# Geometric Definitions of Derived Trigonometric Functions ![[Cotangent.png|300]] We're looking at right triangle $ABC$ whose right angle is at C and its $\theta$ angle at $A$. Per definition, segments $|AC|$ and $|BC|$ correspond to $h \cos \theta$ and $h \sin \theta$, but what isn't immediately obvious is graphing of $\tan \theta$, $\cot \theta$, $\sec \theta$, and $\csc \theta$. To derive those we look at right triangles $BCD$, $ABD$, and $ADE$, all which are similar because they all contain an angle $\theta$. $\angle CBD$ is equal to $\theta$ because $\angle ABD$ is a right angle (per setup), and $\angle ABC = 90\degree - \theta$ : $\angle CBD = 90 \degree - (90 \degree - \theta)$. ## $|BD| = h \tan \theta$ Because two similar triangles have the same ratio of sides we can say that $\frac o a= \frac {|CD|} o$ and $\frac o a = \frac {|BD|} h$. Let's simplify those two statements to derive $|BD|$. $ \begin{align} \frac {|CD|} o &= \frac o a && \text{multiply both sides by $o$} \\ |CD| &= \frac {o^2} a \end{align} $ $ \begin{align} \frac {|BD|} h &= \frac o a = \frac {|CD|} o \\ \frac {|BD|} h &= \frac {|CD|} o && \text{multiply both sides by $h$} \\ |BD| &= \frac {|CD| \cdot h} o && \text {substitute $x$} \\ &= \frac {\frac {o^2} a h} o \\ &= \frac {o^2h} {ao} = \frac o a h \\ |BD| &= h \tan \theta \end{align} $ ## $|BE| = h \cot \theta$ Because two similar triangles have the same ratio of sides we can say that $\frac a o = \frac {|BE|} h$, and so: $ \begin {align} \frac {|BE|} h &= \frac a o &&\text{multiply both sides by $h$} \\ |BE| &= h\frac a o \\ |BE| &= h \cot \theta \end {align} $ ## $|AD| = h \sec \theta$ $ \begin {align} |AD| &= a + |CD| \\ &= a + \frac {o^2} a \\ &= \frac {a^2 + o^2} {a} \\ &= \frac {h^2} {a} = \frac h a h \\ |AD| &= h \sec \theta \end {align} $ ## $|AE| = h \csc \theta$ Because two similar triangles have the same ratio of sides we can say that $\frac o a = \frac {|AD|} {|AE|}$, and $|AD| = h \sec \theta$ (from above), so: $ \begin{align} \frac o a &= \frac {h \sec \theta} {|AE|} && \text{multiply both sides by $|AE|$} \\ \frac{o \cdot |AE|}a &= h \sec \theta && \text{multiply both sides by $\frac a o$} \\ |AE| &= h \frac {a} {o}\sec \theta \\ &= h \frac {a} {o} \frac h a \\ |AE| &= h \csc \theta \end{align} $