## Summary $\sin$ and $\cos$, $\sec$ and $\csc$ and $\tan$ and $\cot$ of the complementary angles of a triangle have a special relationship as follows: $ \begin{align*} \sin \theta &= \cos(90 \textdegree - \theta) \tag{1} \\ \cos \theta &= \sin(90 \textdegree - \theta) \tag{2} \\ \\ \sec \theta &= \csc(90 \textdegree - \theta) \tag{3} \\ \csc \theta &= \sec(90 \textdegree - \theta) \tag{4} \\ \\ \tan \theta &= \cot(90 \textdegree - \theta) \tag{5} \\ \cot \theta &= \tan(90 \textdegree - \theta) \tag{6} \\ \end{align*} $ Because of this relationship we say they are *cofunctions*, which is why one half of each pair has the "co-" prefix. ### Again, but this time more rad As we know that $90 \textdegree = \frac{\pi}{2}\text{rad}$, we can see that the exact same relationships can be expressed in [[Radians]] as follows. $ \begin{align*} \sin \theta &= \cos(\frac{\pi}{2}\text{rad} - \theta) \tag{1} \\ \cos \theta &= \sin(\frac{\pi}{2}\text{rad} - \theta) \tag{2} \\ \\ \sec \theta &= \csc(\frac{\pi}{2}\text{rad} - \theta) \tag{3} \\ \csc \theta &= \sec(\frac{\pi}{2}\text{rad} - \theta) \tag{4} \\ \\ \tan \theta &= \cot(\frac{\pi}{2}\text{rad} - \theta) \tag{5} \\ \cot \theta &= \tan(\frac{\pi}{2}\text{rad} - \theta) \tag{6} \\ \end{align*} $ The derivations below are done using degrees, however the exact same derivations are typically done using the [[Unit Circle Definitions of Trigonometric Functions|unit circle]] in radians. At some point I may repeat the exercise for my own practise. ## Derivation: Sine and Cosine Consider the right triangle $\Delta ABC$ with sides $a$, $b$ and $c$ and a right angle at vertex $B$ as follows: ![[Basic Triangle Diagram.png]] [^1] Let $m \angle CAB = \theta$. What is $m \angle ACB$? Since $m \angle ABC = 90 \textdegree$, it follows that $m \angle ACB = 90 \textdegree - \theta$, as the interior angles must sum to $180 \textdegree$. Since we now know $m \angle ACB$, we can look at the ratios $\frac{a}{b}$ and $\frac{c}{b}$ from the perspective of both angles, and deduce the following extremely important and convenient relationships: $ \begin{align*} \sin \theta &= \cos(90 \textdegree - \theta) = \frac{a}{b} \tag{1} \\ \cos \theta &= \sin(90 \textdegree - \theta) = \frac{c}{b} \tag{2} \\ \end{align*} $ ## Derivation: Secant and Cosecant Given the reciprocal nature of $\sin$ and $\csc$ and $\cos$ and $\sec$ it should not be surprising that the following relationships also hold, however using the same triangle $\Delta ABC$ we can just show them from first principles: $ \begin{align*} \sec \theta &= \csc(90 \textdegree - \theta) = \frac{b}{c} \tag{3} \\ \csc \theta &= \sec(90 \textdegree - \theta) = \frac{b}{a} \tag{4} \\ \end{align*} $ ## Derivation: Tan Finally, considering the same triangle $\Delta ABC$, we notice that $ \begin{align*} \tan \theta &= \frac{a}{c} \\ \tan(90 \textdegree - \theta) &= \frac{c}{a} \\ \\ \text{Therefore } \tan \theta &= \frac{1}{\tan(90 \textdegree - \theta)} = \cot(90 \textdegree - \theta) \tag{5} \\ \\ \cot \theta &= \frac{c}{a} \\ \cot(90 \textdegree - \theta) &= \frac{a}{c} \\ \\ \text{Therefore } \cot \theta &= \frac{1}{\cot(90 \textdegree - \theta)} = \tan(90 \textdegree - \theta) \tag{6} \\ \end{align*} $ ...giving us a third pair of useful properties of the trig ratios of complementary angles. ## Reference [Here's](https://www.khanacademy.org/math/trigonometry/trigonometry-right-triangles/reciprocal-trig-ratios/a/sine-and-cosine-are-cofunctions) the Khan Academy review on this topic. [^1] Diagram available on [geogebra](https://www.geogebra.org/calculator/npw4rmcw)