5.5.1 Reciprocal Trig Functions - Definitions

Reciprocal trigonometric functions are the reciprocals (or multiplicative inverses) of the basic trigonometric functions (sine, cosine, and tangent). These reciprocal functions are commonly used in trigonometry to simplify expressions and solve equations.

1. Definitions of the Reciprocal Trigonometric Functions:

• Cosecant $(\csc or\ \text{cosec} ):$ $\csc \theta = \frac{1}{\sin \theta}$
  • Cosecant is the reciprocal of the sine function.
  • It is undefined when $\sin \theta = 0 , which\ occurs\ at\ \theta = 0^\circ, 180^\circ, 360^\circ$(or multiples of $\pi$radians).
• Secant $(\sec):$ $\sec \theta = \frac{1}{\cos \theta}$
  • Secant is the reciprocal of the cosine function.
  • It is undefined when $\cos \theta$ $= 0$ , which occurs at $\theta = 90^\circ, 270^\circ (or\ odd\ multiples\ of\ \frac{\pi}{2}\ radians).$
• Cotangent ($\cot):$ $\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}$
  • Cotangent is the reciprocal of the tangent function.
  • It is undefined when $\tan \theta = 0$ , which occurs at $\theta = 0^\circ, 180^\circ, 360^\circ$ (or multiples of $\pi$ radians).

2. Domains and Ranges:

• Cosecant ($\csc \theta$):
  • Domain: All real numbers $\theta$ except $\theta \neq n\pi$ , where $n$ is any integer.
  • Range: $\csc \theta$ is undefined between $-1$ and $1$, so the range is $(-\infty, -1] \cup [1, \infty) .$
• Secant $(\sec \theta$):
  • Domain: All real numbers $\theta$ except $\theta \neq \frac{\pi}{2} + n\pi$, where $n$ is any integer.
  • Range: $\sec \theta$ is undefined between $-1$ and $1$, so the range is $(-\infty, -1] \cup [1, \infty) .$
• Cotangent ($\cot \theta$):
  • Domain: All real numbers $\theta$ except $\theta \neq n\pi$, where $n$ is any integer.
  • Range: All real numbers $(-\infty, \infty)$.

3. Graphs of Reciprocal Trigonometric Functions:

• Cosecant ($\csc \theta$):
  • The graph of $\csc \theta$ has vertical asymptotes at points where $\sin \theta$ $= 0$ , because $\csc \theta$ is undefined there.
  • The graph consists of branches that approach these asymptotes and have minimum and maximum points corresponding to the peaks and troughs of the sine graph.
• Secant ($\sec \theta$):
  • The graph of $\sec \theta$ has vertical asymptotes at points where \cos \theta$$ = 0 , because $\sec$ $\theta$ is undefined there.
  • The graph has branches that approach these asymptotes, with similar peaks and troughs corresponding to the cosine graph.
• Cotangent ($\cot \theta$):
  • The graph of $\cot \theta$ has vertical asymptotes where $\sin \theta$ $= 0$ , corresponding to points where $\theta = n\pi .$
  • The graph decreases as $\theta$ increases, creating a series of repeating curves over each interval $(n\pi, (n+1)\pi) .$

4. Key Identities Involving Reciprocal Trigonometric Functions:

• Pythagorean Identity for Cosecant and Cotangent: $\csc^2 \theta = 1 + \cot^2 \theta$
  • Derived from dividing the basic Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ by $\sin^2 \theta .$
• Pythagorean Identity for Secant and Tangent: $\sec^2 \theta = 1 + \tan^2 \theta$
  • Derived from dividing the basic Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ by $\cos^2$ $\theta$ .

5. Example Problems Involving Reciprocal Trig Functions:

Summary:

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Further Reciprocal Trig Functions

SOLUTION

a)

$$\sin^2 \theta + \cos^2 \theta \equiv 1 \quad \text{(Derive relevant trig identity)}$$ $$\div \cos^2\theta$$ $$\tan^2 \theta + 1 \equiv \sec^2 \theta \quad \text{(Identity)}$$ $$\therefore \sec^2 A = \left(\frac{1}{3}\right)^2 + 1 = :success[\frac{10}{9}]$$

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b)

$$\sin^2 \theta + \cos^2 \theta \equiv 1$$ $$\div \sin^2\theta$$ $$1 + \cot^2 \theta = \cosec^2 \theta \quad \text{(Identity)}$$ $$\cot^2 \theta \equiv \cosec^2 - 1$$ $$\cot^2 B = \left (1 + \sqrt{3}\right)^2 - 1$$ $$\cot^2 B = :success[3 + 2\sqrt{3}] \quad \text{(Simplify)}$$

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c)

$$\sec C = \frac{3}{2}$$ $$\sin^2 \theta + \cos^2 \theta \equiv 1$$ $$\div \cos^2\theta$$ $$1 + \tan^2 \theta = \sec^2 \theta \quad \text{(Identity)}$$ $$\tan^2 \theta \equiv \sec^2 \theta -1$$ $$\therefore \tan^2 C = \left(\frac{3}{2}\right)^2-1 = \frac{5}{4}$$ $$\tan^2 C = \frac{9}{4} - 1 = \frac{5}{4}$$ $$\tan C = \pm \sqrt{\frac{5}{4}} = :success[\pm \frac{\sqrt{5}}{2}]$$

Note:

Derive relevant trig identities

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