Reciprocal Trigonometric Functions

Starting point: what you already know

• In a right-angled triangle: $\sin \theta =\frac{O}{H}$, $\cos \theta =\frac{A}{H}$, $\tan \theta =\frac{O}{A}$.
• Exact values, such as $\sin 30^{\circ }=\frac{1}{2}$, $\cos 45^{\circ }=\frac{1}{\sqrt{2}}$, $\tan 60^{\circ }=\sqrt{3}$.
• Radians: $\pi$ radians $=180^{\circ }$.
• The graphs of $y=\sin x$, $y=\cos x$ and $y=\tan x$, and the identity ${\sin }^{2}x+{\cos }^{2}x=1$.
• From the chapter on graphing related functions: zeros of $f$ become vertical asymptotes of $\frac{1}{f}$.

The three new ratios

$$\mathrm{cosec}\theta =\frac{1}{\sin \theta }\sec \theta =\frac{1}{\cos \theta }\cot \theta =\frac{1}{\tan \theta }=\frac{\cos \theta }{\sin \theta }$$

In a triangle: $\mathrm{cosec}\theta =\frac{H}{O}$, $\sec \theta =\frac{H}{A}$, $\cot \theta =\frac{A}{O}$.

The graphs

Use the reciprocal rules: zeros become asymptotes, and heights of $\pm 1$ stay where they are.

%%FIG:sec%%

• $\sec x$ and $\mathrm{cosec}x$ have U-shaped branches that never get between $-1$ and 1.
• $\cot x$ looks like $\tan x$ reflected and shifted, decreasing between each pair of asymptotes. It is zero where $\cos x=0$.

New Pythagorean identities

Start from ${\sin }^{2}x+{\cos }^{2}x=1$.

• Divide every term by ${\cos }^{2}x$: ${\tan }^{2}x+1={\sec }^{2}x$
• Divide every term by ${\sin }^{2}x$: $1+{\cot }^{2}x={\mathrm{cosec}}^{2}x$

Quick summary

• $\mathrm{cosec}\theta =\frac{1}{\sin \theta }$, $\sec \theta =\frac{1}{\cos \theta }$, $\cot \theta =\frac{\cos \theta }{\sin \theta }$.
• Their graphs have vertical asymptotes where the original function is zero.
• $\sec x$ and $\mathrm{cosec}x$ have range $|y|\ge 1$ and period $2\pi$. $\cot x$ has period $\pi$.
• $1+{\tan }^{2}x={\sec }^{2}x$ and $1+{\cot }^{2}x={\mathrm{cosec}}^{2}x$.

Quiz