Emma's dad leans a ladder against the house so that it reaches a gutter 20 ft above the ground. The ladder makes a $60^\circ$ angle with the ground. How long is the ladder? Since $\sin 60^\circ = \dfrac{20}{L}$, we get $L = \dfrac{20}{\sin 60^\circ}$. That "one over sine" shows up so often that it has its own name: the cosecant. Together with the secant and cotangent, it saves you steps in half of the problems in this chapter. We will finish the ladder problem at the end of the page.
Secant, Cosecant, and Cotangent
Sine, cosine, and tangent each have a "partner": their reciprocal (one over the ratio). The three new ratios are defined that way.
In a right triangle, flipping each fraction gives the side ratios. Study this picture, because you will use it all through the page.
Here is a graph of $y = \cos x$ and its reciprocal $y = \dfrac{1}{\cos x}$. Notice what happens to the reciprocal wherever the cosine touches $0$.
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Knowing that $\sin \alpha = \frac{3}{5}$ and that $\alpha$ is an angle in the first quadrant, calculate the exact value of $\csc \alpha$ and $\sec \alpha$.
Knowing that $\sin \alpha = \frac{3}{5}$ and that $\alpha$ is an angle in the first quadrant, calculate the exact value of $\csc \alpha$ and $\sec \alpha$.
A spotlight illuminates a wall from the ground. The horizontal distance from the spotlight to the base of the wall is $10$ ft. The upper light ray forms an angle…
A spotlight illuminates a wall from the ground. The horizontal distance from the spotlight to the base of the wall is $10$ ft. The upper light ray forms an angle $\alpha$ with the ground. The length of the light ray from the spotlight to the highest illuminated point on the wall is $L$. Express $L$ using the secant ratio and calculate its exact value if $\cot \alpha = \frac{3}{4}$.
The exact values of sec, csc, and cot come from flipping the values of cos, sin, and tan that you already know. The two special triangles give them all at once.
A few more values come straight from the unit circle: $\sec 0 = 1$, $\csc\frac{\pi}{2} = 1$, $\cot\frac{\pi}{2} = 0$, $\sec\pi = -1$, and $\csc\frac{3\pi}{2} = -1$.
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Determine the exact value of $\sec 60^\circ$ and $\cot 45^\circ$ without using a calculator.
Determine the exact value of $\sec 60^\circ$ and $\cot 45^\circ$ without using a calculator.
A reciprocal always has the same sign as the original number. So $\sec\alpha$ has the sign of $\cos\alpha$, $\csc\alpha$ has the sign of $\sin\alpha$, and $\cot\alpha$ has the sign of $\tan\alpha$. You do not need to memorize anything new.
To evaluate a reciprocal at a non-acute angle, use the reference angle for the size and the quadrant for the sign.
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An angle $\gamma$ satisfies $\sec \gamma < 0$ and $\csc \gamma > 0$. Reason in which quadrant of the unit circle the angle $\gamma$ is located.
An angle $\gamma$ satisfies $\sec \gamma < 0$ and $\csc \gamma > 0$. Reason in which quadrant of the unit circle the angle $\gamma$ is located.
You know $\sin^2\alpha + \cos^2\alpha = 1$. Divide every term by $\cos^2\alpha$ and the reciprocals appear: $\dfrac{\sin^2\alpha}{\cos^2\alpha} + 1 = \dfrac{1}{\cos^2\alpha}$, that is, $\tan^2\alpha + 1 = \sec^2\alpha$. Dividing by $\sin^2\alpha$ instead gives the third identity.
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If $\tan \beta = \frac{4}{3}$ and $0 < \beta < \frac{\pi}{2}$, use the fundamental relationship $1 + \tan^2 \beta = \sec^2 \beta$ to calculate $\sec \beta$.
If $\tan \beta = \frac{4}{3}$ and $0 < \beta < \frac{\pi}{2}$, use the fundamental relationship $1 + \tan^2 \beta = \sec^2 \beta$ to calculate $\sec \beta$.
If you know one ratio and the quadrant, you can find the other five. The ratios also have a simple description on the coordinate plane. For a point $(x, y)$ on the terminal side of $\alpha$, at distance $r = \sqrt{x^2 + y^2}$ from the origin:
Strategy: (1) turn the given ratio into one you know well, (2) use a Pythagorean identity or a reference triangle to get the other one, (3) pick the signs from the quadrant, (4) flip to get the reciprocals.
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Knowing that $\sec \alpha = -\frac{13}{5}$ and that $\pi < \alpha < \frac{3\pi}{2}$, calculate the exact value of the other trigonometric ratios (sine, cosine, tangent,…
Knowing that $\sec \alpha = -\frac{13}{5}$ and that $\pi < \alpha < \frac{3\pi}{2}$, calculate the exact value of the other trigonometric ratios (sine, cosine, tangent, cosecant and cotangent).
Knowing that $\csc x = 3$ and that the angle $x$ belongs to the second quadrant, calculate the exact value of the expression $E = \frac{\sec x + \tan x}{\cot x}$.
Knowing that $\csc x = 3$ and that the angle $x$ belongs to the second quadrant, calculate the exact value of the expression $E = \frac{\sec x + \tan x}{\cot x}$.
Simplify the following trigonometric expression and express it in terms of a single trigonometric ratio or numerical constant: $\frac{\sec^4 x - 1}{\tan^2 x} - \tan^2 x$.
Simplify the following trigonometric expression and express it in terms of a single trigonometric ratio or numerical constant: $\frac{\sec^4 x - 1}{\tan^2 x} - \tan^2 x$.
To prove an identity, you show that the two sides are equal for every angle where they are defined. The rules of the game: pick one side (usually the more complicated one) and transform it until it looks like the other. Never move terms across the equals sign, because that assumes what you want to prove.
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Prove that for any angle $\alpha$ where the ratios are defined, the following equality holds: $\cot^2 \alpha - \cos^2 \alpha = \cot^2 \alpha \cdot \cos^2 \alpha$.
Prove that for any angle $\alpha$ where the ratios are defined, the following equality holds: $\cot^2 \alpha - \cos^2 \alpha = \cot^2 \alpha \cdot \cos^2 \alpha$.
Two Pythagorean identities, rewritten as differences of squares, give you a very handy fact.
This gives two powerful uses. First, if you know the sum, you instantly know the difference, and then you can solve for each ratio. Second, a stray "$1$" in a proof can be replaced by $\sec^2 x - \tan^2 x$ (or $\csc^2 x - \cot^2 x$) and then factored.
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Prove the following trigonometric identity: $\frac{\tan x + \sec x - 1}{\tan x - \sec x + 1} = \tan x + \sec x$.
Prove the following trigonometric identity: $\frac{\tan x + \sec x - 1}{\tan x - \sec x + 1} = \tan x + \sec x$.
To solve an equation with a reciprocal ratio, do one of two things. If the equation is linear in the ratio (like $2\csc x + 4 = 0$), solve for the ratio and then flip. If you get both a squared and a plain ratio (like $\sec^2 x + \tan x = 3$), use a Pythagorean identity to leave a single ratio, then factor like a quadratic. Finally, find all the angles in the interval, using the quadrants.
For a system, treat the reciprocal ratios as unknowns and use elimination, exactly as in Algebra 1.
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Solve the following trigonometric equation indicating all solutions in the interval $[0, 2\pi]$: $2\csc x + 4 = 0$.
Solve the following trigonometric equation indicating all solutions in the interval $[0, 2\pi]$: $2\csc x + 4 = 0$.
Solve the following system of trigonometric equations for values of $x, y \in [0, \pi]$: $$ \begin{cases} \csc x + \sec y = 3 \\ \csc x - \sec y = 1 \end{cases} $$
Solve the following system of trigonometric equations for values of $x, y \in [0, \pi]$:
$$ \begin{cases} \csc x + \sec y = 3 \\ \csc x - \sec y = 1 \end{cases} $$
Reciprocal ratios appear whenever a length is a hypotenuse that you find by dividing. Back to the ladder:
If $\cot\theta = c$ for an acute angle, then $\csc\theta = \sqrt{1 + c^2}$ and $\sec\theta = \dfrac{\sqrt{1 + c^2}}{c}$. These formulas help when $c$ is not a nice fraction, for example when it involves a cube root.
Back to the beginning: the ladder at $60^\circ$ reaching 20 ft has length $L = 20\csc 60^\circ = 20\cdot\dfrac{2\sqrt3}{3} = \dfrac{40\sqrt3}{3} \approx 23.1$ ft.
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In an optics experiment, a laser beam passes through a glass block with flat, parallel faces. The path of the beam inside the glass has a length…
In an optics experiment, a laser beam passes through a glass block with flat, parallel faces. The path of the beam inside the glass has a length $L = h \cdot \sec \theta_r$, where $h = 5$ in is the thickness of the block and $\theta_r$ is the angle of refraction. If the beam is incident from the air at an angle $\theta_i = 60^\circ$ and the refractive index of the glass is $n = \sqrt{2}$, calculate the exact value of the length $L$ traveled by the laser inside the block. (Fact: Snell's law states that $\sin \theta_i = n \cdot \sin \theta_r$).
An L-shaped corridor has two sections of widths $a = 2$ ft and $b = 3$ ft that intersect at a right angle. A rigid beam of length $L$ gets stuck when trying to turn the…
An L-shaped corridor has two sections of widths $a = 2$ ft and $b = 3$ ft that intersect at a right angle. A rigid beam of length $L$ gets stuck when trying to turn the corner, touching both outer walls and the inner corner. The length of the beam as a function of the angle $\theta$ it forms with one of the outer walls can be modeled as $L(\theta) = a \csc \theta + b \sec \theta$. Calculate the exact value of $L(\theta)$ when $\cot \theta = \sqrt[3]{\frac{a}{b}}$.