Chapter 1: Trigonometric Ratios
Trigonometric Ratios
Beginner (high-school level)
About This Chapter
A trigonometric ratio is a concept that expresses a ratio of side lengths in a right triangle. Mathematicians in ancient Greece, India, and the Islamic world developed it for astronomy and surveying. This chapter covers the definitions and basic properties of sin, cos, and tan, together with the law of sines and the law of cosines.
Definition of Trigonometric Ratios
Definition in a Right Triangle
In a right triangle, for one acute angle $\theta$, the trigonometric ratios are defined as follows.
Definition 1.1 (Trigonometric ratios)
In a right triangle, for an acute angle $\theta$:
\begin{align} \sin\theta &= \frac{\text{opposite}}{\text{hypotenuse}} = \frac{a}{c} \label{eq:sin-def}\\ \cos\theta &= \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{b}{c} \label{eq:cos-def}\\ \tan\theta &= \frac{\text{opposite}}{\text{adjacent}} = \frac{a}{b} \label{eq:tan-def} \end{align}Here the "opposite" is the side opposite angle $\theta$, and the "adjacent" is the side next to $\theta$ (the one that is not the hypotenuse).
Trigonometric Ratios of Special Angles
Let us find the trigonometric ratios of the especially important angles $30°$, $45°$, and $60°$.
Table 1.1 Trigonometric ratios of special angles
| Angle $\theta$ | $\sin\theta$ | $\cos\theta$ | $\tan\theta$ |
|---|---|---|---|
| $30°$ | $\dfrac{1}{2}$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$ |
| $45°$ | $\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$ | $\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$ | $1$ |
| $60°$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{2}$ | $\sqrt{3}$ |
Derivation
Case 45°: Consider a right isosceles triangle (two equal legs, with 45° angles adjacent to the right angle). Taking each leg to be $1$, the hypotenuse is $\sqrt{1^2 + 1^2} = \sqrt{2}$ by the Pythagorean theorem. Hence:
$$\sin 45° = \cos 45° = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}, \quad \tan 45° = \dfrac{1}{1} = 1$$Cases 30° and 60°: Splitting an equilateral triangle of side $2$ in half gives a right triangle with base $1$, height $\sqrt{3}$, and hypotenuse $2$. In this triangle the angle facing the hypotenuse is $60°$ and the angle at the base is $30°$.
Case $60°$:
$$\sin 60° = \dfrac{\sqrt{3}}{2}, \quad \cos 60° = \dfrac{1}{2}, \quad \tan 60° = \dfrac{\sqrt{3}}{1} = \sqrt{3}$$Case $30°$:
$$\sin 30° = \dfrac{1}{2}, \quad \cos 30° = \dfrac{\sqrt{3}}{2}, \quad \tan 30° = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$$Reciprocal Trigonometric Ratios
There are also functions defined as the reciprocals of sin, cos, and tan.
Definition 1.2 (Reciprocal ratios)
\begin{align} \csc\theta &= \frac{1}{\sin\theta} = \frac{c}{a} \quad \text{(cosecant)} \label{eq:csc-def}\\ \sec\theta &= \frac{1}{\cos\theta} = \frac{c}{b} \quad \text{(secant)} \label{eq:sec-def}\\ \cot\theta &= \frac{1}{\tan\theta} = \frac{b}{a} \quad \text{(cotangent)} \label{eq:cot-def} \end{align}These are used mainly in advanced mathematics and physics, and are seldom treated in Japanese high-school mathematics.
Relations Among the Trigonometric Ratios
Derivation from the Pythagorean Theorem
In a right triangle the Pythagorean theorem $a^2 + b^2 = c^2$ holds. Dividing both sides by $c^2$:
$$\dfrac{a^2}{c^2} + \dfrac{b^2}{c^2} = 1$$By $(\ref{eq:sin-def})$ and $(\ref{eq:cos-def})$, $\sin\theta = a/c$ and $\cos\theta = b/c$, so:
Theorem 1.1 (Fundamental trigonometric identity)
\begin{equation} \sin^2\theta + \cos^2\theta = 1 \label{eq:pythagorean-identity} \end{equation}Relation with tan
From the definition $(\ref{eq:tan-def})$:
$$\tan\theta = \dfrac{a}{b} = \dfrac{a/c}{b/c} = \dfrac{\sin\theta}{\cos\theta}$$Theorem 1.2 (Identities for tan)
\begin{align} \tan\theta &= \frac{\sin\theta}{\cos\theta} \label{eq:tan-identity}\\ 1 + \tan^2\theta &= \sec^2\theta = \frac{1}{\cos^2\theta} \label{eq:tan-sec-identity} \end{align}Derivation of $(\ref{eq:tan-sec-identity})$
Dividing both sides of $(\ref{eq:pythagorean-identity})$ by $\cos^2\theta$:
$$\dfrac{\sin^2\theta}{\cos^2\theta} + \dfrac{\cos^2\theta}{\cos^2\theta} = \dfrac{1}{\cos^2\theta}$$ $$\tan^2\theta + 1 = \sec^2\theta$$The Inscribed Angle Theorem
The proof of the law of sines requires the following theorem about circles.
Theorem 1.3 (Inscribed angle theorem)
In a circle, the inscribed angle subtending a given arc is constant and equals half the central angle subtending the same arc.
$$\angle APB = \dfrac{1}{2} \angle AOB$$Proof
Draw the line through the center $O$ and the vertex $P$, and let $Q$ be its other intersection with the circle (so $PQ$ is a diameter; the figure shows the case where the center $O$ lies inside $\angle APB$). Since $O$ is the center and $A,\ P,\ B$ lie on the circle, $OA = OP = OB$ all equal the radius $R$. Hence triangles $OPA$ and $OPB$ are both isosceles.
Set $\angle OPA = \alpha$. The base angles of an isosceles triangle are equal, so $\angle OAP = \alpha$. Since the angles of a triangle sum to $180^\circ$, $\angle AOP = 180^\circ - 2\alpha$, and because $P,\ O,\ Q$ are collinear,
$$\angle AOQ = 180^\circ - \angle AOP = 2\alpha = 2\,\angle OPA.$$In exactly the same way, setting $\angle OPB = \beta$ gives $\angle BOQ = 2\beta = 2\,\angle OPB$. Since the center $O$ lies inside $\angle APB$, we have $\angle APB = \alpha + \beta$, and
$$\angle AOB = \angle AOQ + \angle BOQ = 2(\alpha + \beta) = 2\,\angle APB.$$The fact used here — that the exterior angle $\angle AOQ$ equals the sum of the two non-adjacent interior angles — follows immediately from the $180^\circ$ angle sum of a triangle, as above, so this proof needs no further assumptions. For the statement as a general theorem about triangles, see the exterior angle theorem.
The Law of Sines
Beyond right triangles, there are important theorems using trigonometric ratios that hold for general triangles: the law of sines and the law of cosines.
Theorem 1.4 (Law of sines)
Let the three side lengths of a triangle be $a, b, c$, their opposite angles $\alpha, \beta, \gamma$, and the circumradius $R$. Then:
\begin{equation} \frac{a}{\sin\alpha} = \frac{b}{\sin\beta} = \frac{c}{\sin\gamma} = 2R \label{eq:sine-law} \end{equation}Proof
Acute triangle: Drop a perpendicular $OH$ from the circumcenter $O$ to the side $a = BC$. By the inscribed angle theorem (Theorem 1.3), the central angle subtending arc $BC$ is $2\alpha$.
Since $OH$ is the perpendicular bisector of the chord $BC$, $BH = a/2$. In the right triangle $OBH$:
$$\sin\alpha = \sin\dfrac{2\alpha}{2} = \dfrac{BH}{OB} = \dfrac{a/2}{R}$$Hence $a = 2R\sin\alpha$, that is $\dfrac{a}{\sin\alpha} = 2R$. The same holds for the other sides. $\blacksquare$
Obtuse triangle ($\alpha > 90°$): When $\alpha$ is obtuse, the circumcenter $O$ lies on the same side as side $a$ (the side opposite $A$).
The central angle on arc $BC$ is $2\alpha$, but since $\alpha > 90°$ we have $2\alpha > 180°$, so it is a reflex angle (greater than $180°$). In this case the angle $\angle BOH$ in the right triangle $OBH$ equals $\pi - \alpha$ ($= 180° - \alpha$), so
$$\sin(\pi - \alpha) = \dfrac{BH}{OB} = \dfrac{a/2}{R}$$Here $\sin(\pi - \alpha) = \sin\alpha$. This can be checked on the unit circle: the point for the angle $\pi - \alpha$ is $(-\cos\alpha, \sin\alpha)$, whose $y$-coordinate (the value of $\sin$) is the same as for the angle $\alpha$. Therefore
$$\dfrac{a}{\sin\alpha} = 2R$$holds. $\blacksquare$
Right triangle ($\alpha = 90°$): When $\alpha = 90°$, the inscribed angle theorem (Theorem 1.3) gives a central angle of $2\alpha = 180°$, so arc $BC$ is a semicircle and the side $a = BC$ equals the diameter $2R$ of the circumscribed circle.
Since $\sin 90° = 1$,
$$\dfrac{a}{\sin\alpha} = \dfrac{2R}{1} = 2R$$holds. $\blacksquare$
The Law of Cosines
Theorem 1.5 (Law of cosines)
Let the three side lengths of a triangle be $a, b, c$ with opposite angles $\alpha, \beta, \gamma$. Then:
\begin{align} a^2 &= b^2 + c^2 - 2bc\cos\alpha \label{eq:cosine-law-a}\\ b^2 &= c^2 + a^2 - 2ca\cos\beta \label{eq:cosine-law-b}\\ c^2 &= a^2 + b^2 - 2ab\cos\gamma \label{eq:cosine-law-c} \end{align}The law of cosines is a generalization of the Pythagorean theorem. When $\gamma = 90°$, $\cos\gamma = 0$ and equation $(\ref{eq:cosine-law-c})$ becomes $c^2 = a^2 + b^2$.
Proof
Use coordinates. Place the vertex of angle $\gamma$ at the origin and side $a$ along the positive $x$-axis. Then the other end of side $a$ is $(a, 0)$ and the other end of side $b$ is $(b\cos\gamma, b\sin\gamma)$.
The square of the length of side $c$ is:
\begin{align*} c^2 &= (b\cos\gamma - a)^2 + (b\sin\gamma - 0)^2\\ &= b^2\cos^2\gamma - 2ab\cos\gamma + a^2 + b^2\sin^2\gamma\\ &= b^2(\cos^2\gamma + \sin^2\gamma) + a^2 - 2ab\cos\gamma\\ &= a^2 + b^2 - 2ab\cos\gamma \end{align*}Worked Examples
Example 1.1
In triangle $ABC$, let $a = 7$, $b = 5$, $\gamma = 60°$. Find the length of side $c$.
Solution
By the law of cosines $(\ref{eq:cosine-law-c})$:
\begin{align*} c^2 &= a^2 + b^2 - 2ab\cos\gamma\\ &= 7^2 + 5^2 - 2 \cdot 7 \cdot 5 \cdot \cos 60°\\ &= 49 + 25 - 70 \cdot \frac{1}{2}\\ &= 74 - 35 = 39 \end{align*}Hence $c = \sqrt{39}$.
Example 1.2
In triangle $ABC$, let $a = 8$, $b = 6$, $c = 7$. Find angle $\alpha$.
Solution
By the law of cosines $(\ref{eq:cosine-law-a})$:
\begin{align*} a^2 &= b^2 + c^2 - 2bc\cos\alpha\\ 64 &= 36 + 49 - 84\cos\alpha\\ 84\cos\alpha &= 21\\ \cos\alpha &= \frac{21}{84} = \frac{1}{4} \end{align*}Hence $\alpha = \arccos\left(\dfrac{1}{4}\right) \approx 75.5°$.
Exercises
Problem 1
In a right triangle $ABC$ ($C = 90°$), let $a = 3$, $b = 4$. Find the values of $\sin A$, $\cos A$, and $\tan A$.
Hint
First find the hypotenuse $c$ using the Pythagorean theorem, then apply Definition 1.1.
Sample solution
By the Pythagorean theorem:
$$c = \sqrt{a^2 + b^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$For angle $A$ the opposite side is $a = 3$, the adjacent side is $b = 4$, and the hypotenuse is $c = 5$, so:
$$\sin A = \dfrac{a}{c} = \dfrac{3}{5}, \quad \cos A = \dfrac{b}{c} = \dfrac{4}{5}, \quad \tan A = \dfrac{a}{b} = \dfrac{3}{4}$$Problem 2
In a right triangle $ABC$ ($C = 90°$), let $A = 30°$, $c = 10$. Find the lengths of sides $a$ and $b$.
Hint
Using the values of $\sin 30°$ and $\cos 30°$ (Table 1.1), compute $a = c \sin A$ and $b = c \cos A$.
Sample solution
From $\sin A = a / c$:
$$a = c \sin 30° = 10 \times \dfrac{1}{2} = 5$$From $\cos A = b / c$:
$$b = c \cos 30° = 10 \times \dfrac{\sqrt{3}}{2} = 5\sqrt{3}$$Problem 3
Given $\sin\theta = \dfrac{5}{13}$ ($0° < \theta < 90°$), find the values of $\cos\theta$ and $\tan\theta$.
Hint
Use the identity $\sin^2\theta + \cos^2\theta = 1$ (Theorem 1.1) to find $\cos\theta$. Then $\tan\theta$ follows from Theorem 1.2.
Sample solution
From $\sin^2\theta + \cos^2\theta = 1$:
$$\cos^2\theta = 1 - \sin^2\theta = 1 - \dfrac{25}{169} = \dfrac{144}{169}$$For $0° < \theta < 90°$ we have $\cos\theta > 0$, so:
$$\cos\theta = \dfrac{12}{13}$$Therefore:
$$\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{5/13}{12/13} = \dfrac{5}{12}$$Problem 4
In triangle $ABC$, let $b = 5$, $c = 8$, $\alpha = 60°$. Find the length of side $a$ and the area $S$ of the triangle.
Hint
Side $a$ is found from the law of cosines (Theorem 1.5). The area is $S = \dfrac{1}{2}bc\sin\alpha$.
Sample solution
By the law of cosines:
\begin{align*} a^2 &= b^2 + c^2 - 2bc\cos\alpha\\ &= 25 + 64 - 2 \cdot 5 \cdot 8 \cdot \cos 60°\\ &= 89 - 80 \cdot \frac{1}{2} = 89 - 40 = 49 \end{align*}Hence $a = 7$.
The area is:
$$S = \dfrac{1}{2}bc\sin\alpha = \dfrac{1}{2} \cdot 5 \cdot 8 \cdot \sin 60° = 20 \cdot \dfrac{\sqrt{3}}{2} = 10\sqrt{3}$$References
Frequently Asked Questions
Q1: What is a trigonometric ratio?
A: A trigonometric ratio is a ratio of side lengths in a right triangle. For an angle $\theta$ it is defined by $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$, $\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}$. Because similar triangles share the same side ratios, a trigonometric ratio depends only on the angle.
Q2: What is the law of sines?
A: The law of sines relates the sides and angles of a triangle. With side lengths $a, b, c$, opposite angles $\alpha, \beta, \gamma$, and circumradius $R$, it states $\dfrac{a}{\sin\alpha} = \dfrac{b}{\sin\beta} = \dfrac{c}{\sin\gamma} = 2R$. The ratio of each side to the sine of its opposite angle is always constant (the diameter of the circumscribed circle).
Q3: What is the law of cosines?
A: The law of cosines relates the three sides of a triangle to one of its angles: $a^2 = b^2 + c^2 - 2bc\cos A$. When $A = 90°$, $\cos A = 0$ and it reduces to the Pythagorean theorem $a^2 = b^2 + c^2$. It is used to find the third side from two sides and the included angle, or to find an angle from the three sides.