Trigonometric ratios and measuring angles

This course takes trigonom­e­try beyond right-angled tri­an­gles. Angles can now be any size, even neg­a­tive or larger than a full turn, and the trigono­met­ric ratios become func­tions defined for every such angle. This first les­son lays the foun­da­tion: the six ratios defined from a point on the plane, the iden­ti­ties that con­nect them, and the three ways of mea­sur­ing an angle (degrees, grades and radi­ans), with the radian val­ues of the angles you will use most.

The six ratios from a point#

Draw the $x$- and $y$-axes with ori­gin $O$. Let $P(x, y)$ be any point other than $O$, and let $OP = r$, so that $r = \sqrt{x^2 + y^2}$ and $r > 0$. Let $\theta$ be the angle that $OP$ makes with the pos­i­tive $x$-axis, mea­sured anti­clock­wise. Then

$$\displaystyle \sin\theta = \frac{y}{r}, \qquad \cos\theta = \frac{x}{r}, \qquad \tan\theta = \frac{y}{x}$$

$$\displaystyle \csc\theta = \frac{r}{y}, \qquad \sec\theta = \frac{r}{x}, \qquad \cot\theta = \frac{x}{y}$$

(each defined when­ever its denom­i­na­tor is not zero).

If $\theta$ is acute ($0° < \theta < 90°$), then $P$ is in the first quad­rant and drop­ping a per­pen­dic­u­lar from $P$ to the $x$-axis makes a right tri­an­gle with hypotenuse $r$, side $x$ next to $\theta$ and side $y$ oppo­site $\theta$. The def­i­n­i­tions then agree with the ones you already know:

$$\displaystyle \sin\theta = \frac{\text{opposite side}}{\text{hypotenuse}}, \qquad \cos\theta = \frac{\text{adjacent side}}{\text{hypotenuse}}, \qquad \tan\theta = \frac{\text{opposite side}}{\text{adjacent side}}$$

The point-based def­i­n­i­tion is bet­ter because it works for every angle: $x$ and $y$ may be neg­a­tive, and the ratios then take signs.

Exam­ple 1#

The point $P(-3, 4)$ lies on the arm of an angle $\theta$. Find all six ratios.

$r = \sqrt{(-3)^2 + 4^2} = \sqrt{25} = 5$. So

$$\displaystyle \sin\theta = \frac{4}{5}, \quad \cos\theta = -\frac{3}{5}, \quad \tan\theta = -\frac{4}{3}, \quad \csc\theta = \frac{5}{4}, \quad \sec\theta = -\frac{5}{3}, \quad \cot\theta = -\frac{3}{4}$$

Iden­ti­ties con­nect­ing the ratios#

Rec­i­p­ro­cal rela­tions.

$$\displaystyle \csc\theta = \frac{1}{\sin\theta} \implies \sin\theta \cdot \csc\theta = 1$$

$$\displaystyle \sec\theta = \frac{1}{\cos\theta} \implies \cos\theta \cdot \sec\theta = 1$$

$$\displaystyle \cot\theta = \frac{1}{\tan\theta} \implies \tan\theta \cdot \cot\theta = 1$$

Quo­tient rela­tions.

$$\displaystyle \tan\theta = \frac{\sin\theta}{\cos\theta}, \qquad \cot\theta = \frac{\cos\theta}{\sin\theta}$$

Pythagorean iden­ti­ties. Since $x^2 + y^2 = r^2$, divid­ing by $r^2$ gives $\displaystyle \left(\frac{x}{r}\right)^2 + \left(\frac{y}{r}\right)^2 = 1$, that is,

$$\sin^2\theta + \cos^2\theta = 1, \quad \text{so} \quad \sin^2\theta = 1 - \cos^2\theta, \quad \cos^2\theta = 1 - \sin^2\theta$$

Divid­ing $x^2 + y^2 = r^2$ by $x^2$ instead gives $1 + \tan^2\theta = \sec^2\theta$, and divid­ing by $y^2$ gives $\cot^2\theta + 1 = \csc^2\theta$:

$$\sec^2\theta - \tan^2\theta = 1, \quad \text{so} \quad \sec^2\theta = 1 + \tan^2\theta, \quad \tan^2\theta = \sec^2\theta - 1$$

$$\csc^2\theta - \cot^2\theta = 1, \quad \text{so} \quad \csc^2\theta = 1 + \cot^2\theta, \quad \cot^2\theta = \csc^2\theta - 1$$

Exam­ple 2#

If $\displaystyle \tan\theta = \frac{8}{15}$ and $\theta$ is acute, find $\sec\theta$ and $\sin\theta$.

$\displaystyle \sec^2\theta = 1 + \tan^2\theta = 1 + \frac{64}{225} = \frac{289}{225}$, and $\sec\theta > 0$ for an acute angle, so $\displaystyle \sec\theta = \frac{17}{15}$. Then $\displaystyle \cos\theta = \frac{15}{17}$ and $\displaystyle \sin\theta = \tan\theta \cos\theta = \frac{8}{15} \times \frac{15}{17} = \frac{8}{17}$.

Mea­sur­ing angles#

Imag­ine the arm $OP$ turn­ing about $O$. One com­plete turn is called a com­plete angle or a com­plete rota­tion. There are three com­mon units.

• Degree. $1°$ is $\displaystyle \frac{1}{360}$ of a com­plete angle. So a com­plete angle is $360°$.
• Grade. $1^g$ is $\displaystyle \frac{1}{400}$ of a com­plete angle. So a com­plete angle is $400^g$.
• Radian. $1^c$ (one radian) is the angle at the cen­tre of a cir­cle made by an arc whose length equals the radius.

How many radi­ans make a com­plete angle? An arc of length $r$ makes $1^c$; the whole cir­cum­fer­ence, of length $2\pi r$, makes $2\pi$ radi­ans. So

$$\text{complete angle} = 360° = 400^g = 2\pi^c$$

and, halv­ing,

$$180° = 200^g = \pi^c$$

The radian is the unit of higher math­e­mat­ics, and when no unit is writ­ten, radi­ans are meant. So $\pi$ on its own stands for $180°$. Also $\displaystyle 1^c = \frac{180°}{\pi} \approx 57.3°$, a fig­ure you will use to place an angle such as $1$ or $2$ radi­ans.

To con­vert: degrees to radi­ans, mul­ti­ply by $\displaystyle \frac{\pi}{180}$; radi­ans to degrees, mul­ti­ply by $\displaystyle \frac{180}{\pi}$ (that is, replace $\pi$ by $180°$).

Radian val­ues to know#

$$\displaystyle 360° = 2\pi, \quad 180° = \pi, \quad 90° = \frac{\pi}{2}, \quad 60° = \frac{\pi}{3}, \quad 45° = \frac{\pi}{4}, \quad 30° = \frac{\pi}{6}, \quad 0° = 0$$

Each of $30°$, $45°$ and $60°$ has a "fam­ily": the angles in the other quad­rants with the same ref­er­ence angle.

Whole turns are also worth know­ing:

$$1 \times 360° = 360° = 2\pi, \quad 2 \times 360° = 720° = 4\pi, \quad 3 \times 360° = 1080° = 6\pi$$

$$4 \times 360° = 1440° = 8\pi, \quad 5 \times 360° = 1800° = 10\pi$$

Exam­ple 3#

Con­vert (a) $150°$ to radi­ans (b) $\displaystyle \frac{7\pi}{4}$ to degrees (c) $45°$ to grades.

(a) $\displaystyle 150 \times \frac{\pi}{180} = \frac{5\pi}{6}$.

(b) $\displaystyle \frac{7 \times 180°}{4} = 315°$.

(c) $180° = 200^g$, so $\displaystyle 1° = \frac{10}{9}^g$ and $\displaystyle 45° = 45 \times \frac{10}{9} = 50^g$.

Prac­tice#

1. The point $(5, 12)$ lies on the arm of $\theta$. Find $\sin\theta$, $\cos\theta$ and $\tan\theta$.
2. The point $(-8, -6)$ lies on the arm of $\theta$. Find all six ratios.
3. Con­vert $210°$ to radi­ans.
4. Con­vert $\displaystyle \frac{5\pi}{3}$ to degrees.
5. Con­vert $135°$ to grades.
6. Con­vert $50^g$ to degrees.
7. Express $72°$ in radi­ans.
8. If $\displaystyle \sin\theta = \frac{3}{5}$ and $\theta$ is acute, find $\cos\theta$ and $\cot\theta$.
9. Sim­plify $(1 - \cos^2\theta)\csc^2\theta$.
10. Find the value of $\sec^2 37° - \tan^2 37°$.
11. Write the mem­bers of the $45°$ fam­ily in radi­ans.
12. Express $1440°$ in terms of $\pi$.

Answers#