3. Trigonometry

b. Triangle Definitions

In the right triangle at the right, the angle at the base has been denoted by \(\theta\), the opposite side is the altitude \(a\), the adjacent side is the base \(b\) and the hypotenuse is the diagonal side \(c\). In terms of these the six trig functions Sine  Cosine Tangent  Cotangent Secant  Cosecant are defined to be: \[\begin{aligned} \sin\theta&=\dfrac{\text{Opp}}{\text{Hyp}}=\dfrac{a}{c}\quad &\cos\theta&=\dfrac{\text{Adj}}{\text{Hyp}}=\dfrac{b}{c} \\[8pt] \tan\theta&=\dfrac{\text{Opp}}{\text{Adj}}=\dfrac{a}{b}\quad &\cot\theta&=\dfrac{\text{Adj}}{\text{Opp}}=\dfrac{b}{a} \\[8pt] \sec\theta&=\dfrac{\text{Hyp}}{\text{Adj}}=\dfrac{c}{b}\quad &\csc\theta&=\dfrac{\text{Hyp}}{\text{Opp}}=\dfrac{c}{a} \end{aligned}\]

Memorize this!

def_labeledTriangle

There is a Mnemonic to remember the first three of these:

Mnemonic: SOH CAH TOA which stands for Sin=Opp/Hyp   Cos=Adj/Hyp   Tan=Opp/Adj

Based on these we have the standard quotient and reciprocal identities:

\[\begin{aligned} \tan\theta&=\dfrac{\sin\theta}{\cos\theta}\quad &\cot\theta&=\dfrac{\cos\theta}{\sin\theta}=\dfrac{1}{\tan\theta} \\[8pt] \sec\theta&=\dfrac{1}{\cos\theta}\quad &\csc\theta&=\dfrac{1}{\sin\theta} \end{aligned}\]

Given the right triangle shown with adjacent side \(5\) and opposite side \(12\), find the hypotenuse and the six trig functions for the angle \(\theta\).

eg_5-12-13

By the Pythagorean Theorem, the hypotenuse is \(\text{Hyp}\,=\sqrt{5^2+12^2}=\sqrt{169}=13\). So, the six trig functions are: \[\begin{aligned} \sin\theta&=\dfrac{\text{Opp}}{\text{Hyp}}=\dfrac{12}{13}\quad &\cos\theta&=\dfrac{\text{Adj}}{\text{Hyp}}=\dfrac{5}{13} \\[8pt] \tan\theta&=\dfrac{\text{Opp}}{\text{Adj}}=\dfrac{12}{5}\quad &\cot\theta&=\dfrac{\text{Adj}}{\text{Opp}}=\dfrac{5}{12} \\[8pt] \sec\theta&=\dfrac{\text{Hyp}}{\text{Adj}}=\dfrac{13}{5}\quad &\csc\theta&=\dfrac{\text{Hyp}}{\text{Opp}}=\dfrac{13}{12} \end{aligned}\]

Given the right triangle shown with adjacent side \(0.9\) and opposite side \(1.2\), find the hypotenuse and the six trig functions for the angle \(\theta\).

ex_9-12-15

\(\begin{aligned} \sin\theta&=\dfrac{4}{5}\quad &\cos\theta&=\dfrac{3}{5} \\ \tan\theta&=\dfrac{4}{3}\quad &\cot\theta&=\dfrac{3}{4} \\ \sec\theta&=\dfrac{5}{3}\quad &\csc\theta&=\dfrac{5}{4} \end{aligned}\)

By the Pythagorean Theorem, the hypotenuse is \(\text{Hyp}\,=\sqrt{.9^2+1.2^2}=\sqrt{.81+1.44}=\sqrt{2.25}=1.5\). So, the six trig functions are: \[\begin{aligned} \sin\theta&=\dfrac{\text{Opp}}{\text{Hyp}}=\dfrac{1.2}{1.5}=\dfrac{4}{5}\quad &\cos\theta&=\dfrac{\text{Adj}}{\text{Hyp}}=\dfrac{0.9}{1.5}=\dfrac{3}{5} \\[8pt] \tan\theta&=\dfrac{\text{Opp}}{\text{Adj}}=\dfrac{1.2}{0.9}=\dfrac{4}{3}\quad &\cot\theta&=\dfrac{\text{Adj}}{\text{Opp}}=\dfrac{0.9}{1.2}=\dfrac{3}{4} \\[8pt] \sec\theta&=\dfrac{\text{Hyp}}{\text{Adj}}=\dfrac{1.5}{0.9}=\dfrac{5}{3}\quad &\csc\theta&=\dfrac{\text{Hyp}}{\text{Opp}}=\dfrac{1.5}{1.2}=\dfrac{5}{4} \end{aligned}\]

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