The trigonometric ratios

The six trigono­met­ric ratios, sine, cosine, tan­gent and their rec­i­p­ro­cals, con­nect an angle to lengths. They are defined using a point on the ter­mi­nal side of the angle, which makes them work for angles of any size, and for an acute angle they reduce to the famil­iar ratios of the sides of a right-angled tri­an­gle. This les­son gives both def­i­n­i­tions, the rela­tions between the six ratios, and the method for find­ing all six when one is known.

The def­i­n­i­tion#

Let $P(x, y)$ be a point in the plane, at dis­tance $OP = r$ from the ori­gin ($r > 0$), and let $\theta$ be the angle that $OP$ makes with the pos­i­tive x-axis, mea­sured anti­clock­wise. Then

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

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

(The cose­cant is also writ­ten $\text{cosec}\,\theta$.) Each ratio is defined when­ever its denom­i­na­tor is not zero. The val­ues depend only on the angle, not on which point of the ter­mi­nal side we choose, because mov­ing $P$ along the ray scales $x$, $y$ and $r$ by the same fac­tor. Since $r = \sqrt{x^2 + y^2}$, the dis­tance for­mula links the three num­bers.

In a right-angled tri­an­gle#

When $\theta$ is acute ($\displaystyle 0 < \theta < \frac{\pi}{2}$), let $M$ be the foot of the per­pen­dic­u­lar from $P$ to the x-axis. Tri­an­gle $OMP$ is right-angled at $M$, with $OM = x$ adja­cent to $\theta$, $MP = y$ oppo­site $\theta$, and hypotenuse $OP = r$. So

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

and the other three are their rec­i­p­ro­cals.

Rela­tions between the ratios#

Rec­i­p­ro­cal rela­tions.

$$\displaystyle \csc\theta = \frac{1}{\sin\theta}, \qquad \sec\theta = \frac{1}{\cos\theta}, \qquad \cot\theta = \frac{1}{\tan\theta}$$

so $\sin\theta \csc\theta = 1$, $\cos\theta \sec\theta = 1$ and $\tan\theta \cot\theta = 1$.

Quo­tient rela­tions.

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

These fol­low at once from the def­i­n­i­tions: $\displaystyle \frac{y/r}{x/r} = \frac{y}{x}$.

Find­ing all six ratios from one#

Draw (or imag­ine) a right tri­an­gle with the given ratio, find the third side by Pythago­ras' the­o­rem, then read off the rest. It helps to recog­nise the com­mon whole-num­ber right tri­an­gles:

$$3, 4, 5 \qquad 5, 12, 13 \qquad 6, 8, 10 \qquad 7, 24, 25 \qquad 8, 15, 17 \qquad 9, 40, 41 \qquad 11, 60, 61 \qquad 16, 63, 65$$

Exam­ple 1#

In tri­an­gle $ABC$, $\angle B = 90^\circ$, $AB = 3$ cm and $BC = 4$ cm. Find $AC$, and $\sin A$, $\cos A$, $\sin C$, $\tan C$.

$AC = \sqrt{3^2 + 4^2} = 5$ cm. For angle $A$, the oppo­site side is $BC = 4$ and the adja­cent side is $AB = 3$; for angle $C$ it is the other way round.

$$\displaystyle \sin A = \frac{4}{5}, \quad \cos A = \frac{3}{5}, \quad \sin C = \frac{3}{5}, \quad \tan C = \frac{3}{4}$$

Notice that $\sin A = \cos C$: the side oppo­site one acute angle is adja­cent to the other.

Exam­ple 2#

If $\displaystyle \sin\theta = \frac{3}{5}$ and $\theta$ is acute, find the other five ratios.

Oppo­site $= 3$, hypotenuse $= 5$, so adja­cent $= \sqrt{25 - 9} = 4$.

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

Exam­ple 3#

If $\theta$ is acute and $\displaystyle \tan\theta = \frac{1}{\sqrt{7}}$, find $\displaystyle \frac{\csc^2\theta - \sec^2\theta}{\csc^2\theta + \sec^2\theta}$.

Oppo­site $= 1$, adja­cent $= \sqrt{7}$, so hypotenuse $= \sqrt{1 + 7} = \sqrt{8}$. Then $\displaystyle \csc^2\theta = \frac{8}{1} = 8$ and $\displaystyle \sec^2\theta = \frac{8}{7}$.

$$\displaystyle \frac{8 - \frac{8}{7}}{8 + \frac{8}{7}} = \frac{\frac{48}{7}}{\frac{64}{7}} = \frac{48}{64} = \frac{3}{4}$$

Exam­ple 4#

The ter­mi­nal side of an angle $\theta$ passes through $P(-8, 15)$. Find $\sin\theta$, $\cos\theta$ and $\tan\theta$.

$r = \sqrt{64 + 225} = \sqrt{289} = 17$.

$$\displaystyle \sin\theta = \frac{15}{17}, \qquad \cos\theta = \frac{-8}{17}, \qquad \tan\theta = \frac{15}{-8} = -\frac{15}{8}$$

Here $\theta$ is not acute, and the def­i­n­i­tion with $x$ and $y$ gives some ratios neg­a­tive signs. A later les­son stud­ies these signs.

Exam­ple 5#

If $3\tan\theta = 4$, find $\displaystyle \frac{3\sin\theta + 2\cos\theta}{3\sin\theta - 2\cos\theta}$.

Divide the top and bot­tom by $\cos\theta$, which turns every term into $\tan\theta$:

$$\displaystyle \frac{3\tan\theta + 2}{3\tan\theta - 2} = \frac{4 + 2}{4 - 2} = 3$$

Prac­tice#

1. In tri­an­gle $PQR$, $\angle Q = 90^\circ$, $PQ = 5$ cm and $QR = 12$ cm. Find $PR$, $\sin P$, $\cos P$ and $\tan R$.
2. If $\displaystyle \cos\theta = \frac{8}{17}$ and $\theta$ is acute, find the other five ratios.
3. If $\displaystyle \tan\theta = \frac{7}{24}$ and $\theta$ is acute, find $\sin\theta$ and $\cos\theta$.
4. If $\displaystyle \sec\theta = \frac{13}{12}$ and $\theta$ is acute, find $\sin\theta$ and $\cot\theta$.
5. If $\cot\theta = 3$, find $\displaystyle \frac{(1 + \sin\theta)(1 - \sin\theta)}{(1 + \cos\theta)(1 - \cos\theta)}$.
6. If $\theta$ is acute and $\displaystyle \tan\theta = \frac{1}{\sqrt{3}}$, find $\displaystyle \frac{\csc^2\theta - \sec^2\theta}{\csc^2\theta + \sec^2\theta}$.
7. The ter­mi­nal side of $\theta$ passes through $(5, -12)$. Find $\sin\theta$, $\cos\theta$ and $\tan\theta$.
8. If $4\tan\theta = 3$, find $\displaystyle \frac{4\sin\theta - \cos\theta}{4\sin\theta + \cos\theta}$.

Answers#