}{\cos \theta} = \dfrac{ \dfrac{1}{\csc \theta} }{\dfrac{1}{\sec \theta}} = \dfrac{1}{\csc \theta} \cdot \dfrac{\sec \theta}{1}

This yields the relation:

tan⁡θ=sec⁡θcsc⁡θ \tan \theta = \dfrac{\sec \theta}{\csc \theta}

Cotangent-cosecant-secant Relation (CCSR)

Because cotangent is the inverse of tangent, TCSR tells us that:

cot⁡θ=1θ=1sec⁡θcsc⁡θ \cot \theta = \dfrac{1}{\theta} = \dfrac{1}{\dfrac{\sec \theta}{\csc \theta}}

Thus, we have the relation:

cot⁡θ=csc⁡θsec⁡θ \cot \theta = \dfrac{\csc \theta}{\sec \theta}

Summary of Ratio Identities

Summarizing, we have the following ratio identities:

sin⁡θ=1csc⁡θ \sin \theta = \dfrac{1}{\csc \theta}
csc⁡θ=1sin⁡θ \csc \theta = \dfrac{1}{\sin \theta}
cos⁡θ=1sec⁡θ \cos \theta = \dfrac{1}{\sec \theta}
sec⁡θ=1cos⁡θ \sec \theta = \dfrac{1}{\cos \theta}
tan⁡θ=1cot⁡θ \tan \theta = \dfrac{1}{\cot \theta}
cot⁡θ=1tan⁡θ \cot \theta = \dfrac{1}{\tan \theta}
tan⁡θ=sin⁡θcos⁡θ \tan \theta = \dfrac{\sin \theta}{\cos \theta}
cot⁡θ=cos⁡θsin⁡θ \cot \theta = \dfrac{\cos \theta}{\sin \theta}
sec⁡θ=1cos⁡θ \sec \theta = \dfrac{1}{\cos \theta}
tan⁡θ=csc⁡θsec⁡θ \tan \theta = \dfrac{\csc \theta}{\sec \theta}

Radians

With this understanding of triangles, consider the the following plot:

This is a circle of radius r=1.{r = 1.} In mathematics, this is called the unit circle.1 The unit circle can be algebraically defined with the equation:

x2+y2=1 x^2 + y^2 = 1

We can express a point on this circle with the familiar form (x,y).{(x,y).} For example, the points (1,0),{(1,0),} (0,−1),{(0,-1),} (−1,0),{(-1,0),} and (0,1){(0,1)} all lie on the circle:

However, we can also express the a point with the notation:

(cos⁡θ,sin⁡θ) (\cos \theta, \sin \theta)

Because the graph is a unit circle, the point can also be thought of as lying at the edge of a triangle right trangle. And if we can think of the point in this manner, the x{x} and y{y} coordinates can be expressed as:

x=cos⁡θ,    y=sin⁡θ x = \cos \theta,~~~~y = \sin \theta

Remember that cosine and sine are just ratios:

cos⁡θ=adjacenthypotenuse    sin⁡θ=oppositehypotenuse \cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} ~~~~ \sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}

And since the radius of the circle is 1,{1,} that hypotenuse is 1:

cos⁡θ=adjacent1    sin⁡θ=opposite1 \cos \theta = \dfrac{\text{adjacent}}{1} ~~~~ \sin \theta = \dfrac{\text{opposite}}{1}

The side adjacent to the angle can be measured in the amount of steps we take along the x{x} axis, and the side opposite to the angle can be measured in the amount of y{y} steps we take:

cos⁡θ=x1    sin⁡θ=y1 \cos \theta = \dfrac{x}{1} ~~~~ \sin \theta = \dfrac{y}{1}

It should now be clear why:

(x,y)=(cos⁡θ,sin⁡θ) (x,y) = (\cos \theta, \sin \theta)

Now here's the real kicker: We can express any point on the Cartesian plane using sine and cosine. Why? Because of that seemingly trivial law we saw earlier — the law of similar triangles:

The only difference is, the hypotenuse (the circle's radius) changes:

PointRadiuscos⁡θ{\cos \theta}sin⁡θ{\sin \theta}
a{a}0.5x0.5{\dfrac{x}{0.5}}y0.5{\dfrac{y}{0.5}}
b{b}1x1{\dfrac{x}{1}}y1{\dfrac{y}{1}}
c{c}1.5x1.5{\dfrac{x}{1.5}}y1.5{\dfrac{y}{1.5}}

We now have a relationship between the Cartesian coordinates, cosine, and sine. But how can we make this relationship useful? So far, all we've done is tie cosine and sine to x{x} and y.{y.} With Cartesian coordinates, we can write (1,1),{(1,1),} (3,−1),{(3,-1),} (0,0),{(0,0),} or any other pair of real numbers. All we have with the trigonometric ratios is that:

(x,y)=(cos⁡θ,sin⁡θ) (x,y) = (\cos \theta, \sin \theta)

We can fix this by passing in values for θ,{\theta,} just as we'd pass in values for x{x} and y.{y.} But what values do we pass for θ?{\theta?} Radians. If we look closely at the last diagram, we'd see that we marked the angle θ{\theta} with a tiny curve. That curve is called an arc — a portion of a circle. This is a common convention in geometry, and it reveals what exactly the radian is.

What's the length of that arc? Well, it's a portion of the circumference. The circumference C{C} of a circle is given by the formula:

C=2πr C = 2 \pi r

where r{r} is the radius and π{\pi} is the constant irrational pi. Since the unit circle has a radius of 1,{1,} the unit circle's circumference is:

C=2π C = 2 \pi

Thus, a radian is simply a fraction of 2π.{2 \pi.} Notice that this completely removes the radius. It doesn't matter how big or how small a circle is, we can always measure some angle whose vertex is at the center of the circle using some fraction of 2π:{2 \pi:}

DegreesRadians
−180∘{-180^\circ}−π{- \pi}
−135∘{-135^\circ}−3π/4{-{3 \pi}/{4}}
−90∘{-90^\circ}−π/2{- \pi/2}
−45∘{-45^\circ}−π/4{- \pi/4}
0∘{0^\circ}0{0}
30∘{30^\circ}π/6{\pi/6}
45∘{45^\circ}π/4{\pi/4}
60∘{60^\circ}π/3{\pi/3}
90∘{90^\circ}π/2{\pi/2}
120∘{120^\circ}2π/3{{2 \pi}/{3}}
135∘{135^\circ}3π/4{{3 \pi}/{4}}
150∘{150^\circ}5π/6{{5 \pi}/{6}}
180∘{180^\circ}π{\pi}
270∘{270^\circ}3π/2{{3\pi}/{2}}
360∘{360^\circ}2π{2 \pi}

The First Pythagorean Identity

From the fact that x=cos⁡θ{x = \cos \theta} and y=sin⁡θ,{y = \sin \theta,} we can make numerous inferences about the trigonometric ratios. Collectively, these inferences are called the trigonometric identities.

Since we've established that x=cos⁡θ{x = \cos \theta} and y=sin⁡θ,{y = \sin \theta,} we can use simple substitution to rewrite the equation of the unit circle:

x2+y2=1⇒(cos⁡θ)2+(sin⁡θ)2=1 x^2 + y^2 = 1 \nc (\cos \theta)^2 + (\sin \theta)^2 = 1

first pythagorean identity. Let θ{\theta} be an subtended by two radii on the unit circle. Then:

sin⁡2θ+cos⁡2θ=1 \sin^2 \theta + \cos^2 \theta = 1

Note that the expression:

sin⁡2θ+cos⁡2θ=1 \sin^2 \theta + \cos^2 \theta = 1

is really:

(sin⁡θ)(sin⁡θ)+(cos⁡θ)(cos⁡θ)=1 (\sin \theta)(\sin \theta) + (\cos \theta)(\cos \theta) = 1

In deriving the remaining identities, we will use this expanded form, but express our results in the more conventional notation contained in the definition above.

The Second Pythogrean Identity

Since the trigonometric ratios are just ratios, we can divide each by another. Let's divide all of the terms of the first pythogorean identity by cos⁡2θ:{\cos^2 \theta:}

(sin⁡θ)cos⁡θ(sin⁡θ)cos⁡θ+(cos⁡θ)cos⁡θ(cos⁡θ)cos⁡θ=1cos⁡θ1cos⁡θ \dfrac{(\sin \theta)}{\cos \theta} \dfrac{(\sin \theta)}{\cos \theta} + \dfrac{(\cos \theta)}{\cos \theta} \dfrac{(\cos \theta)}{\cos \theta} = \dfrac{1}{\cos \theta} \dfrac{1}{\cos \theta}

We see some ratios in there that we derived as ratio identities:

(sin⁡θ)cos⁡θ⏞tan⁡θ(sin⁡θ)cos⁡θ⏞tan⁡θ+(cos⁡θ)cos⁡θ⏞1(cos⁡θ)cos⁡θ⏞1=1cos⁡θ⏞sec⁡θ1cos⁡θ⏞sec⁡θ \tnote{\dfrac{(\sin \theta)}{\cos \theta}}{$\tan \theta$} \tnote{\dfrac{(\sin \theta)}{\cos \theta}}{$\tan \theta$} + \tnote{\dfrac{(\cos \theta)}{\cos \theta}}{1} \tnote{\dfrac{(\cos \theta)}{\cos \theta}}{1} = \tnote{\dfrac{1}{\cos \theta}}{$\sec \theta$} \tnote{\dfrac{1}{\cos \theta}}{$\sec \theta$}

Thus, we have:

(tan⁡θ)(tan⁡θ)+(1)(1)=(sec⁡θ)(sec⁡θ) (\tan \theta)(\tan \theta) + (1)(1) = (\sec \theta)(\sec \theta)

This gives us the second pythagorean identity:

second pythagorean identity. Let θ{\theta} be an subtended by two radii on the unit circle. Then:

tan⁡2θ+1=sec⁡2θ \tan^2 \theta + 1 = \sec^2 \theta

The Third Pythagorean Identity

Just as we did for the second pythogrean identity, let's divide all of the terms of the first pythagorean identity by sin⁡2θ:{\sin^2 \theta:}

(sin⁡θ)sin⁡θ⏞1(sin⁡θ)sin⁡θ⏞1+(cos⁡θ)sin⁡θ⏞cot⁡θ(cos⁡θ)sin⁡θ⏞cot⁡θ=1sin⁡θ⏞csc⁡θ1sin⁡θ⏞csc⁡θ \tnote{\dfrac{(\sin \theta)}{\sin \theta}}{1} \tnote{\dfrac{(\sin \theta)}{\sin \theta}}{1} + \tnote{\dfrac{(\cos \theta)}{\sin \theta}}{$\cot \theta$} \tnote{\dfrac{(\cos \theta)}{\sin \theta}}{$\cot \theta$} = \tnote{\dfrac{1}{\sin \theta}}{$\csc \theta$} \tnote{\dfrac{1}{\sin \theta}}{$\csc \theta$}

We now have the third pythagorean identity:

third pythagorean identity. Let θ{\theta} be an subtended by two radii on the unit circle. Then:

1+cot⁡2θ=csc⁡2θ 1 + \cot^2 \theta = \csc^2 \theta

Summary of Pythagorean Identities

In sum, we have the following Pythogrean identities (the additional identities are simple rearrangements of the three we derived):

sin⁡2θ+cos⁡2θ=1{\sin^2 \theta + \cos^2 \theta = 1}
tan⁡2θ+1=sec⁡2θ{\tan^2 \theta + 1 = \sec^2 \theta}
1+cot⁡2θ=csc⁡2θ{1 + \cot^2 \theta = \csc^2 \theta}
sin⁡2θ=1−cos⁡2θ{\sin^2 \theta = 1 - \cos^2 \theta}
cos⁡2θ=1−sin⁡2θ{\cos^2 \theta = 1 - \sin^2 \theta}

Root Identities

Given the equation x2+y2=1,{x^2 + y^2 = 1,} if we solve for x{x} and y,{y,} we get the following:

x=±1−y2y=±1−x2 x = \pm \sqrt{1 - y^2} \\[1em] y = \pm \sqrt{1 - x^2}

Since we know that x=cos⁡θ{x = \cos \theta} and y=sin⁡θ,{y = \sin \theta,} we can plug these in two get:

cos⁡θ=±1−cos⁡2θsin⁡θ=±1−sin⁡2θ \cos \theta = \pm \sqrt{1 - \cos^2 \theta} \\[1em] \sin \theta = \pm \sqrt{1 - \sin^2 \theta}

This gives us the root identity of sine:

root identity of sine. Let θ{\theta} be an angle subtended by two radii on the unit circle. Then:

sin⁡θ=±1−cos⁡2θ \sin \theta = \pm \sqrt{1 - \cos^2 \theta}

and the root identity of cosine:

root identity of cosine. Let θ{\theta} be an angle subtended by two radii on the unit circle. Then:

cos⁡θ=±1−sin⁡2θ \cos \theta = \pm \sqrt{1 - \sin^2 \theta}

Co-function Identities

Having both radians and the Pythagorean identities allows us to establish a relationship between the trigonometric identities and the constant π.{\pi.}

From our discussion of the unit circle, we know that cos⁡θ{\cos \theta} gives us the number of x{x} steps, and sin⁡θ{\sin \theta} gives us the number of y{y} steps. x{x} and y,{y,} therefore, correspond to the two legs of a right triangle on the unit circle.

As we likely remember from elementary school, the angles of a right triangle sum to 180∘.{180^{\circ}.} Now that we know what radians are, we can reformulate this familiar idea:

sum of triangle angles. The sum of the interior angles of a triangle is π.{\pi.}

Now, we know that all right triangles have a single 90∘{90^\circ} angle, by definition. In terms of radians, this means that all right-triangles have a single angle with the measure π2.{\dfrac{\pi}{2}.} If we so happen to have a right-triangle whose legs are of equal length, then we the law complementary angles appies: The sum of those two angles is π2.{\dfrac{\pi}{2.}} And if the law of complementary angles applies, we get the following lemma:

cofunction lemma. Let α{\alpha} and β{\beta} be interior angles of a right-triangle. If α=β,{\alpha = \beta,} then:

sin⁡α=cos⁡βsin⁡β=cos⁡α \eqs { \sin \alpha &= \cos \beta \\ \sin \beta &= \cos \alpha \\ }

Thus, the magic number (perhaps not so magical) is π2.{\dfrac{\pi}{2}.} Using the ratio identities we saw earlier, we have the following cofunction identities:

sin⁡θ=cos⁡(π2−θ) \sin \theta = \cos \ar{\dfrac{\pi}{2} - \theta}
sec⁡θ=csc⁡(π2−θ) \sec \theta = \csc \ar{\dfrac{\pi}{2} - \theta}
tan⁡θ=cot⁡(π2−θ) \tan \theta = \cot \ar{\dfrac{\pi}{2} - \theta}
cos⁡θ=sin⁡(π2−θ) \cos \theta = \sin \ar{\dfrac{\pi}{2} - \theta}
csc⁡θ=sec⁡(π2−θ) \csc \theta = \sec \ar{\dfrac{\pi}{2} - \theta}
cot⁡θ=tan⁡(π2−θ) \cot \theta = \tan \ar{\dfrac{\pi}{2} - \theta}

Sum and Difference Formulas

While the identities we've derived thus far have taken us a long way, we want more. To propel us even further, let's see what else we can get out of the unit circle.

Suppose there are two points on the graph of x2+y2=1,{x^2 + y^2 = 1,} separated by a distance d.{d.}

circle 1

Keeping the points separated by the distance d{d} and changing the angle to α−β,{\alpha - \beta,} we get:

circle 2

Applying the distance formula for the two points on the first graph, we have:

d1=(x1−x0)2+(y1−y0)2=(cos⁡α−cos⁡β)2+(sin⁡α−sin⁡β)2 \eqs { d_1 &= \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} \\ &= \sqrt{(\cos \alpha - \cos \beta)^2 + (\sin \alpha - \sin \beta)^2} }

and for the second graph, we have:

d2=(x1−x0)2+(y1−y0)2=(cos⁡(α−β)−1)2+(sin⁡(α−β)−0)2 \eqs { d_2 &= \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} \\ &= \sqrt{ (\cos(\alpha - \beta) - 1)^2 + (\sin(\alpha - \beta) - 0)^2 } }

Since we maintained the distance, we have:

d1=d2(cos⁡α−cos⁡β)2+(sin⁡α−sin⁡β)2=(cos⁡(α−β)−1)2+(sin⁡(α−β)−0)2 \eqs{ d_1 &= d_2 \\ \sqrt{(\cos \alpha - \cos \beta)^2 + (\sin \alpha - \sin \beta)^2} &= \sqrt{ (\cos(\alpha - \beta) - 1)^2 + (\sin(\alpha - \beta) - 0)^2 } }

Squaring both sides of the equation, we get:

(cos⁡α−cos⁡β)2+(sin⁡α−sin⁡β)2=(cos⁡(α−β)−1)2+(sin⁡(α−β)−0)2 (\cos \alpha - \cos \beta)^2 + (\sin \alpha - \sin \beta)^2 = (\cos(\alpha - \beta) - 1)^2 + (\sin(\alpha - \beta) - 0)^2

Let's denote the first term on the left-hand side as A:{A:}

A=(cos⁡α−cos⁡β)2 A = (\cos \alpha - \cos \beta)^2

Expanding A,{A,} we get:

A=(cos⁡α−cos⁡β)(cos⁡α−cos⁡β)=cos⁡2α−2cos⁡αcos⁡β+cos⁡2β \eqs{ A &= (\cos \alpha - \cos \beta)(\cos \alpha - \cos \beta) \\ &= \cos^2 \alpha - 2 \cos \alpha \cos \beta + \cos^2 \beta }

Now the second term:

B=(sin⁡α−sin⁡β)2 B = (\sin \alpha - \sin \beta)^2

Here, we have:

B=(sin⁡α−sin⁡β)(sin⁡α−sin⁡β)=sin⁡2α−2sin⁡αsin⁡β+sin⁡2β \eqs{ B &= (\sin \alpha - \sin \beta)(\sin \alpha - \sin \beta) \\ &= \sin^2 \alpha - 2 \sin \alpha \sin \beta + \sin^2 \beta }

Putting the two together, we get A+B:{A + B:}

cos⁡2α−2cos⁡αcos⁡β+cos⁡2β+sin⁡2α−2sin⁡αsin⁡β+sin⁡2β \cos^2 \alpha - 2 \cos \alpha \cos \beta + \cos^2 \beta + \sin^2 \alpha - 2 \sin \alpha \sin \beta + \sin^2 \beta

Using the law of commutativity, we can rearrange some of these terms:

cos⁡2α+sin⁡2α−2cos⁡αcos⁡β−2sin⁡αsin⁡β+cos⁡2β+sin⁡2β \cos^2 \alpha + \sin^2 \alpha - 2 \cos \alpha \cos \beta - 2 \sin \alpha \sin \beta + \cos^2 \beta + \sin^2 \beta

Rearranged this way, we can see that some terms can be reduced using the first Pythagorean identity:

cos⁡2α+sin⁡2α⏞1−2cos⁡αcos⁡β−2sin⁡αsin⁡β+cos⁡2β+sin⁡2β⏞1 \tnote{\cos^2 \alpha + \sin^2 \alpha}{1} - 2 \cos \alpha \cos \beta - 2 \sin \alpha \sin \beta + \tnote{\cos^2 \beta + \sin^2 \beta}{1}

Applying that identity and factoring the two in the second and third terms, we get:

1−2(cos⁡αcos⁡β−sin⁡αsin⁡β)+1=2−2(cos⁡αcos⁡β−sin⁡αsin⁡β) 1 - 2(\cos \alpha \cos \beta - \sin \alpha \sin \beta) + 1 = 2 - 2(\cos \alpha \cos \beta - \sin \alpha \sin \beta)

That's about all we can do for the lefthand side. Let's look at the righthand. We'll denote the first term as C:{C:}

C=(cos⁡(α−β)−1)2 C = (\cos(\alpha - \beta) - 1)^2

Expanding this term, we have:

C=(cos⁡(α−β)−1)(cos⁡(α−β)−1)=cos⁡2(α−β)−2cos⁡(α−β)+1 \eqs{ C &= (\cos(\alpha - \beta) - 1)(\cos(\alpha - \beta) - 1) \\ &= \cos^2(\alpha - \beta) - 2 \cos(\alpha - \beta) + 1 }

Finally, for the righthand's second term:

D=(sin⁡(α−β)−0)2 D = (\sin(\alpha - \beta) - 0)^2

Expanding:

D=sin⁡2(α−β) D = \sin^2(\alpha - \beta)

Putting the two together, we get C+D:{C + D:}

cos⁡2(α−β)−2cos⁡(α−β)+1+sin⁡2(α−β)\cos^2(\alpha - \beta) - 2 \cos(\alpha - \beta) + 1 + \sin^2(\alpha - \beta)

Once again, let's the commutative law to rearrange some of the terms:

cos⁡2(α−β)+sin⁡2(α−β)−2cos⁡(α−β)+1\cos^2(\alpha - \beta) + \sin^2(\alpha - \beta) - 2 \cos(\alpha - \beta) + 1

Once again we see terms that can be reduced with the first Pythagorean identity:

cos⁡2(α−β)+sin⁡2(α−β)⏞1−2cos⁡(α−β)+1\tnote{\cos^2(\alpha - \beta) + \sin^2(\alpha - \beta)}{1} - 2 \cos(\alpha - \beta) + 1

Thus, we have:

1−2cos⁡(α−β)+1=2−2cos⁡(α−β) 1 - 2 \cos(\alpha - \beta) + 1 = 2 - 2 \cos(\alpha - \beta)

Let's bring the left and righthand sides back together:

2−2(cos⁡αcos⁡β−sin⁡αsin⁡β)=2−2cos⁡(α−β)2 - 2(\cos \alpha \cos \beta - \sin \alpha \sin \beta) = 2 - 2 \cos(\alpha - \beta)

Dividing two from both sides:

1−cos⁡αcos⁡β+sin⁡αsin⁡β=1−cos⁡(α−β)1 - \cos \alpha \cos \beta + \sin \alpha \sin \beta = 1 - \cos(\alpha - \beta)

Subtracting the 1:

−cos⁡αcos⁡β+sin⁡αsin⁡β=−cos⁡(α−β)- \cos \alpha \cos \beta + \sin \alpha \sin \beta = - \cos(\alpha - \beta)

Rearranging:

cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta

We now have our first double-angle formula.

cosine angle difference formula. Let α{\alpha} and β{\beta} be the interior angles of a right trangle with legs of length 1. Then:

cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta

Let's draw another formula. If we replace β{\beta} with −β,{-\beta,} the ratios remain the same, since we're on the unit circle:

cos⁡(α−−β)=cos⁡(α+β)=cos⁡αcos⁡(−β)+sin⁡αsin⁡(−β)=cos⁡αcos⁡(β)−sin⁡αsin⁡β \eqs{ \cos(\alpha - -\beta) &= \cos(\alpha + \beta) \\ &= \cos \alpha \cos (-\beta) + \sin \alpha \sin (-\beta) \\ &= \cos \alpha \cos(\beta) - \sin \alpha \sin \beta }

This gives us the second double-angle formula.

cosine angle sum formula. Let α{\alpha} and β{\beta} be the interior angles of a right trangle with legs of length 1. Then:

cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta

Using this new formula, let's replace α{\alpha} with π2−α.{\dfrac{\pi}{2} - \alpha.} This yields:

cos⁡(π2−α+β)=cos⁡(π2−α)cos⁡β−sin⁡(π2−α)sin⁡β \cos\ar{\frac{\pi}{2} - \alpha + \beta} = \cos \ar{\frac{\pi}{2} - \alpha} \cos \beta - \sin \ar{\frac{\pi}{2} - \alpha} \sin \beta

Once again, we can apply identities:

cos⁡(π2−α+β)⏞sin⁡(α−β)=cos⁡(π2−α)⏞sin⁡αcos⁡β−sin⁡(π2−α)⏞cos⁡αsin⁡β \small \tnote{\cos\ar{\frac{\pi}{2} - \alpha + \beta}}{$\sin(\alpha - \beta)$} = \tnote{\cos \ar{\frac{\pi}{2} - \alpha}}{$\sin \alpha$} \cos \beta - \tnote{\sin \ar{\frac{\pi}{2} - \alpha}}{$\cos \alpha$} \sin \beta

Thus, we have:

sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta

This gives us the third double-angle formula:

sine angle difference formula. Let α{\alpha} and β{\beta} be the interior angles of a right trangle with legs of length 1. Then:

sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta

We get a fourth double-angle formula by replacing β{\beta} with −β{-\beta} in the formula we just derived:

sin⁡(α−(−β))=sin⁡(α+β)=sin⁡αcos⁡(−β)−cos⁡αsin⁡(−β)=sin⁡αcos⁡β+cos⁡αsin⁡β \eqs{ \sin(\alpha -(- \beta)) &= \sin(\alpha + \beta) \\ &= \sin \alpha \cos(-\beta) - \cos \alpha \sin(-\beta) \\ &= \sin \alpha \cos \beta + \cos \alpha \sin \beta }

We now have the fourth double-angle formula:

sine angle sum formula. Let α{\alpha} and β{\beta} be the interior angles of a right trangle with legs of length 1. Then: Then:

sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta

Summary: Sum & Difference Formulas

Summarizing, we now have the following formulas:

cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta
cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta
sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta
sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta

The Double Angle Formulas

The sum and difference formulas provide a powerful arsenal for drawing even more formulas. We know that:

sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta

There's nothing stopping us from saying α=β,{\alpha = \beta,} so let's see what happens if we do that:

sin⁡(α+α)=sin⁡αcos⁡α+cos⁡αsin⁡α \sin(\alpha + \alpha) = \sin \alpha \cos \alpha + \cos \alpha \sin \alpha

Simplifying the lefthand side:

sin⁡(2α)=sin⁡αcos⁡α+cos⁡αsin⁡α=sin⁡αcos⁡α+sin⁡αcos⁡α=2sin⁡αcos⁡α \eqs{ \sin(2\alpha) &= \sin \alpha \cos \alpha + \cos \alpha \sin \alpha \\ &= \sin \alpha \cos \alpha + \sin \alpha \cos \alpha \\ &= 2 \sin \alpha \cos \alpha \\ }

This gives us our first double angle formula:

double-angle formula i. Let α{\alpha} and β{\beta} be the interior angles of a right trangle with legs of length 1. Then:

sin⁡(2α)=2sin⁡αcos⁡α \sin(2\alpha) = 2 \sin \alpha \cos \alpha

Let's do the same for the cosine angle sum formula:

cos⁡(α+α)=cos⁡αcos⁡α−sin⁡αsin⁡α=cos⁡2α−sin⁡2α \eqs{ \cos(\alpha + \alpha) &= \cos \alpha \cos \alpha - \sin \alpha \sin \alpha \\ &= \cos^2 \alpha - \sin^2 \alpha \\ }

We now have the second double-angle formula:

double-angle formula ii. Let α{\alpha} and β{\beta} be the interior angles of a right trangle with legs of length 1. Then:

cos⁡(2α)=cos⁡2α−sin⁡2α \cos(2\alpha) = \cos^2 \alpha - \sin^2 \alpha

We can manipulate the second double-angle formula to derive a third, using the Pythagorean identity:

sin⁡2α=1−cos⁡2α \sin^2 \alpha = 1 - \cos^2 \alpha

Plugging this identity into the second formula, we get:

cos⁡(2α)=cos⁡2α−sin⁡2α=cos⁡2α−(1−cos⁡2α)=cos⁡2α−1+cos⁡2α=2cos⁡2α−1 \eqs{ \cos(2\alpha) &= \cos^2 \alpha - \sin^2 \alpha \\ &= \cos^2 \alpha - (1 - \cos^2 \alpha) \\ &= \cos^2 \alpha - 1 + \cos^2 \alpha \\ &= 2 \cos^2 \alpha - 1 }

This gives us our third double-angle formula:

double-angle formula iii. Let α{\alpha} and β{\beta} be the interior angles of a right trangle with legs of length 1. Then:

cos⁡(2α)=2cos⁡2α−1 \cos(2\alpha) = 2 \cos^2 \alpha - 1

Let's use the same process, only this time we'll use the identity:

cos⁡2α=1−sin⁡2α \cos^2 \alpha = 1 - \sin^2 \alpha

Doing so, we get:

cos⁡(2α)=cos⁡2α−sin⁡2α=(1−sin⁡2α)−sin⁡2α=1−2sin⁡2α \eqs{ \cos(2 \alpha) &= \cos^2 \alpha - \sin^2 \alpha \\ &= (1 - \sin^2 \alpha) - \sin^2 \alpha \\ &= 1 - 2 \sin^2 \alpha }

Now we have a fourth double-angle formula:

double-angle formula iv. Let α{\alpha} and β{\beta} be the interior angles of a right trangle with legs of length 1. Then:

cos⁡(2α)=1−2sin⁡2α \cos(2\alpha) = 1 - 2 \sin^2 \alpha

Summary: Double-angle Formulas

Summarizing the double-angle formulas we've derived:

sin⁡(2α)=2sin⁡αcos⁡α \sin(2\alpha) = 2 \sin \alpha \cos \alpha
cos⁡(2α)=cos⁡2α−sin⁡2α \cos(2\alpha) = \cos^2 \alpha - \sin^2 \alpha
cos⁡(2α)=2cos⁡2α−1 \cos(2\alpha) = 2 \cos^2 \alpha - 1
cos⁡(2α)=1−2sin⁡2α \cos(2\alpha) = 1 - 2 \sin^2 \alpha

The Half-angle Formulas

Having derived the double-angle formulas, let's play with the values of α{\alpha} a little. What if we instead defined:

2α=θ 2 \alpha = \theta

Let's plug this in to the third double-angle formula:

cos⁡(θ)=2cos⁡2(θ2)−1 \cos(\theta) = 2 \cos^2\ar{\dfrac{\theta}{2}} - 1

Let's transpose 1:

1+cos⁡(θ)=2cos⁡2(θ2)−1+1=2cos⁡2θ2 \eqs{ 1 + \cos(\theta) &= 2 \cos^2\ar{\dfrac{\theta}{2}} - 1 + 1 \\[1em] &= 2 \cos^2 \dfrac{\theta}{2} \\ }

We now have:

1+cos⁡(θ)=2cos⁡2(θ2) 1 + \cos(\theta) = 2 \cos^2 \ar{ \dfrac{\theta}{2} }

Let's flip the equation:

2cos⁡2(θ2)=1+cos⁡θ 2 \cos^2 \ar{ \dfrac{\theta}{2} } = 1 + \cos \theta

Now let's divide by 2:{2:}

cos⁡2(θ2)=1+cos⁡θ2 \cos^2 \ar{ \dfrac{\theta}{2} } = \dfrac{1 + \cos \theta}{2}

Taking the square root of both sides:

cos⁡2(θ2)=±1+cos⁡θ2cos⁡(θ2)=±1+cos⁡θ2\eqs{ \sqrt{\cos^2\ar{\dfrac{\theta}{2}}} &= \pm \sqrt{\dfrac{1 + \cos \theta}{2}} \\[1em] \cos \ar{\dfrac{\theta}{2}} &= \pm \sqrt{\dfrac{1 + \cos \theta}{2}} }

This gives us the half-angle formula of cosine:

half-angle formula of cosine. Let θ{\theta} be an angle. Then:

cos⁡(θ2)=±1+cos⁡θ2 \cos \ar{\dfrac{\theta}{2}} = \pm \sqrt{\dfrac{1 + \cos \theta}{2}}

If we set 2α=θ,{2\alpha = \theta,} in the formula:

cos⁡2α=1−2sin⁡2α \cos 2 \alpha = 1 - 2 \sin^2 \alpha

we get the half-angle formula for sine:

half-angle formula of sine. Let θ{\theta} be an angle. Then:

sin⁡(θ2)=±1−cos⁡θ2 \sin \ar{\dfrac{\theta}{2}} = \pm \sqrt{\dfrac{1 - \cos \theta}{2}}

Footnotes

  1. The form "unit S,{S,}" where S{S} is some shape is often used in mathematics to describe some simple, easy-to-reason-about object. This object is then used as a starting point for inferences. Other instances include the unit square (a square whose sides are all of length 1), the unit triangle (likewise), the unit pentagon (likewise), and so on. ↩