Trigonometric Functions of Allied Angles

Allied angles

Definition. If $\theta$ is the measure of an angle, then $-\theta,\ \pi/2\pm\theta,\ \pi\pm\theta, \ 3\pi/2\pm\theta,\ 2\pi\pm\theta$ are called its allied angles — angles whose sum or difference with

$\theta$ is an integral multiple of $\pi/2$.

The results (all proved from the compound-angle formulas of section 3.1)

Already shown in 3.1: $\sin(\pi/2-\theta)=\cos\theta$, $\cos(\pi/2-\theta)=\sin\theta$, $\tan(\pi/2-\theta)=\cot\theta$;

and $\sin(\pi/2+\theta)=\cos\theta$, $\cos(\pi/2+\theta)=-\sin\theta$, $\tan(\pi/2+\theta)=-\cot\theta$.

The remaining five pairs are proved the same way, substituting into Theorems 1–4 of section 3.1:

The working rule these results embody: write the given angle as $n\times90°\pm\theta$ for the smallest

acute $\theta$. If $n$ is even, the ratio's NAME is unchanged (sin stays sin); if $n$ is odd, the ratio changes

to its CO-ratio (sin becomes cos, tan becomes cot). The SIGN is whatever the original ratio would have in the

quadrant where the angle $n\times90°\pm\theta$ actually lies, treating $\theta$ as acute.

These are collected in a single reference table (see the Allied Angles table note attached to this section).

Solved Examples

Ex.1 — Find $\sin495°,\cos930°,\tan840°$.

$\sin495°=\sin(360°+135°)=\sin135°=\sin(\pi/2+45°)=\cos45°=\dfrac{1}{\sqrt2}$.

$\cos930°=\cos(2\times360°+210°)=\cos210°=\cos(180°+30°)=-\cos30°=-\dfrac{\sqrt3}{2}$.

$\tan840°=\tan(2\times360°+120°)=\tan120°=\tan(\pi/2+30°)=-\cot30°=-\sqrt3$.

Ex.2(i) — Show $\cos24°+\cos55°+\cos125°+\cos204°+\cos300°=\dfrac12$. Rewrite $\cos125°=\cos(180°-55°)= -\cos55°$, $\cos204°=\cos(180°+24°)=-\cos24°$, $\cos300°=\cos(360°-60°)=\cos60°$. The sum becomes

$\cos24°+\cos55°-\cos55°-\cos24°+\cos60°=\cos60°=\dfrac12$.

Ex.2(ii) — Show $\sec840°\cdot\cot(-945°)+\sin600°\cdot\tan(-690°)=\dfrac32$. Reduce each: $\sec840°=\sec120° =-\text{cosec}\,30°=-2$; $\cot(-945°)=-\cot945°=-\cot225°=-\cot45°=-1$; $\sin600°=\sin240°=-\sin60°=-\dfrac{\sqrt3}{2}$;

$\tan(-690°)=-\tan(-30°)=\tan30°=\dfrac{1}{\sqrt3}$. Substituting: $(-2)(-1)+\left(-\dfrac{\sqrt3}{2}\right)\left(\dfrac{1}{\sqrt3}\right) =2-\dfrac12=\dfrac32$.

Ex.2(iii) — Show $\dfrac{\text{cosec}(90°-\theta)\sin(180°-\theta)\cot(360°-\theta)}{\sec(180°+\theta)\tan(90°+\theta)\sin(-\theta)}=1$.

Reduce every factor: $\text{cosec}(90°-\theta)=\sec\theta$, $\sin(180°-\theta)=\sin\theta$, $\cot(360°-\theta)=-\cot\theta$,

$\sec(180°+\theta)=-\sec\theta$, $\tan(90°+\theta)=-\cot\theta$, $\sin(-\theta)=-\sin\theta$. The numerator and

denominator become identical products, so the ratio is $1$.

Ex.2(iv) — Show $\cot(\pi/2+\theta)\sin(\pi-\theta)\cot(\pi-\theta)\div[\cos(2\pi-\theta)\sin(-\pi-\theta)\tan(\pi/2-\theta)]=-\text{cosec}\,\theta$.

Reduces each allied-angle factor similarly and cancels common terms to reach $-\text{cosec}\,\theta$.

Ex.3(i) — Show $\sin\dfrac{\pi}{15}+\sin\dfrac{4\pi}{15}-\sin\dfrac{14\pi}{15}-\sin\dfrac{11\pi}{15}=0$.

Since $\dfrac{14\pi}{15}=\pi-\dfrac{\pi}{15}$ and $\dfrac{11\pi}{15}=\pi-\dfrac{4\pi}{15}$, use $\sin(\pi-\theta) =\sin\theta$ to see the four terms cancel in pairs.

Ex.3(ii) — Show $\sin^2(\pi/4-x)+\sin^2(\pi/4+x)=1$. Substitute $y=\pi/4-x$, so $\pi/4+x=\pi/2-y$; the sum …