Graphing tangent and the reciprocal functions
A camera on a pivot follows a car along a straight road. As the camera turns toward pointing along the road, the car it aims at runs off toward the horizon. What does the graph of tangent look like, with its gaps, and how do the graphs of secant, cosecant and cotangent come out of sine and cosine?
Where you are 8 in Trigonometric functions and their graphs
Why this exists
The sine and cosine graphs are smooth waves with a value at every input. The other four trigonometric functions each divide by a coordinate, and at some angles that coordinate is 0, so the function has no value there. A list of missing inputs says where the graph breaks, but not what the graph does just before and just after a break.
That is what this lesson draws. Near each missing input the graph shoots up or down without bound, hugging a vertical line it never touches: a vertical asymptote, the same kind of line that Asymptotes describes for rational functions. Knowing where the asymptotes stand and which way each branch runs is most of what there is to know about these four graphs.
This lesson takes two days. Part 1 draws tangent and cotangent. Part 2 draws secant and cosecant from the cosine and sine curves.
Definition
A vertical asymptote of the graph of f is a vertical line x = a such that, as x moves toward a from one side or both, the values f(x) grow without bound, positive or negative.
For each trigonometric function, with k standing for any integer:
- \displaystyle y = \tan x = \frac{\sin x}{\cos x} has a vertical asymptote at every zero of cosine, \displaystyle x = \frac{\pi}{2} + k\pi.
- \displaystyle y = \cot x = \frac{\cos x}{\sin x} has a vertical asymptote at every zero of sine, x = k\pi.
- \displaystyle y = \sec x = \frac{1}{\cos x} has the same asymptotes as tangent, \displaystyle x = \frac{\pi}{2} + k\pi.
- \displaystyle y = \csc x = \frac{1}{\sin x} has the same asymptotes as cotangent, x = k\pi.
In each case the numerator is not 0 where the denominator is 0 (\sin x = \pm 1 where \cos x = 0, and the other way round), so each gap in the domain is a true asymptote, never a single missing point.
Intuition
Part 1: the camera and the tangent
Tihana runs a tracking camera on a pivot, a distance d from a straight road. Measure the camera’s turn x from the perpendicular to the road, positive when it turns right. The perpendicular, the road and the line of sight make a right triangle, so the car the camera aims at is d\tan x along the road from the nearest point, and d\sec x from the camera.
Straight ahead, x = 0 and the car is at the nearest point. As the camera turns right the car moves right, slowly at first and then faster and faster: each extra small turn swings the line of sight across more and more road. As x comes close to \displaystyle \frac{\pi}{2} the line of sight is nearly parallel to the road and the car is very far off. At \displaystyle x = \frac{\pi}{2} the camera points along the road and meets no car at all. Turning left does the same with negative positions. That is the tangent graph between \displaystyle -\frac{\pi}{2} and \displaystyle \frac{\pi}{2}: rising through 0, racing up toward the asymptote on the right and down toward the asymptote on the left.
Past \displaystyle \frac{\pi}{2}, the angle x + \pi has its terminal point opposite that of x, so sine and cosine both change sign and their quotient does not: \tan(x + \pi) = \tan x. The branch on \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) repeats every \pi, which is the period of tangent.
Cotangent is cosine over sine, so the roles flip: it breaks where sine is 0 and is 0 where cosine is 0. Each of its branches falls instead of rising.
Part 2: secant and cosecant from the waves
A number and its reciprocal have the same sign. Where one is near 1 in size the other is too, and where one is small the other is large. So the secant graph can be read off the cosine wave. Where cosine peaks at 1, secant is 1; as cosine falls toward 0, secant climbs without bound; where cosine is 0, secant has an asymptote; where cosine is negative, so is secant, with a downward U around each trough at -1.
The cosecant graph comes from the sine wave in exactly the same way.
Back at the road, Tihana’s line of sight is d\sec x long. It is shortest, d, when the camera points straight at the road, and grows without bound as the camera turns toward parallel: the bottom of the U at x = 0 and its two arms.
Results
Tangent and cotangent. With k any integer,
| y = \tan x | y = \cot x | |
|---|---|---|
| Asymptotes | \displaystyle x = \frac{\pi}{2} + k\pi | x = k\pi |
| Zeros | x = k\pi | \displaystyle x = \frac{\pi}{2} + k\pi |
| Period | \pi | \pi |
| Range | all real numbers | all real numbers |
| On each branch | increasing | decreasing |
| Symmetry | odd: \tan(-x) = -\tan x | odd: \cot(-x) = -\cot x |
Just left of each tangent asymptote the values are large and positive; just right, large and negative. Cotangent does the reverse. The two graphs are related by \displaystyle \cot x = -\tan\left(x - \frac{\pi}{2}\right).
Stretching tangent. For B \ne 0 and A \ne 0, the graph of y = A\tan(Bx) has period \displaystyle \frac{\pi}{|B|} and asymptotes where \displaystyle Bx = \frac{\pi}{2} + k\pi. The factor A stretches it vertically and puts the points of height \pm A halfway between a zero and an asymptote. Its range is still all real numbers. The same rule for B gives cotangent a period of \displaystyle \frac{\pi}{|B|}.
Secant and cosecant.
| y = \sec x | y = \csc x | |
|---|---|---|
| Asymptotes | \displaystyle x = \frac{\pi}{2} + k\pi | x = k\pi |
| Period | 2\pi | 2\pi |
| Symmetry | even | odd |
Both have range (-\infty, -1] \cup [1, \infty) and neither has a zero. The turning points of secant are (2k\pi, 1) and ((2k + 1)\pi, -1), at the peaks and troughs of cosine; those of cosecant are \displaystyle \left(\frac{\pi}{2} + 2k\pi, 1\right) and \displaystyle \left(\frac{3\pi}{2} + 2k\pi, -1\right), at the peaks and troughs of sine. For A > 0, the graph of y = A\sec x touches its guide y = A\cos x exactly at the guide’s peaks and troughs, has asymptotes at the guide’s zeros, and has range (-\infty, -A] \cup [A, \infty). The same holds for y = A\csc x and y = A\sin x.
Why it is true
Period of tangent. Adding \pi turns the terminal point to the opposite point of the circle, so \sin(x + \pi) = -\sin x and \cos(x + \pi) = -\cos x, and
No smaller positive number p works: \tan p would have to equal \tan 0 = 0, but tangent is 0 only where sine is, at multiples of \pi.
Shape of a branch. On \displaystyle \left[0, \frac{\pi}{2}\right) sine grows and cosine shrinks, both positive, so their quotient grows. As x comes close to \displaystyle \frac{\pi}{2} the numerator comes close to 1 and the positive denominator comes close to 0, so the quotient passes any bound. Tangent is odd, since \sin(-x) = -\sin x and \cos(-x) = \cos x, so the left half of the branch is the right half turned half a turn about the origin. Just right of \displaystyle \frac{\pi}{2}, cosine is small and negative while sine is near 1, which gives large negative values.
Cotangent. Since \displaystyle \sin\left(x - \frac{\pi}{2}\right) = -\cos x and \displaystyle \cos\left(x - \frac{\pi}{2}\right) = \sin x,
So the cotangent graph is the tangent graph moved right by \displaystyle \frac{\pi}{2} and reflected across the x-axis. Rising branches become falling ones, and the asymptotes move from \displaystyle \frac{\pi}{2} + k\pi to k\pi.
Stretching. The point (x, y) is on y = A\tan(Bx) exactly when \displaystyle (Bx, \frac{y}{A}) is on y = \tan x. So every feature of the tangent graph at input u appears at \displaystyle x = \frac{u}{B}, and every height is multiplied by A.
Secant from cosine. Where \cos x \ne 0, \displaystyle \sec x = \frac{1}{\cos x} has the same sign as \cos x, and since 0 < |\cos x| \le 1, |\sec x| \ge 1, with equality exactly where \cos x = \pm 1. On an interval where cosine is positive and shrinking, its reciprocal is growing, so the peak of cosine is the bottom of a U. Secant inherits period 2\pi and evenness from cosine. For A \sec x and A\cos x, both are multiplied by the same A, so they still agree where \cos x = \pm 1.
Worked examples
Examples 1 to 4 go with Part 1, and Examples 5 to 7 with Part 2.
Reading one interval of the tangent graph
On [0, 2\pi], find the asymptotes and zeros of y = \tan x, the set where \tan x < 0, and every solution of \tan x = -1.
SolutionAsymptotes stand where cosine is 0: \displaystyle x = \frac{\pi}{2} and \displaystyle x = \frac{3\pi}{2}. Zeros are where sine is 0: x = 0, \pi and 2\pi.
Tangent is negative where sine and cosine have opposite signs, in Quadrants II and IV. On the graph, each branch is negative from its asymptote up to its zero, so \tan x < 0 on
Each branch takes every value exactly once, so \tan x = -1 has one solution on each negative piece. Since \displaystyle \tan\frac{\pi}{4} = 1 and tangent is odd, \displaystyle \tan\left(-\frac{\pi}{4}\right) = -1; adding the period \pi once and twice gives \displaystyle x = \frac{3\pi}{4} and \displaystyle x = \frac{7\pi}{4}.
A tangent with half the input
Graph \displaystyle y = \tan\left(\frac{x}{2}\right): give its period, asymptotes and zeros, two points at heights 1 and -1, and its value at \displaystyle x = \frac{2\pi}{3}.
SolutionHere \displaystyle B = \frac{1}{2}, so the period is \displaystyle \frac{\pi}{1/2} = 2\pi. The asymptotes are where the input to tangent is \displaystyle \frac{\pi}{2} + k\pi:
so they stand at x = \pm\pi, \pm 3\pi and so on. The zeros are where \displaystyle \frac{x}{2} = k\pi, that is x = 2k\pi. Height 1 needs \displaystyle \frac{x}{2} = \frac{\pi}{4}, so \displaystyle \left(\frac{\pi}{2}, 1\right) is on the graph, and by oddness so is \displaystyle \left(-\frac{\pi}{2}, -1\right). One full branch runs from the asymptote at -\pi to the one at \pi through these three points.
At \displaystyle x = \frac{2\pi}{3} the input is \displaystyle \frac{\pi}{3}, so \displaystyle y = \tan\frac{\pi}{3} = \sqrt{3}.
Cotangent from tangent
On (0, \pi), find the asymptotes, zero and the points of height \pm 1 of y = \cot x. Check \displaystyle \cot x = -\tan\left(x - \frac{\pi}{2}\right) at \displaystyle x = \frac{\pi}{3}, and solve \displaystyle \cot x = \sqrt{3} on (-\pi, \pi).
SolutionCotangent is \displaystyle \frac{\cos x}{\sin x}, so its asymptotes are at the zeros of sine, x = 0 and x = \pi, and its zero is at the zero of cosine, \displaystyle x = \frac{\pi}{2}. At \displaystyle x = \frac{\pi}{4} sine and cosine are equal, so \displaystyle \cot\frac{\pi}{4} = 1; at \displaystyle x = \frac{3\pi}{4} they are opposite, so \displaystyle \cot\frac{3\pi}{4} = -1. The branch falls from far above near 0, through \displaystyle \left(\frac{\pi}{4}, 1\right), \displaystyle \left(\frac{\pi}{2}, 0\right) and \displaystyle \left(\frac{3\pi}{4}, -1\right), to far below near \pi.
For the check, \displaystyle \cot\frac{\pi}{3} = \frac{1/2}{\sqrt{3}/2} = \frac{\sqrt{3}}{3}, and
using that tangent is odd. For the equation, \displaystyle \cot x = \sqrt{3} means \displaystyle \tan x = \frac{\sqrt{3}}{3}, so \displaystyle x = \frac{\pi}{6} is one solution. Cotangent has period \pi, so the only other one in (-\pi, \pi) is \displaystyle \frac{\pi}{6} - \pi = -\frac{5\pi}{6}.
The car running away
Tihana’s camera is 30 meters from the road, so the car is s = 30\tan x meters along the road when the camera has turned x radians. Find the turn that aims at a car 30 meters to the right, and where the car is when x = 1.25, x = 1.45 and x = 1.54. How far does the car travel while the camera turns from 1.45 to 1.54? Why can the camera never reach \displaystyle x = \frac{\pi}{2}?
SolutionA car 30 meters right needs 30\tan x = 30, so \tan x = 1 and \displaystyle x = \frac{\pi}{4} \approx 0.785. With a calculator in radian mode,
all in meters. Turning from 1.45 to 1.54, less than a tenth of a radian, the car goes about 973.8 - 247.1 \approx 727 meters. This is the branch steepening as it approaches the asymptote \displaystyle x = \frac{\pi}{2} \approx 1.571. At \displaystyle \frac{\pi}{2} itself the line of sight is parallel to the road, so there is no car to follow and no value of \tan x.
Secant from the cosine curve
On [0, 2\pi], find the asymptotes and turning points of y = \sec x, the values at \displaystyle x = \frac{\pi}{3} and \displaystyle x = \frac{2\pi}{3}, and the set where \sec x \ge 2.
SolutionThe asymptotes are at the zeros of cosine, \displaystyle x = \frac{\pi}{2} and \displaystyle x = \frac{3\pi}{2}. The turning points sit at cosine’s peaks and trough: (0, 1), (\pi, -1) and (2\pi, 1). Reciprocals give \displaystyle \sec\frac{\pi}{3} = \frac{1}{1/2} = 2 and \displaystyle \sec\frac{2\pi}{3} = \frac{1}{-1/2} = -2.
For \sec x \ge 2 the secant must be positive, so cosine is positive, and \displaystyle \frac{1}{\cos x} \ge 2 means \displaystyle 0 < \cos x \le \frac{1}{2}. On the left arm of the graph, rising from (0, 1) to the asymptote, secant reaches 2 at \displaystyle \frac{\pi}{3} and stays above it. The right arm, falling from the asymptote at \displaystyle \frac{3\pi}{2} to (2\pi, 1), passes 2 at \displaystyle \frac{5\pi}{3}. So
The ends at the asymptotes are open because secant has no value there.
Graphing three times the secant
Graph y = 3\sec x using a guide curve. Give its asymptotes, turning points, period and range, and its values at \displaystyle x = \frac{\pi}{3} and \displaystyle x = \frac{3\pi}{4}.
SolutionDraw the guide y = 3\cos x lightly. Both 3\sec x and 3\cos x are 3 times a number and its reciprocal, so they agree where \cos x = \pm 1, at the turning points (2k\pi, 3) and ((2k + 1)\pi, -3). Where the guide is 0, 3\sec x has asymptotes, at \displaystyle x = \frac{\pi}{2} + k\pi. Between them sit upward U shapes with bottom 3 and downward U shapes with top -3.
The period is 2\pi, the same as cosine’s, and the range is (-\infty, -3] \cup [3, \infty). Since the graph has no top and no bottom, the 3 is not an Amplitude and reflection; it sets how close to the axis the U shapes come. The values are \displaystyle 3\sec\frac{\pi}{3} = 3 \cdot 2 = 6 and
Cosecant at the roadside
Dubravka measures her camera’s angle \theta from the road itself, with 0 < \theta < \pi. The camera is 30 meters from the road. Show that the car is 30\csc\theta meters from the camera, describe this function’s graph on (0, \pi), and find when the car is 60 meters away and when it is 45 meters away.
SolutionThe perpendicular from the camera to the road, 30 meters, the road and the line of sight make a right triangle with hypotenuse L, the distance to the car. The line of sight meets the road at the angle \theta (alternate angles, since the road runs parallel to the direction \theta is measured from), and the perpendicular is the side opposite that angle. So \displaystyle \sin\theta = \frac{30}{L} and \displaystyle L = \frac{30}{\sin\theta} = 30\csc\theta.
On (0, \pi) sine is positive with its peak 1 at \displaystyle \frac{\pi}{2}, so 30\csc\theta is one upward U with bottom \displaystyle \left(\frac{\pi}{2}, 30\right) and asymptotes at \theta = 0 and \theta = \pi, where the camera points along the road. The nearest the car comes is 30 meters, straight across.
A horizontal line above 30 meets the U twice. Distance 60 needs \csc\theta = 2, so \displaystyle \sin\theta = \frac{1}{2} and \displaystyle \theta = \frac{\pi}{6} or \displaystyle \frac{5\pi}{6}. Distance 45 needs \displaystyle \sin\theta = \frac{30}{45} = \frac{2}{3}, so \theta \approx 0.730 or \theta \approx \pi - 0.730 \approx 2.412. Each pair is symmetric about \displaystyle \frac{\pi}{2}: the car is as far on one side of the nearest point as on the other.
Boundaries
- An asymptote is not part of the graph. At \displaystyle x = \frac{\pi}{2} + k\pi tangent and secant have no value; the dashed line marks where the graph is missing, and nothing is plotted on it.
- Cotangent is not always one over tangent. At \displaystyle x = \frac{\pi}{2}, \displaystyle \cot x = \frac{0}{1} = 0 while \displaystyle \frac{1}{\tan x} has no meaning, because \displaystyle \tan\frac{\pi}{2} does not exist. Graphing \displaystyle \frac{1}{\tan x} gives the cotangent graph with holes at its zeros.
- No amplitude. Tangent, cotangent, secant and cosecant have no greatest or least value, so “amplitude” does not apply; the A in y = A\tan x or y = A\sec x is a vertical stretch.
- The guide-curve trick needs a reciprocal. It works for A\sec(Bx) against A\cos(Bx) because their product is A^2. It does not turn a cosine curve into a tangent graph.
- The road model is one branch. The camera formulas use turns strictly between \displaystyle -\frac{\pi}{2} and \displaystyle \frac{\pi}{2} (or \theta strictly between 0 and \pi); the other branches of the graphs describe no aim at the road.
Where this goes next
Shifting these graphs left, right, up and down, alongside the sine waves, is the work of Amplitude, period and phase shift, and reading B from a period is Finding the period from the equation. Because tangent is increasing on each branch, the branch on \displaystyle \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) takes each value once, which is the slice that the inverse tangent is built on. In calculus the steepening near an asymptote and the slope of each branch become exact.
Exercises
Part 1: tangent and cotangent
- On [-2\pi, 0], list the asymptotes and the zeros of y = \tan x.
- Solve \displaystyle \tan x = \sqrt{3} on [-2\pi, 2\pi].
- Find where \tan x > 0 on (-\pi, \pi).
- Give the period, asymptotes and zeros of y = \tan(2x), and the point of height 1 nearest the origin on the right.
- Give the period, the asymptotes and the points of height \pm 1 on the middle branch of \displaystyle y = \tan\left(\frac{x}{3}\right).
- For y = 2\tan x, state the asymptotes and range, and find the values at \displaystyle x = \frac{\pi}{4}, \displaystyle x = -\frac{\pi}{4} and \displaystyle x = \frac{\pi}{3}.
- On [-\pi, \pi], list the asymptotes and zeros of y = \cot x and find where \cot x < 0.
- Solve \displaystyle \cot x = -\sqrt{3} on (0, 2\pi).
- Give the period, asymptotes and zeros of y = \cot(\pi x).
- Krunoslav’s camera is 50 meters from the road, so the car is 50\tan x meters along it. Where is the car when \displaystyle x = \frac{\pi}{6}? At what turn is the car 120 meters to the right? Where is it when x = 1.52?
- Find B > 0 so that y = \tan(Bx) has its two asymptotes nearest the origin at \displaystyle x = \pm\frac{\pi}{6}.
Part 2: secant and cosecant
- On [-2\pi, 0], list the asymptotes and turning points of y = \sec x.
- On [0, 2\pi], list the asymptotes and turning points of y = \csc x, and find where \csc x < 0.
- Find \displaystyle \sec\frac{3\pi}{4}, \displaystyle \csc\left(-\frac{\pi}{6}\right), \displaystyle \sec\frac{5\pi}{6} and \displaystyle \csc\frac{4\pi}{3}.
- Solve \displaystyle \sec x = \sqrt{2} on [0, 2\pi].
- Solve \csc x = -2 on [0, 2\pi].
- Describe the graph of y = 2\csc x on [0, 2\pi] using a guide curve: asymptotes, turning points and range.
- Give the period, asymptotes and the turning points on [0, \pi] of y = \sec(2x).
- Find where \sec x \le -2 on [0, 2\pi].
- Dubravka’s second camera is 40 meters from the road, with \theta measured from the road. When is the car \displaystyle 40\sqrt{2} meters from the camera, and what is the least distance?
- Krunoslav’s line of sight is 50\sec x meters long for \displaystyle -\frac{\pi}{2} < x < \frac{\pi}{2}. For which turns is it at most 100 meters?
Reasoning
- Explain why the tangent graph has no highest point and no lowest point, even on a single branch.
- Show that the graphs of y = \sec x and y = \cos x meet only where \cos x = \pm 1.
- Prove that tangent is an odd function, and say what this means for its graph.
- Explain why the graph of y = 3\sec x has no point strictly between heights -3 and 3, while the graph of y = 3\tan x crosses every horizontal line.
Answers
- Asymptotes \displaystyle x = -\frac{3\pi}{2}, -\frac{\pi}{2}; zeros x = -2\pi, -\pi, 0. Cosine is 0 at the first two, sine at the last three.
- \displaystyle x = -\frac{5\pi}{3}, -\frac{2\pi}{3}, \frac{\pi}{3}, \frac{4\pi}{3}. Since \displaystyle \tan\frac{\pi}{3} = \sqrt{3} and the period is \pi, the solutions are \displaystyle \frac{\pi}{3} + k\pi.
- \displaystyle \left(-\pi, -\frac{\pi}{2}\right) \cup \left(0, \frac{\pi}{2}\right). Sine and cosine share a sign in Quadrants I and III.
- Period \displaystyle \frac{\pi}{2}; asymptotes \displaystyle x = \frac{\pi}{4} + \frac{k\pi}{2}; zeros \displaystyle x = \frac{k\pi}{2}; the point is \displaystyle \left(\frac{\pi}{8}, 1\right), since \displaystyle 2 \cdot \frac{\pi}{8} = \frac{\pi}{4}.
- Period 3\pi; asymptotes \displaystyle x = \frac{3\pi}{2} + 3k\pi; points \displaystyle \left(\frac{3\pi}{4}, 1\right) and \displaystyle \left(-\frac{3\pi}{4}, -1\right), from \displaystyle \frac{x}{3} = \pm\frac{\pi}{4}.
- Asymptotes \displaystyle x = \frac{\pi}{2} + k\pi, range all real numbers. Values 2, -2 and \displaystyle 2\sqrt{3}, twice the tangent values 1, -1, \displaystyle \sqrt{3}.
- Asymptotes x = -\pi, 0, \pi; zeros \displaystyle x = -\frac{\pi}{2}, \frac{\pi}{2}. \cot x < 0 on \displaystyle \left(-\frac{\pi}{2}, 0\right) \cup \left(\frac{\pi}{2}, \pi\right), Quadrants IV and II.
- \displaystyle x = \frac{5\pi}{6}, \frac{11\pi}{6}. \displaystyle \cot\frac{5\pi}{6} = \frac{-\sqrt{3}/2}{1/2} = -\sqrt{3}, then add the period \pi.
- Period \displaystyle \frac{\pi}{\pi} = 1; asymptotes where \pi x = k\pi, at every integer x = k; zeros where \displaystyle \pi x = \frac{\pi}{2} + k\pi, at \displaystyle x = \frac{1}{2} + k.
- At \displaystyle \frac{\pi}{6}: \displaystyle 50 \cdot \frac{\sqrt{3}}{3} = \frac{50\sqrt{3}}{3} \approx 28.87 meters right. \displaystyle \tan x = \frac{120}{50} = 2.4 gives x \approx 1.176. At 1.52: 50\tan 1.52 \approx 983.5 meters.
- B = 3. The asymptotes nearest the origin are where \displaystyle Bx = \pm\frac{\pi}{2}, at \displaystyle x = \pm\frac{\pi}{2B}, and \displaystyle \frac{\pi}{2B} = \frac{\pi}{6}.
- Asymptotes \displaystyle x = -\frac{3\pi}{2}, -\frac{\pi}{2}; turning points (-2\pi, 1), (-\pi, -1), (0, 1).
- Asymptotes x = 0, \pi, 2\pi; turning points \displaystyle \left(\frac{\pi}{2}, 1\right) and \displaystyle \left(\frac{3\pi}{2}, -1\right); \csc x < 0 on (\pi, 2\pi), where sine is negative.
- \displaystyle \sec\frac{3\pi}{4} = -\sqrt{2}, \displaystyle \csc\left(-\frac{\pi}{6}\right) = -2, \displaystyle \sec\frac{5\pi}{6} = -\frac{2\sqrt{3}}{3}, \displaystyle \csc\frac{4\pi}{3} = -\frac{2\sqrt{3}}{3}. Each is one over the cosine or sine.
- \displaystyle x = \frac{\pi}{4}, \frac{7\pi}{4}, where \displaystyle \cos x = \frac{\sqrt{2}}{2}.
- \displaystyle x = \frac{7\pi}{6}, \frac{11\pi}{6}, where \displaystyle \sin x = -\frac{1}{2}.
- Guide y = 2\sin x. Asymptotes x = 0, \pi, 2\pi; turning points \displaystyle \left(\frac{\pi}{2}, 2\right) and \displaystyle \left(\frac{3\pi}{2}, -2\right); range (-\infty, -2] \cup [2, \infty).
- Period \pi; asymptotes \displaystyle x = \frac{\pi}{4} + \frac{k\pi}{2}; turning points (0, 1), \displaystyle \left(\frac{\pi}{2}, -1\right), (\pi, 1), where 2x is a multiple of \pi.
- \displaystyle \left(\frac{\pi}{2}, \frac{2\pi}{3}\right] \cup \left[\frac{4\pi}{3}, \frac{3\pi}{2}\right). It needs \displaystyle -\frac{1}{2} \le \cos x < 0.
- \displaystyle 40\csc\theta = 40\sqrt{2} gives \displaystyle \sin\theta = \frac{\sqrt{2}}{2}, so \displaystyle \theta = \frac{\pi}{4} or \displaystyle \frac{3\pi}{4}. The least distance is 40 meters, at \displaystyle \theta = \frac{\pi}{2}.
- \displaystyle -\frac{\pi}{3} \le x \le \frac{\pi}{3}. 50\sec x \le 100 with \cos x > 0 means \displaystyle \cos x \ge \frac{1}{2}.
- Each branch is increasing, so no point on it is highest or lowest, and near its asymptotes it passes every bound upward and downward. The other branches are copies of it.
- Where both exist, \displaystyle \frac{1}{\cos x} = \cos x means \cos^2 x = 1, so \cos x = \pm 1; and at those x both equal \pm 1.
- \displaystyle \tan(-x) = \frac{\sin(-x)}{\cos(-x)} = \frac{-\sin x}{\cos x} = -\tan x, using that sine is odd and cosine is even. The graph is unchanged by a half turn about the origin.
- \displaystyle |3\sec x| = \frac{3}{|\cos x|} \ge 3 because |\cos x| \le 1. Each tangent branch is increasing and passes every bound up and down, so it takes every value, and so does 3\tan x.