Chapter 3Measurement of Figures — Trigonometric Ratios, the Law of Sines, and the Law of Cosines
In Chapter 2, we used the fact that corresponding sides of similar triangles have equal ratios to investigate properties of triangles such as the centroid and the circumcenter. However, the idea of similarity has its limits. For example, similarity alone is powerless against a problem such as “Given a triangle whose side lengths are all known, find the size of one of its angles.” What we need is the fact that once the shape of a right triangle (that is, the size of an acute angle) is determined, the ratios of its side lengths are also determined in exactly one way. Putting a name to the relation “once the shape is determined, the ratios are determined” gives us the trigonometric ratios, the stars of this chapter—$\sin\theta$ (sine), $\cos\theta$ (cosine), and $\tan\theta$ (tangent).
In this chapter, we first prove three basic relations connecting the trigonometric ratios. Next we prove two powerful tools that can be used for any triangle, not only right triangles: the law of sines and the law of cosines. The law of sines states that “the ratio of a side to the $\sin$ of its opposite angle equals the diameter of the circumcircle of the triangle,” and it can be proved from properties of inscribed angles. The law of cosines is “a theorem that extends the Pythagorean theorem so that it can be used for triangles other than right triangles”; we prove it by comparing areas, treating the three cases of an acute, a right, and an obtuse angle separately. Finally, we study radian measure, a new unit for measuring angles. For now it is nothing more than the conversion rule “$180°=\pi\,\mathrm{rad}$,” but here we plant the seed of an idea that will become essential when we deal with trigonometric functions in calculus in the future.
- The definitions of the trigonometric ratios $\sin\theta,\cos\theta,\tan\theta$, and proofs of the three basic relations that follow from the Pythagorean theorem (such as $\sin^2\theta+\cos^2\theta=1$)
- The meaning of the law of sines $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$, and its proof using properties of inscribed angles
- The meaning of the law of cosines $a^2=b^2+c^2-2bc\cos A$, and its proof from areas, split into the three cases of an acute, a right, and an obtuse angle
- The definition of radian measure (measuring an angle as the ratio of arc length to radius), and the derivation of $180°=\pi\,\mathrm{rad}$
- The computational skill to choose between the law of sines and the law of cosines in order to find unknown side lengths and angles of a triangle
Source notes: Yasuhide Mochizuki, Mathematics Notes: High School Mathematics (handwritten lecture notes, in Japanese), pp. 7–10.
3.1 Basic Relations among the Trigonometric Ratios
Once we fix the size $\theta$ (theta, a Greek letter often used to denote an angle) of one acute angle of a right triangle, the ratios of the lengths of the three sides are determined in exactly one way (because all similar right triangles have equal ratios of sides). The following definition gives a name to these ratios.
Definition 3.1: Trigonometric ratios
As in Figure 3.1, let $\theta$ be one acute angle of a right triangle, let $a$ be the length of the side adjacent to $\theta$ (the one that is not the hypotenuse), let $b$ be the length of the side opposite $\theta$, and let $c$ be the length of the hypotenuse. Then
$$ \sin\theta=\frac{b}{c}\ (\text{opposite}/\text{hypotenuse}),\qquad \cos\theta=\frac{a}{c}\ (\text{adjacent}/\text{hypotenuse}),\qquad \tan\theta=\frac{b}{a}\ (\text{opposite}/\text{adjacent}) $$are called the sine, cosine, and tangent of $\theta$, respectively, and are collectively called the trigonometric ratios of $\theta$. From the definition, $a=c\cos\theta$ and $b=c\sin\theta$ follow immediately.
Intuition: Trigonometric ratios depend only on “shape,” not on “size”
If we enlarge or shrink the right triangle while keeping $\theta$ fixed, the corresponding side ratios of similar figures are equal, so the values of $\sin\theta,\cos\theta,\tan\theta$ do not change. In other words, the trigonometric ratios do not depend on the size of the triangle; they are functions determined by the angle $\theta$ alone. For example, in a right triangle with $\theta=30°$, no matter how long the hypotenuse is, the opposite side is always exactly half of the hypotenuse ($\sin30°=1/2$).
Definition 3.1 makes sense only when $0°\lt\theta\lt90°$ (an acute angle), since the angle in question is an acute angle of a right triangle. In the second half of this chapter (Section 3.3), however, there will be occasions to use $\cos\theta$ and so on for obtuse angles ($90°\lt\theta\lt180°$) as well. So let us extend the definition of the trigonometric ratios to obtuse angles, in a way that does not contradict the acute case.
Definition 3.2: Trigonometric ratios of a general angle ($0°\le\theta\le180°$)
In the coordinate plane, draw the circle of radius 1 centered at the origin $O$ (the unit circle). Let $P$ be the point on the unit circle reached by rotating counterclockwise by the angle $\theta$ from the positive direction of the $x$-axis; the coordinates of $P$ are given by exactly one pair of real numbers $(x,y)$. We then define
$$ \cos\theta = x,\qquad \sin\theta = y,\qquad \tan\theta=\frac{y}{x}\ (x\ne0) $$When $0°\lt\theta\lt90°$, this agrees with Definition 3.1 (just consider the right triangle formed by dropping a perpendicular from $P$ to the $x$-axis; since the hypotenuse has length $1$, $\cos\theta=$ adjacent side $/1$ and $\sin\theta=$ opposite side $/1$). When $\theta=90°$, we have $P=(0,1)$, so $\cos90°=0,\ \sin90°=1$ ($\tan90°$ is not defined, since $x=0$). When $90°\lt\theta\le180°$ (an obtuse angle), the $x$-coordinate of $P$ is negative, so $\cos\theta\lt0$.
Why?: $\cos(180°-\theta)=-\cos\theta$, $\sin(180°-\theta)=\sin\theta$
The point on the unit circle corresponding to the angle $180°-\theta$ is the point $(-\cos\theta,\sin\theta)$ obtained by reflecting the point $(\cos\theta,\sin\theta)$ corresponding to the angle $\theta$ across the $y$-axis (because the angle measured from the positive direction of the $x$-axis is, in effect, measured again on the opposite side of the $y$-axis). Reflecting a point in this way reverses only the sign of its $x$-coordinate, while the $y$-coordinate does not change, so from Definition 3.2 we immediately obtain
$$ \begin{equation} \cos(180°-\theta)=-\cos\theta,\qquad \sin(180°-\theta)=\sin\theta \label{eq:3-obtuse-conv} \end{equation} $$This relation will actually be used in Section 3.3 (the law of cosines, the obtuse case), so let us remember it.
Now we prove the three basic relations that hold among the trigonometric ratios. All three are proved for an acute angle $\theta$ (the situation of Definition 3.1), but in fact they hold unchanged for the general angles of Definition 3.2 as well (the proof follows by the same argument as in the acute case, from the fact that the coordinates $(\cos\theta,\sin\theta)$ on the unit circle lie on the circle at distance 1 from the origin, that is, from $x^2+y^2=1$).
Theorem 3.1: $\sin^2\theta+\cos^2\theta=1$
For any $\theta$ ($0°\lt\theta\lt90°$), $\sin^2\theta+\cos^2\theta=1$ holds.
Proof
Applying the Pythagorean theorem (a relation among the lengths of the three sides; already learned in junior high school mathematics) to the right triangle in Figure 3.1, we have $a^2+b^2=c^2$. Since $a=c\cos\theta$ and $b=c\sin\theta$ by Definition 3.1, substituting these gives
$$ (c\cos\theta)^2+(c\sin\theta)^2=c^2 $$Expanding the left side and factoring out $c^2$ ($c^2\cos^2\theta+c^2\sin^2\theta=c^2(\cos^2\theta+\sin^2\theta)$), we obtain
$$ c^2\cos^2\theta+c^2\sin^2\theta=c^2(\sin^2\theta+\cos^2\theta)=c^2 $$Since the length of a side of a triangle is nonzero, $c\ne0$, so we may divide both sides by $c^2$ ($\ne0$). Dividing gives
$$ \sin^2\theta+\cos^2\theta=\frac{c^2}{c^2}=1 $$which is what we wanted to show.
∎
Theorem 3.2: $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$
For any $\theta$ ($0°\lt\theta\lt90°$, $\cos\theta\ne0$), $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$ holds.
Proof
By Definition 3.1, $\tan\theta=\dfrac{b}{a}$. Substituting $a=c\cos\theta$ and $b=c\sin\theta$ into the denominator and the numerator of this fraction, respectively, we get
$$ \tan\theta=\frac{b}{a}=\frac{c\sin\theta}{c\cos\theta} $$Canceling the factor $c$ ($\ne0$) common to the numerator and the denominator, we find that
$$ \tan\theta=\frac{\sin\theta}{\cos\theta} $$holds.
∎
Theorem 3.3: $1+\tan^2\theta=\dfrac{1}{\cos^2\theta}$
For any $\theta$ ($0°\lt\theta\lt90°$, $\cos\theta\ne0$), $1+\tan^2\theta=\dfrac{1}{\cos^2\theta}$ holds.
Proof
The starting point is $\sin^2\theta+\cos^2\theta=1$ from Theorem 3.1. We multiply both sides of this equation by $\dfrac{1}{\cos^2\theta}$ (since $\cos\theta\ne0$, this does not involve division by zero, so it is a permitted operation). Multiplying each term on the left side separately, we get
$$ \frac{\sin^2\theta}{\cos^2\theta}+\frac{\cos^2\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta} $$The second term on the left side is $\dfrac{\cos^2\theta}{\cos^2\theta}=1$, and the first term can be rewritten as $\dfrac{\sin^2\theta}{\cos^2\theta}=\left(\dfrac{\sin\theta}{\cos\theta}\right)^2$, so we obtain
$$ \left(\frac{\sin\theta}{\cos\theta}\right)^2+1=\frac{1}{\cos^2\theta} $$Now we use Theorem 3.2 ($\tan\theta=\sin\theta/\cos\theta$) to replace the expression inside the parentheses by $\tan\theta$, which gives
$$ \tan^2\theta+1=\frac{1}{\cos^2\theta} $$That is, $1+\tan^2\theta=\dfrac{1}{\cos^2\theta}$ has been shown.
∎
Caution: When $\tan\theta$ is undefined
$\tan\theta=\sin\theta/\cos\theta$ cannot be defined when $\cos\theta=0$ (this never happens for acute angles, but for the general angle of Definition 3.2 it happens when $\theta=90°$). This reflects a rule common to all of mathematics: we do not consider expressions in which the denominator of a division is 0.
Example 3.1: Numerical check of the basic relations among the trigonometric ratios
When $\theta=30°$, we have $\sin\theta=\dfrac12$ and $\cos\theta=\dfrac{\sqrt3}{2}$. Using these values, verify that the three equations in Theorems 3.1–3.3 actually hold.
Solution. First we compute $\tan\theta$ from its definition. $\tan30°=\dfrac{\sin30°}{\cos30°}=\dfrac{1/2}{\sqrt3/2}=\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{3}$ (we rationalized the denominator).
(Check of Theorem 3.1) $\sin^2 30°+\cos^2 30°=\left(\dfrac12\right)^2+\left(\dfrac{\sqrt3}{2}\right)^2=\dfrac14+\dfrac34=\dfrac44=1$. It does indeed hold.
(Check of Theorem 3.2) $\dfrac{\sin30°}{\cos30°}=\dfrac{1/2}{\sqrt3/2}=\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{3}=\tan30°$. It does indeed hold.
(Check of Theorem 3.3) $1+\tan^2 30°=1+\left(\dfrac{\sqrt3}{3}\right)^2=1+\dfrac{3}{9}=1+\dfrac13=\dfrac43$. On the other hand, $\dfrac{1}{\cos^2 30°}=\dfrac{1}{3/4}=\dfrac43$. Both sides equal $\dfrac43$, so they agree.
(Cross-check: computing the three equations with Python's sympy, keeping exact fractions, gives the same results.)
3.2 The Law of Sines
From here on we consider a triangle $\triangle ABC$ and write $A,B,C$ for the sizes of $\angle A,\angle B,\angle C$ themselves. We also denote the length of the side opposite an angle (the side facing it) by the same letter as that angle, written in lowercase—that is, $a$ is the length of the side $BC$ opposite $\angle A$, $b$ is the length of the side $CA$ opposite $\angle B$, and $c$ is the length of the side $AB$ opposite $\angle C$. We use this notation throughout this chapter. (In Figure 3.1 of Section 3.1 we also used $a,b,c$ as the names of the sides of a right triangle, but note that the meaning there is different. The same letters are reused because “denoting an angle and its opposite side by the same letter” is a widely established convention for triangles.)
Every triangle has exactly one circumcircle (the circle passing through all the vertices; its center is the circumcenter studied in Chapter 2). The law of sines is a beautiful theorem that connects the ratios among the three sides and the three angles to the size of this circumcircle.
Theorem 3.4: The law of sines
If $R$ is the circumradius of a triangle $\triangle ABC$, then
$$ \begin{equation} \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R \label{eq:3-sine-rule} \end{equation} $$holds.
Proof
First, let us recall the inscribed angle theorem learned in junior high school (the theorem that all inscribed angles standing on the same arc are equal—that is, the angle subtended by a chord is constant regardless of where on the arc the vertex lies). As in Figure 3.3 (panel (i)), inscribe $\triangle ABC$ in its circumcircle of radius $R$. $\angle A$ is an inscribed angle subtending the arc $BC$ (the arc not containing $A$).
To investigate the relation between this $\angle A$ and the side $a=BC$, we construct an auxiliary right triangle as in Figure 3.4 (panel (ii)). Draw the diameter of the circumcircle through the point $C$, and let $D$ be its endpoint on the opposite side of the circle from C ($CD$ is a diameter, so its length is $2R$). Then join $B$ and $D$ by a line segment.
Since $CD$ is a diameter, $\angle DBC=90°$ for the point $B$ on the circle (a special case of the inscribed angle theorem: an inscribed angle subtending a diameter is a right angle). That is, $\triangle DBC$ is a right triangle whose right angle is $\angle B$ ($=\angle DBC$).
On the other hand, $\angle BDC$ and $\angle BAC$ ($=\angle A$) are both inscribed angles subtending the same arc $BC$, so by the inscribed angle theorem, $\angle BDC=\angle BAC=A$ (the point of this auxiliary construction is that “moving” the position from which the arc $BC$ is viewed, from the vertex $A$ to the vertex $D$, does not change the size of the angle).
Now we apply Definition 3.1 to the right triangle $DBC$. The right angle is at $B$, so the hypotenuse is $DC=2R$, and the side opposite $\angle BDC=A$ is $BC=a$. Hence we have
$$ \sin A=\sin(\angle BDC)=\frac{BC}{DC}=\frac{a}{2R} $$Multiplying both sides by $2R$ and then dividing by $\sin A$ (since $\sin A\ne0$ when $0°\lt A\lt180°$, the division is allowed) gives
$$ \frac{a}{\sin A}=2R $$For $\angle B$ and $\angle C$ as well, if we replace the diameter by the one through $A$ and the one through $B$, respectively, and repeat exactly the same argument, we likewise show that $\dfrac{b}{\sin B}=2R$ and $\dfrac{c}{\sin C}=2R$. Putting all this together, we find that
$$ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R $$holds.
∎
Intuition: The law of sines is a formula that makes “the larger the angle subtended, the longer the opposite side” precise
Intuitively, within the same circumcircle, the longer a chord (side) is, the larger the inscribed angle subtending it. The law of sines makes this intuition precise as the equation “$a=2R\sin A$.” Since $\sin A$ attains its maximum value 1 at $A=90°$, the chord whose length comes closest to the diameter ($2R$) is the chord subtended by a right angle—this actually corresponds to the fact that the diameter itself is the hypotenuse of a right triangle (which is exactly the case for DC in Figure 3.4).
Example 3.2: Finding the circumradius with the law of sines
In $\triangle ABC$, suppose $a=8$, $B=45°$, and $C=75°$. Find the circumradius $R$.
Solution. The sum of the interior angles of a triangle is $180°$, so $A=180°-B-C=180°-45°-75°=60°$. By the law of sines $\dfrac{a}{\sin A}=2R$, we have
$$ R=\frac{a}{2\sin A}=\frac{8}{2\sin60°}=\frac{8}{2\times\dfrac{\sqrt3}{2}}=\frac{8}{\sqrt3}=\frac{8\sqrt3}{3} $$(In the last step we rationalized the denominator.) Hence $R=\dfrac{8\sqrt3}{3}\approx4.62$.
3.3 The Law of Cosines — Proof by Cases: Acute, Right, and Obtuse Angles
The Pythagorean theorem $a^2=b^2+c^2$ can be used only for a right triangle ($\angle A=90°$). If $\angle A$ were not a right angle, by how much would $a^2$ deviate from $b^2+c^2$? The answer is given by the law of cosines, a theorem that extends the Pythagorean theorem to an arbitrary triangle.
Theorem 3.5: The law of cosines
In a triangle $\triangle ABC$,
$$ \begin{equation} a^2=b^2+c^2-2bc\cos A \label{eq:3-cosine-rule} \end{equation} $$holds. If the labels of the vertices are interchanged, then $b^2=c^2+a^2-2ca\cos B$ and $c^2=a^2+b^2-2ab\cos C$ also hold, as formulas with the same meaning.
The idea of the proof is that “thinking in terms of area makes it easy.” Draw, on the outside of the triangle, a square on each side of the triangle (their areas are $a^2,b^2,c^2$, respectively). Drop the perpendicular from vertex $A$ to the opposite side $BC$, and let $H$ be its foot. Extend this perpendicular straight on into the square on side $BC$ (area $a^2$). The perpendicular is perpendicular to $BC$, and the remaining two sides of the square are also perpendicular to $BC$, so the extended line divides the square into exactly two rectangles. Finding the areas of these two rectangles is the heart of the proof. How the areas of the rectangles are found changes slightly depending on whether $\angle A$ is acute, right, or obtuse, so we consider three cases (an obtuse triangle has exactly one obtuse angle, so we may simply call that obtuse angle $\angle A$; for an acute triangle, the same argument holds for any of $\angle A,\angle B,\angle C$).
[1] When $\angle A$ is acute (acute triangle)
Consider a triangle all three of whose angles are acute. Since $\angle B,\angle C$ are both acute, the foot $H$ of the perpendicular lies in the interior of side $BC$. In the right triangle $ABH$, the side adjacent to $\angle B$ is $BH$ and the hypotenuse is $c$, so by Definition 3.1
$$ BH=c\cos B $$Similarly, in the right triangle $ACH$, the side adjacent to $\angle C$ is $HC$ and the hypotenuse is $b$, so we have
$$ HC=b\cos C $$The two rectangles formed by extending the perpendicular both have height equal to the side length $a$ of the square, and their bases are $BH,HC$, respectively, so their areas are
$$ (\text{area of the left rectangle})=a\cdot BH=a\cdot c\cos B=ca\cos B,\qquad (\text{area of the right rectangle})=a\cdot HC=a\cdot b\cos C=ab\cos C $$The sum of the areas of the two rectangles equals the area $a^2$ of the original square, so we obtain
$$ \begin{equation} a^2=ca\cos B+ab\cos C \label{eq:3-square-a} \end{equation} $$If we apply exactly the same argument to the case where we drop the perpendicular from vertex $B$ to the opposite side $CA$ (since the triangle is acute, the foot of this perpendicular also lies in the interior of side $CA$), the square on side $CA$ (area $b^2$) is divided into two rectangles, and we obtain
$$ \begin{equation} b^2=bc\cos A+ab\cos C \label{eq:3-square-b} \end{equation} $$(This is derived in the same way, from the fact that the side $AH'$ adjacent to $\angle A$ has length $c\cos A$ and the side $H'C$ adjacent to $\angle C$ has length $a\cos C$. Here $H'$ is the foot of the perpendicular from vertex $B$.) Furthermore, if we drop the perpendicular from vertex $C$ to the opposite side $AB$, then for the square on side $AB$ (area $c^2$) as well,
$$ \begin{equation} c^2=ca\cos B+bc\cos A \label{eq:3-square-c} \end{equation} $$is obtained in the same way. Now, solving Eq. \eqref{eq:3-square-b} for $ab\cos C$ and Eq. \eqref{eq:3-square-c} for $ca\cos B$, we get
$$ ab\cos C=b^2-bc\cos A,\qquad ca\cos B=c^2-bc\cos A $$Substituting these into Eq. \eqref{eq:3-square-a} shows that
$$ a^2=ca\cos B+ab\cos C=(c^2-bc\cos A)+(b^2-bc\cos A)=b^2+c^2-2bc\cos A $$holds.
∎ (acute triangle case)
Erratum: Swapped formulas in the original notes
In the original notes (p. 9, [1]), it is written that $b^2=S_2+S_3=ca\cos B+bc\cos A$ and $c^2=S_4+S_5=bc\cos A+ab\cos C$. However, when we check by substituting specific numerical values ($a=5,b=6,c=7$), we find that these two formulas have the values of $b^2$ and $c^2$ exactly interchanged (with these values we should have $b^2=36$ and $c^2=49$, but the formulas in the notes give the opposite values). In case [3], the obtuse case (p. 10), of the same law of cosines, formulas of exactly the same form are written correctly, with no interchange, so this is considered to be a slip confined to [1]. Eqs. \eqref{eq:3-square-b} and \eqref{eq:3-square-c} above are shown in the correct form, with the subscripts interchanged.
In the original notes: Area correspondence via $S_1$ through $S_6$
In the original notes, the formulas for $b^2,c^2$ are obtained not by dropping new perpendiculars from vertices $B,C$, but as follows: the areas of the two rectangles found above ($ca\cos B$ and $ab\cos C$) are called $S_1,S_6$, respectively, within the square $a^2$, and it is shown in a single figure that these have the same areas as the corresponding parts ($S_2,S_3,S_4,S_5$) that appear in the squares on sides $b,c$ (namely $S_1=S_2$, $S_3=S_4$, $S_5=S_6$) (in Figure 3.5, the notes' symbols $S_1,S_6$ are attached to the regions $ca\cos B$ and $ab\cos C$). This correspondence of areas follows from the fact that parallelograms with the same base and the same height have equal areas (the area does not change when a vertex is slid along a parallel line), but explaining the basis for that fact without omitting steps would require one more stage of preparation. So in this book we also drop new perpendiculars from vertices $B,C$, and derive $b^2,c^2$ in a form in which every step of the algebra can be followed. The conclusions finally obtained (Eqs. \eqref{eq:3-square-b} and \eqref{eq:3-square-c}) agree with those of the notes.
[2] When $\angle A=90°$ (right triangle)
In this case, since $\angle A$ itself is a right angle, there is no need to drop a “perpendicular” from vertex $A$ to the opposite side $BC$ in the first place, and the construction that measures the distance from $A$ to $BC$ is already contained in the triangle as it stands—one would like to say that actually $H=A$, but that is not so: $H$ is still a different point on side $BC$ (since $\angle B,\angle C$ are both acute, $H$ lies in the interior of side $BC$, just as in case [1]). The argument of case [1] can be used as it is: $BH=c\cos B$ and $HC=b\cos C$ hold, and
$$ a^2=ca\cos B+ab\cos C $$holds (this has the same form as Eq. \eqref{eq:3-square-a} of case [1]). Here, noting that $\cos90°=0$, the term “$bc\cos A$” used in the proofs of Eqs. \eqref{eq:3-square-b} and \eqref{eq:3-square-c} becomes 0, and we get
$$ b^2=bc\cos A+ab\cos C=ab\cos C,\qquad c^2=ca\cos B+bc\cos A=ca\cos B $$so $b^2,c^2$ are each exactly equal to the area of a single rectangle. Hence
$$ b^2+c^2-2bc\cos A=ab\cos C+ca\cos B-2bc\times0=ca\cos B+ab\cos C=a^2 $$holds, and this agrees with the Pythagorean theorem $a^2=b^2+c^2$ (because $-2bc\cos90°=0$). In other words, the Pythagorean theorem is the special case $\angle A=90°$ of the law of cosines.
∎ (right triangle case)
[3] When $\angle A$ is obtuse (obtuse triangle)
Even when $\angle A$ is obtuse ($90°\lt A\lt180°$), $\angle B,\angle C$ are both acute (because the sum of the interior angles of a triangle is $180°$ and there is only one obtuse angle). Therefore the foot $H$ of the perpendicular from vertex $A$ still lies in the interior of side $BC$, for the same reason as in case [1], and $BH=c\cos B$ and $HC=b\cos C$ also hold by the same argument. Hence
$$ a^2=ca\cos B+ab\cos C $$also takes exactly the same form as in cases [1] and [2] (this formula is always correct, regardless of the size of $\angle A$).
For the formulas for $b^2,c^2$, however, the situation changes. When we drop a perpendicular from vertex $B$ to the opposite side $CA$ and let $H'$ be its foot, $H'$ does not fall inside side $CA$—because $\angle A$ is obtuse, $H'$ lies on the extension of side $CA$ from $C$ toward $A$, beyond $A$ (see “Why?” below). Similarly, the foot $H''$ of the perpendicular dropped from vertex $C$ to the opposite side $AB$ also lies on the extension of side $AB$, beyond $A$.
Why?: Why does the foot of the perpendicular fall outside the side?
When we drop a perpendicular from vertex $B$ to line $CA$, its foot falls inside segment $CA$ only when $\angle A$ and $\angle C$ are both acute (if $\angle A$ is obtuse, then the line bends at $A$ on line $CA$ and approaches the direction of the perpendicular, so the foot of the perpendicular goes past $A$). When $\angle A$ is obtuse, it helps to consider the exterior angle of $\angle BAC$ ($180°-A$, the angle that together with $\angle BAC$ forms a straight line). The foot $H'$ of the perpendicular and $C,A$ lie on one straight line, and $H'$ lies beyond $A$, so in the right triangle $BH'A$ we have $\angle H'AB=180°-A$ (the exterior angle of $\angle BAC$).
So we focus on the right triangle $BH'A$. Since $\angle H'AB=180°-A$ and the hypotenuse is $AB=c$, by Definition 3.1 and Eq. \eqref{eq:3-obtuse-conv} we have
$$ AH'=c\cos(180°-A)=-c\cos A $$(since $\cos A\lt0$ when $A$ is obtuse, $-c\cos A\gt0$, and this is meaningful as a length). Because $H'$ lies beyond $A$, on the side opposite $C$, the distance from $C$ to $H'$ is $CA=b$ plus $AH'$:
$$ CH'=CA+AH'=b+(-c\cos A)=b-c\cos A $$On the other hand, looking at the right triangle $BH'C$ (the right angle is at $H'$), we have $\angle BCH'=\angle BCA=C$ ($H'$ lies on line $CA$, so $\angle C$ itself does not change), and the hypotenuse is $BC=a$, so $CH'=a\cos C$ as well. Setting the two expressions for $CH'$ equal, we obtain
$$ a\cos C=b-c\cos A \quad\Longrightarrow\quad b=c\cos A+a\cos C $$Multiplying both sides by $b$ gives
$$ \begin{equation} b^2=bc\cos A+ab\cos C \label{eq:3-square-b-obtuse} \end{equation} $$Thus an equation of exactly the same form as Eq. \eqref{eq:3-square-b} has been obtained in the obtuse case as well. In exactly the same way, considering the foot $H''$ of the perpendicular dropped from vertex $C$ to the extension of side $AB$, we obtain
$$ \begin{equation} c^2=ca\cos B+bc\cos A \label{eq:3-square-c-obtuse} \end{equation} $$(this has the same form as Eq. \eqref{eq:3-square-c}). All that remains is exactly the same algebraic computation as in case [1]—form $ab\cos C=b^2-bc\cos A$ and $ca\cos B=c^2-bc\cos A$ from Eqs. \eqref{eq:3-square-b-obtuse} and \eqref{eq:3-square-c-obtuse}, and substitute them into $a^2=ca\cos B+ab\cos C$—and this shows that
$$ a^2=(c^2-bc\cos A)+(b^2-bc\cos A)=b^2+c^2-2bc\cos A $$holds in the obtuse case as well.
∎ (obtuse triangle case)
Example 3.3: Finding a side length with the law of cosines (when the cosine of an acute angle is used)
In $\triangle ABC$, if $a=5$, $b=7$, and $C=60°$, find the length of side $c$.
Solution. We use the form of Theorem 3.5 for $C$, namely $c^2=a^2+b^2-2ab\cos C$.
$$ c^2=5^2+7^2-2\times5\times7\times\cos60°=25+49-70\times\frac12=74-35=39 $$Since $c\gt0$, we have $c=\sqrt{39}$. (Here $\sqrt{39}\approx6.24$5, and the triangle inequalities $a+b\gt c$, $a+c\gt b$, $b+c\gt a$ are all satisfied as well, so a triangle with these sides can actually be constructed.)
Example 3.4: Using the law of cosines to test for an obtuse angle
Suppose the three sides of $\triangle ABC$ are $a=7$, $b=3$, $c=5$. Find the cosine of $\angle A$, and verify that $\angle A$ is obtuse.
Solution. Solving Theorem 3.5 for $\cos A$ gives $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$. Substituting the numerical values,
$$ \cos A=\frac{3^2+5^2-7^2}{2\times3\times5}=\frac{9+25-49}{30}=\frac{-15}{30}=-\frac12 $$Since $\cos A=-\dfrac12\lt0$, $A$ is obtuse. (In the range $0°\lt A\lt180°$, the cosine is negative only for an obtuse angle—recall the unit circle from Definition 3.2: the $x$-coordinate was negative exactly for angles greater than $90°$.) In fact, $\cos A=-\dfrac12$ gives $A=120°$.
Example 3.5: Consistency of the law of cosines with the Pythagorean theorem
Suppose the three sides of $\triangle ABC$ are $a=5$, $b=3$, $c=4$ (the $3$-$4$-$5$ right triangle). Use the law of cosines to compute the cosine of $\angle A$, and verify that $\angle A=90°$.
Solution. $\cos A=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{3^2+4^2-5^2}{2\times3\times4}=\dfrac{9+16-25}{24}=\dfrac{0}{24}=0$. In the range $0°\lt A\lt180°$, $\cos A=0$ only when $A=90°$, so $\angle A=90°$. This is a confirmation, by way of the law of cosines, that $b^2+c^2=9+16=25=a^2$—that is, that the Pythagorean theorem holds just as it is (a concrete instance of the fact, shown in case [2], that “the law of cosines contains the Pythagorean theorem”).
Example 3.6: Proving the median theorem from the law of cosines (a proof problem)
Let $M$ be the midpoint of side $BC$ of $\triangle ABC$ (the segment $AM$ is called the median from vertex $A$). Then show that
$$ AB^2+AC^2=2AM^2+2BM^2 $$holds, using the law of cosines (this is a theorem called the median theorem, or Apollonius's theorem).
Solution. Let $\angle AMB=\varphi$ (phi). Since $M$ lies on side $BC$, $\angle AMB$ and $\angle AMC$ are angles that together form a straight line, so $\angle AMC=180°-\varphi$. Also, since $M$ is the midpoint of $BC$, we have $BM=CM$.
Applying the law of cosines to triangle $ABM$ (in the form that focuses on the two sides $AM,BM$ enclosing $\angle AMB=\varphi$), we obtain
$$ c^2=AB^2=AM^2+BM^2-2\cdot AM\cdot BM\cos\varphi $$Similarly, applying the law of cosines to triangle $ACM$ (in the form that focuses on the two sides $AM,CM$ enclosing $\angle AMC=180°-\varphi$), we get
$$ b^2=AC^2=AM^2+CM^2-2\cdot AM\cdot CM\cos(180°-\varphi) $$Now, using $CM=BM$ and Eq. \eqref{eq:3-obtuse-conv} ($\cos(180°-\varphi)=-\cos\varphi$), this can be rewritten as
$$ AC^2=AM^2+BM^2+2\cdot AM\cdot BM\cos\varphi $$Finally, when we add the two equations (the one for $AB^2$ and the one for $AC^2$) side by side, the terms containing $\cos\varphi$ cancel exactly, since they differ only in sign, and we obtain
$$ AB^2+AC^2=2AM^2+2BM^2 $$(Cross-check: if $A(0,0),B(9,0),C(3,5)$, then $M=(6,2.5)$, and $AB^2+AC^2=81+34=115$ while $2AM^2+2BM^2=2\times42.25+2\times15.25=84.5+30.5=115$, so the two sides agree.)
3.4 Radian Measure
Until now we have measured angles in the unit “$°$” (degree). However, there is no mathematical necessity behind the convention of dividing one full turn into $360$ equal parts (the number $360$ is said to derive from the base-60 system used in ancient Babylonia and from the fact that a year has about 360 days, so it is largely due to historical reasons). Here we introduce a more natural way of measuring angles—one that is especially convenient when we handle trigonometric functions in calculus later on—radian measure.
Definition 3.3: Radian measure
In a circle of radius $1$ (the unit circle), the method of measuring angles so that the arc length corresponding to a central angle $\theta$ is exactly $\theta$ is called radian measure, and the unit of angle measurement in that case is called the radian (symbol $\mathrm{rad}$). In other words, in radian measure we measure the size of an angle by the ratio “(length) : (angle).”
Derivation of $180°=\pi\,\mathrm{rad}$
The circumference of a circle of radius $1$ is $2\pi$, by the circumference formula $2\pi r$ (with $r=1$). On the other hand, the angle for one full turn around the circle is $360°$. By the definition of radian measure (arc length and angle are proportional), the ratio “length of the whole circumference : angle of one full turn” and the ratio “arc length $1$ : angle $1\,\mathrm{rad}$” must be the same, so the proportion
$$ 2\pi:360°=1:1\,\mathrm{rad} $$holds. Multiplying the outer terms together and the inner terms together (the product of the means equals the product of the extremes), we obtain $2\pi\times1\,\mathrm{rad}=360°\times1$, that is,
$$ 360°=2\pi\,\mathrm{rad} $$Dividing both sides by $2$ gives
$$ \begin{equation} 180°=\pi\,\mathrm{rad} \label{eq:3-radian} \end{equation} $$This relation, Eq. \eqref{eq:3-radian}, is the basis for converting between degree measure and radian measure ($1°=\dfrac{\pi}{180}\,\mathrm{rad}$ and $1\,\mathrm{rad}=\dfrac{180°}{\pi}$). As Figure 3.8 also shows, radian measure has the effect of converting an angle into a length (with the radius as the unit)—this is a major advantage of radian measure that degree measure does not have.
Why?: Radian measure can be used even in circles whose radius is not $1$
Consider the arc length $l$ corresponding to a central angle of $\theta\,\mathrm{rad}$ in a circle of radius $r$. A circle of radius $1$ and a circle of radius $r$ are similar (related by enlargement or reduction), and the similarity ratio is $r:1$. Since arc length also stretches or shrinks by the same ratio as the similarity ratio, it suffices to multiply the arc length $\theta$ for radius $1$ (which is just Definition 3.3 itself) by $r$, so we have
$$ l=r\theta $$Similarly, the area of a circle of radius $r$ is $\pi r^2$ and its central angle is $2\pi\,\mathrm{rad}$ (the full circle), and the area is proportional to the central angle, so for the area $S$ of a sector with central angle $\theta\,\mathrm{rad}$ we can set up the proportion $S:\pi r^2=\theta:2\pi$. Solving this for $S$ gives
$$ S=\pi r^2\times\frac{\theta}{2\pi}=\frac12 r^2\theta $$The arc length formula $l=r\theta$ and the sector area formula $S=\dfrac12 r^2\theta$ are both important formulas that we will use repeatedly in later chapters.
Example 3.7: Converting between degree measure and radian measure
(a) Express $135°$ in radian measure. (b) Express $\dfrac{5}{6}\pi\,\mathrm{rad}$ in degree measure.
Solution. We use Eq. \eqref{eq:3-radian} ($180°=\pi\,\mathrm{rad}$, that is, $1°=\dfrac{\pi}{180}\,\mathrm{rad}$).
(a) $135°=135\times\dfrac{\pi}{180}\,\mathrm{rad}=\dfrac{135}{180}\pi\,\mathrm{rad}=\dfrac{3}{4}\pi\,\mathrm{rad}$ (we reduced the fraction: the greatest common divisor of $135$ and $180$ is $45$).
(b) Since $1\,\mathrm{rad}=\dfrac{180°}{\pi}$, we have $\dfrac56\pi\,\mathrm{rad}=\dfrac56\pi\times\dfrac{180°}{\pi}=\dfrac56\times180°=150°$ (the $\pi$ cancels out).
Example 3.8: Arc length and area of a sector
In a circle of radius $1$, find the arc length and the area of a sector whose central angle is $2\,\mathrm{rad}$.
Solution. We substitute $r=1,\theta=2$ into the formulas $l=r\theta$ and $S=\dfrac12 r^2\theta$. The arc length is $l=1\times2=2$, and the area is $S=\dfrac12\times1^2\times2=1$. (Note also that we can read off directly from the definition of radian measure itself, Definition 3.3, that an arc with radius $1$ and central angle $2\,\mathrm{rad}$ has length $2$.)
3.5 Summary and Exercises
3.5.1 Summary
- For an acute angle $\theta$ of a right triangle, we define $\sin\theta=$ opposite side $/$ hypotenuse, $\cos\theta=$ adjacent side $/$ hypotenuse, and $\tan\theta=$ opposite side $/$ adjacent side (Definition 3.1). Using the unit circle, the definition can be extended to obtuse angles ($90°\lt\theta\le180°$) as well (Definition 3.2).
- Among the trigonometric ratios, the three basic relations $\sin^2\theta+\cos^2\theta=1$, $\tan\theta=\sin\theta/\cos\theta$, and $1+\tan^2\theta=1/\cos^2\theta$ hold (Theorems 3.1–3.3). All of them follow from the Pythagorean theorem.
- Law of sines: $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$ ($R$ is the circumradius). It can be proved from the inscribed angle theorem and the fact that an inscribed angle subtending a diameter is a right angle (Theorem 3.4).
- Law of cosines: $a^2=b^2+c^2-2bc\cos A$. It is a theorem that extends the Pythagorean theorem to an arbitrary triangle, and it holds whether $\angle A$ is acute, right, or obtuse (Theorem 3.5). When $\angle A=90°$, it coincides with the Pythagorean theorem itself.
- Radian measure: a method of measuring the arc length and the central angle by the same number in a circle of radius 1. $180°=\pi\,\mathrm{rad}$ (Eq. \eqref{eq:3-radian}). In a sector of radius $r$, the arc length is $l=r\theta$ and the area is $S=\dfrac12 r^2\theta$.
- The law of sines connects “the ratio of the side lengths and the ratio of the $\sin$ of their opposite angles,” and the law of cosines connects “the cosine of one angle and the lengths of the three sides.” When the given conditions involve a mixture of sides and angles, rather than sides alone or angles alone, it is a good idea to first check which of the law of cosines and the law of sines can be used (the law of cosines is easier to use when the three sides are known; the law of sines is easier to use when two angles and one side are known, or when a relation between two sides and the angle opposite one of them is known).
(Addendum) Organizing which of the law of sines and the law of cosines to use according to the kind of information we know gives the following table.
| Known information | Theorem to use | What can be found |
|---|---|---|
| Three sides ($a,b,c$) | Law of cosines | All the angles |
| Two sides and the angle between them ($a,b,C$, etc.) | Law of cosines | The remaining side, and then the other angles |
| One side and the angles at both its ends ($c,A,B$, etc.) | Law of sines | The remaining two sides |
| Two sides and the angle opposite one of them ($a,b,A$, etc.) | Law of sines | The angle opposite the other side (determine the angle from the value of $\sin$; note that there may be two possibilities, an acute angle and an obtuse angle) |
3.5.2 Exercises
Exercise 3.1: Basic relations among the trigonometric ratios
Suppose that $\theta$ is an acute angle and $\sin\theta=\dfrac13$. Find $\cos\theta$ and $\tan\theta$, and verify that $1+\tan^2\theta=1/\cos^2\theta$ holds.
Hint: Find $\cos\theta$ from Theorem 3.1 ($\sin^2\theta+\cos^2\theta=1$); since $\theta$ is acute, choose $\cos\theta\gt0$. Then find $\tan\theta$ using Theorem 3.2.
Exercise 3.2: The law of sines
In $\triangle ABC$, if $a=10$ and $A=30°$, find the circumradius $R$. Furthermore, if $B=45°$, find the length of side $b$.
Hint: First find $R$ from $\dfrac{a}{\sin A}=2R$. Next use $b=2R\sin B$.
Exercise 3.3: The law of cosines and determining whether an angle is acute or obtuse
If the three sides of $\triangle ABC$ are $a=7$, $b=8$, and $c=5$, find the cosine of the angle opposite the longest side by the law of cosines, and determine whether that angle is acute or obtuse.
Hint: The largest angle in a triangle is the angle opposite the longest side. First check which side is the longest, and then use the law of cosines.
Exercise 3.4: Using the law of cosines and the law of sines together
In $\triangle ABC$, if $a=6$, $b=4$, and $C=60°$, first find side $c$ by the law of cosines, and then find $\sin A$ by the law of sines.
Hint: Once $c$ is found, $\sin A$ can be found from $\dfrac{a}{\sin A}=\dfrac{c}{\sin C}$.
Exercise 3.5: Radian measure
In a circle of radius $6$, if the arc length is $4\pi$, find the central angle in both radians and degrees. Also find the area of the sector in that case.
Hint: Find $\theta$ (in radians) from $l=r\theta$, and convert it to degrees by Eq. \eqref{eq:3-radian}. The area is $S=\dfrac12 r^2\theta$.
Exercise 3.6: Deriving the Pythagorean theorem in general from the law of cosines
Show that, when $\angle A=90°$, the Pythagorean theorem $a^2=b^2+c^2$ follows from the law of cosines $a^2=b^2+c^2-2bc\cos A$, not with specific numerical values but with the letters $b,c$ left as they are (that is, in general).
Hint: It suffices to substitute the value of $\cos90°$.
3.5.3 References
- Yasuhide Mochizuki, Mathematics Notes: High School Mathematics (handwritten lecture notes, in Japanese). The source of this chapter.
- I. M. Gelfand, M. Saul, Trigonometry, Birkhäuser, 2001.