Allied Angles

Definition. Two angles are called allied angles if their sum or their difference is an integer multiple of $\dfrac{\pi}{2}$ (i.e. $90^\circ$). So, relative to a given angle $\theta$, every angle of the form

is allied to $\theta$.

Ratios of $-\theta$ (reflection in the $x$-axis). Let $P(a,b)$ lie on the terminal side of $\theta$, at distance $OP$ from the origin. The terminal side of $-\theta$ is the mirror image of the terminal side of $\theta$ in the $x$-axis, so the corresponding point is $P'(a,-b)$ with $OP'=OP$. From the coordinate definition, $\sin\theta=b/OP$, $\cos\theta=a/OP$, so

and dividing/inverting these two gives every other one: $\tan(-\theta)=-\tan\theta$, $\cot(-\theta)=-\cot\theta$, $\operatorname{cosec}(-\theta)=-\operatorname{cosec}\theta$, $\sec(-\theta)=\sec\theta$. In words: flipping a point to the other side of the $x$-axis negates its $y$-coordinate and keeps its $x$-coordinate — so sine (built from $y$) is negated while cosine (built from $x$) is unchanged. These two negative-angle identities are also exactly what tells us $\cos$ is an even function and $\sin$ is an odd function (developed fully in §3.4.4).

Ratios of $\left(\dfrac{\pi}{2}+\theta\right)$, $0<\theta<\dfrac{\pi}{2}$ (a quarter-turn rotation). By a similar congruent-triangle argument — if $P(a,b)$ corresponds to $\theta$, the point rotated a further $90^\circ$ turns out to be $P'(-b,a)$ — one finds

and hence $\tan\left(\frac{\pi}{2}+\theta\right)=-\cot\theta$, $\cot\left(\frac{\pi}{2}+\theta\right)=-\tan\theta$, $\operatorname{cosec}\left(\frac{\pi}{2}+\theta\right)=\sec\theta$, $\sec\left(\frac{\pi}{2}+\theta\right)=-\operatorname{cosec}\theta$. (The familiar complementary-angle relations for $\left(\dfrac{\pi}{2}-\theta\right)$ — $\sin(90^\circ-\theta)=\cos\theta$, $\cos(90^\circ-\theta)=\sin\theta$, etc. — are the special case already known from the earlier classes' right-triangle work.)

The ratios of the remaining allied angles $\pi\pm\theta,\ \dfrac{3\pi}{2}\pm\theta,\ 2\pi\pm\theta$ are obtained the same way (a rotation/reflection argument each time), and all nine cases are summarised below for $0<\theta<\dfrac{\pi}{2}$:

(The matching cosecant/secant/cotangent rows follow immediately since they are just reciprocals of sine/cosine/tangent.)

Two rules that make the table easy to reconstruct instead of memorise:

1. 'Keep vs. co-change.' For allied angles that are an even multiple of $\dfrac{\pi}{2}$ away from $\theta$ — i.e. $-\theta,\ \pi\pm\theta,\ 2\pi\pm\theta$ (of the form $2n\cdot\dfrac{\pi}{2}\pm\theta$) — the name of the function is unchanged (sine stays sine, cosine stays cosine, and so on). For allied angles that are an odd multiple of $\dfrac{\pi}{2}$ away — i.e. $\dfrac{\pi}{2}\pm\theta,\ \dfrac{3\pi}{2}\pm\theta$ (of the form $(2n+1)\dfrac{\pi}{2}\pm\theta$) — the function changes to its co-function (sine ↔ cosine, tangent ↔ cotangent, secant ↔ cosecant).
2. The sign follows ASTC. Once the function name (or co-name) is settled, attach a $+$ or $-$ sign by treating $\theta$ as a small acute angle and asking which quadrant the allied angle itself falls into — then read the sign straight off the ASTC table of §3.4.1.

Worked examples of the technique (angle reduction to a first-quadrant acute angle, then a table lookup):

• $\sin150^\circ=\sin(180^\circ-30^\circ)=\sin30^\circ=\dfrac12$ (equally, $\sin150^\circ=\sin(90^\circ+60^\circ)=\cos60^\circ=\dfrac12$ — both routes agree).
• $\cos135^\circ=\cos(180^\circ-45^\circ)=-\cos45^\circ=-\dfrac{1}{\sqrt2}$.
• $\tan120^\circ=\tan(180^\circ-60^\circ)=-\tan60^\circ=-\sqrt3$. …