Allied angles

How do we find $\sin 150^\circ$, $\cos 225^\circ$ or $\tan 1230^\circ$? We do not need new tables. Angles such as $-\theta$, $90^\circ \pm \theta$, $180^\circ \pm \theta$, $270^\circ \pm \theta$ and $360^\circ \pm \theta$ are called allied angles of $\theta$, and their ratios are always equal to a ratio of $\theta$ itself, up to a sign. This les­son derives these results and turns them into one short rule that reduces any angle to an acute one.

Neg­a­tive angles#

Turn­ing by $-\theta$ is the mir­ror image, in the x-axis, of turn­ing by $\theta$. If $\theta$ takes $P(x, y)$ to the ter­mi­nal posi­tion, then $-\theta$ takes it to $P'(x, -y)$, at the same dis­tance $r$. So

$$\sin(-\theta) = -\sin\theta, \qquad \cos(-\theta) = \cos\theta, \qquad \tan(-\theta) = -\tan\theta$$

$$\csc(-\theta) = -\csc\theta, \qquad \sec(-\theta) = \sec\theta, \qquad \cot(-\theta) = -\cot\theta$$

Only cosine (and secant) keep their sign.

Com­ple­men­tary angles: 90° − θ#

If $P(x, y)$ is at angle $\theta$, the point $Q(y, x)$, the mir­ror image in the line $y = x$, is at angle $90^\circ - \theta$. Swap­ping $x$ and $y$ in the def­i­n­i­tions gives

$$\sin(90^\circ - \theta) = \cos\theta, \quad \cos(90^\circ - \theta) = \sin\theta, \quad \tan(90^\circ - \theta) = \cot\theta$$

$$\cot(90^\circ - \theta) = \tan\theta, \quad \sec(90^\circ - \theta) = \csc\theta, \quad \csc(90^\circ - \theta) = \sec\theta$$

The other allied angles#

In the same way (by rotat­ing the point $P$ through $90^\circ$ or $180^\circ$), we get:

Adding a whole num­ber of com­plete turns does not change the ter­mi­nal side, so for any inte­ger $n$:

$$\sin(n \cdot 360^\circ + \theta) = \sin\theta, \quad \cos(n \cdot 360^\circ + \theta) = \cos\theta, \quad \tan(n \cdot 360^\circ + \theta) = \tan\theta$$

and $\sin(n \cdot 360^\circ - \theta) = -\sin\theta$, $\cos(n \cdot 360^\circ - \theta) = \cos\theta$, $\tan(n \cdot 360^\circ - \theta) = -\tan\theta$. The rec­i­p­ro­cal ratios fol­low the same pat­tern.

The rule that replaces the table#

Treat $\theta$ as if it were acute.

1. Which ratio? For $90^\circ \pm \theta$ and $270^\circ \pm \theta$ the ratio changes to its part­ner ($\sin \leftrightarrow \cos$, $\tan \leftrightarrow \cot$, $\sec \leftrightarrow \csc$). For $180^\circ \pm \theta$ and $360^\circ \pm \theta$ the ratio stays the same.
2. Which sign? Find the quad­rant in which the whole angle lies, and use the sign that the orig­i­nal ratio has in that quad­rant (All Sil­ver Tea Cups).

For exam­ple, $\cos(90^\circ + \theta)$: the ratio changes to $\sin$; the angle $90^\circ + \theta$ is in quad­rant II, where cosine is neg­a­tive. So $\cos(90^\circ + \theta) = -\sin\theta$.

Exam­ple 1#

Find $\sin 150^\circ$, $\sin 210^\circ$ and $\sin 330^\circ$.

$$\displaystyle \sin 150^\circ = \sin(180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}$$

$$\displaystyle \sin 210^\circ = \sin(180^\circ + 30^\circ) = -\sin 30^\circ = -\frac{1}{2}$$

$$\displaystyle \sin 330^\circ = \sin(360^\circ - 30^\circ) = -\sin 30^\circ = -\frac{1}{2}$$

Exam­ple 2#

Find $\cos 225^\circ$ and $\tan 300^\circ$.

$\displaystyle \cos 225^\circ = \cos(180^\circ + 45^\circ) = -\cos 45^\circ = -\frac{1}{\sqrt{2}}$ (quad­rant III, cosine neg­a­tive).

$\tan 300^\circ = \tan(360^\circ - 60^\circ) = -\tan 60^\circ = -\sqrt{3}$ (quad­rant IV, tan­gent neg­a­tive).

Exam­ple 3#

Find $\sin 1230^\circ$ and $\cos(-480^\circ)$.

Remove com­plete turns first: $1230^\circ = 3 \times 360^\circ + 150^\circ$, so $\displaystyle \sin 1230^\circ = \sin 150^\circ = \frac{1}{2}$.

$\cos(-480^\circ) = \cos 480^\circ$ (cosine ignores the sign of the angle), and $480^\circ = 360^\circ + 120^\circ$, so it equals $\displaystyle \cos 120^\circ = \cos(180^\circ - 60^\circ) = -\cos 60^\circ = -\frac{1}{2}$.

Exam­ple 4#

Sim­plify $\sin(180^\circ + \theta)\cos(90^\circ - \theta) + \cos(180^\circ - \theta)\sin(90^\circ + \theta)$.

$$(-\sin\theta)(\sin\theta) + (-\cos\theta)(\cos\theta) = -(\sin^2\theta + \cos^2\theta) = -1$$

When is a ratio zero?#

From the graphs of the ratios (next les­son) or from the points on the axes:

$$\displaystyle \sin\theta = 0 \iff \theta = n\pi, \qquad \cos\theta = 0 \iff \theta = (2n + 1)\frac{\pi}{2}, \qquad \tan\theta = 0 \iff \theta = n\pi$$

where $n$ is any inte­ger. So sine is zero at $0, \pm\pi, \pm 2\pi, \dots$ and cosine at $\displaystyle \pm\frac{\pi}{2}, \pm\frac{3\pi}{2}, \dots$

Prac­tice#

1. Find $\sin 120^\circ$, $\cos 135^\circ$ and $\tan 150^\circ$.
2. Find $\cos 240^\circ$, $\sin 300^\circ$ and $\tan 225^\circ$.
3. Find $\sin(-30^\circ)$, $\cos(-60^\circ)$ and $\tan(-45^\circ)$.
4. Find $\sin 780^\circ$.
5. Find $\cos 1020^\circ$.
6. Find $\tan(-945^\circ)$.
7. Eval­u­ate $\sin 600^\circ \cos 330^\circ + \cos 120^\circ \sin 150^\circ$.
8. Find all angles $\theta$ between $0^\circ$ and $360^\circ$ for which $\displaystyle \sin\theta = \frac{1}{2}$.
9. Find all val­ues of $\theta$ in $[0, 2\pi]$ for which $\cos\theta = 0$.

Answers#