Reciprocal Trigonometric Functions

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Define the reciprocal trigonometric function cosecant ($\csc\theta$) in terms of side lengths of a right triangle.

$$\csc\theta = \frac{1}{\sin\theta} = \frac{\text{Hypotenuse}}{\text{Opposite}}$$

Define the reciprocal trigonometric function secant ($\sec\theta$) in terms of side lengths of a right triangle.

$$\sec\theta = \frac{1}{\cos\theta} = \frac{\text{Hypotenuse}}{\text{Adjacent}}$$

Define the reciprocal trigonometric function cotangent ($\cot\theta$) in terms of side lengths of a right triangle.

$$\cot\theta = \frac{1}{\tan\theta} = \frac{\text{Adjacent}}{\text{Opposite}}$$

What is the common misconception regarding $\sin^{-1} x$ versus $(\sin x)^{-1}$?

$\sin^{-1} x$ (or $\arcsin x$) is the inverse trigonometric function used to find an angle, whereas $(\sin x)^{-1} = \frac{1}{\sin x} = \csc x$ is the reciprocal ratio.

If $\sin\theta = \frac{5}{13}$, evaluate $\csc\theta$.

$$\csc\theta = \frac{1}{\sin\theta} = \frac{13}{5}$$

If $\cos\theta = \frac{8}{17}$, evaluate $\sec\theta$.

$$\sec\theta = \frac{1}{\cos\theta} = \frac{17}{8}$$

In a right triangle, the opposite leg is $9$ and the adjacent leg is $40$. Find $\cot\theta$.

$$\cot\theta = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{40}{9}$$

What are the values of $\sec\theta$ and $\csc\theta$ relative to $1$ for any acute angle $\theta$?

Both $\sec\theta > 1$ and $\csc\theta > 1$, because the hypotenuse is strictly longer than either leg.

State the cofunction identity pairing $\sec\theta$ and $\csc\theta$ for an acute angle $\theta$.

$$\sec\theta = \csc(90^{\circ} - \theta)$$ $$\csc\theta = \sec(90^{\circ} - \theta)$$

State the cofunction identity for cotangent in degrees.

$$\cot\theta = \tan(90^{\circ} - \theta)$$

State the Pythagorean identity connecting $\tan\theta$ and $\sec\theta$.

$$1 + \tan^2\theta = \sec^2\theta$$

State the Pythagorean identity connecting $\cot\theta$ and $\csc\theta$.

$$1 + \cot^2\theta = \csc^2\theta$$

Given $\tan\theta = 2$ for an acute angle $\theta$, find $\sec\theta$.

$$\sec^2\theta = 1 + \tan^2\theta = 1 + 2^2 = 5$$ $$\sec\theta = \sqrt{5}$$

If $\csc\theta = \frac{5}{3}$, determine $\cos\theta$ for acute angle $\theta$.

Since $\sin\theta = \frac{3}{5}$, the opposite side is $3$ and hypotenuse is $5$. The adjacent leg is $\sqrt{5^2 - 3^2} = 4$. $$\cos\theta = \frac{4}{5}$$

Reciprocal Trigonometric Functions

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