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Trigonometric Functions Review

Try each problem first. You don’t need to finish everything.

Core path ~55 min Full sheet ~1 h 45 min

Recommended: K1–K5, then one A and one T problem

K = Knowledge & Understanding · A = Application · T = Thinking · Boss = Challenge

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Core Skills (K)

Radians · Graphs · Transformations · Reciprocals · Equations

K1 Radians & Exact Values

  1. a)
    Convert 210∘210^\circ to radians and 7π4\dfrac{7\pi}{4} to degrees.
  2. b)
    Evaluate exactly: sin⁡7π6\sin\dfrac{7\pi}{6}, cos⁡(−π4)\cos\left(-\dfrac{\pi}{4}\right), tan⁡5π3\tan\dfrac{5\pi}{3}.
  3. c)
    For θ=17π4\theta=\dfrac{17\pi}{4}, determine the principal angle and one negative coterminal angle.
Stuck? Peek at a hint

Place each angle in its quadrant first. The related acute angle gives the value, the quadrant gives the sign.

K2 Key Features

  1. a)
    For y=tan⁡xy=\tan x, state the period, the domain and the equations of the vertical asymptotes.
  2. b)
    For y=cos⁡xy=\cos x on 0≤x≤2π0\le x\le 2\pi, state the zeros and the coordinates of every maximum and minimum point.
  3. c)
    Determine the smallest positive value of dd such that cos⁡x=sin⁡(x+d)\cos x=\sin(x+d) for every real xx.
Stuck? Peek at a hint

Write tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x} and ask where the denominator is 00.

K3 Transformations

Let f(x)=−3sin⁡(2x−π3)+1f(x)=-3\sin\left(2x-\dfrac{\pi}{3}\right)+1.

  1. a)
    Write ff in the form y=asin⁡(k(x−d))+cy=a\sin(k(x-d))+c. State the amplitude, period, phase shift and the equation of the midline.
  2. b)
    State the range of ff and the coordinates of its first maximum point with x>0x>0.
  3. c)
    Determine an equation of the form y=acos⁡(k(x−d))+cy=a\cos(k(x-d))+c for the graph below. Each grid square is 1 unit.
    A wave-shaped curve oscillating between y = -3 and y = 1, with maximum points at (1, 1) and (7, 1) and a minimum point at (4, -3).xy−2−112345678−4−312
Stuck? Peek at a hint

Factor kk out of the bracket before reading the phase shift. For the graph, the midline is halfway between the maximum and the minimum.

K4 Reciprocal Functions

  1. a)
    Evaluate exactly: csc⁡5π4\csc\dfrac{5\pi}{4}, sec⁡2π3\sec\dfrac{2\pi}{3}, cot⁡π6\cot\dfrac{\pi}{6}.
  2. b)
    State the domain, range and period of y=sec⁡xy=\sec x.
  3. c)
    State the period and the equations of the vertical asymptotes of y=csc⁡(2x)y=\csc(2x).
Stuck? Peek at a hint

Rewrite each function as a reciprocal, such as csc⁡x=1sin⁡x\csc x=\dfrac{1}{\sin x}. Asymptotes sit where that denominator is 00.

K5 Trig Equations

  1. a)
    Solve 2sin⁡x+3=02\sin x+\sqrt{3}=0 for 0≤x≤2π0\le x\le 2\pi.
  2. b)
    Solve 2cos⁡2x−cos⁡x−1=02\cos^2x-\cos x-1=0 for 0≤x≤2π0\le x\le 2\pi.
  3. c)
    Determine the general solution of csc⁡x=−2\csc x=-2.
Stuck? Peek at a hint

Isolate the trig ratio first; in (b), factor as a quadratic in cos⁡x\cos x. Then find the related acute angle and use CAST to place every solution.

Real World (A)

Sinusoidal Models · Rates of Change · Secant in Context

A1 Bay of Fundy Tide

A tide table gives the depth of water DD, in metres, at a wharf on the Bay of Fundy, tt hours after midnight.

t (h)t\text{ (h)} 0033669912121515
D (m)D\text{ (m)} 8814148822881414
  1. a)
    Determine an equation of the form D(t)=asin⁡(k(t−d))+cD(t)=a\sin(k(t-d))+c that models the depth.
  2. b)
    A fishing boat needs at least 5 m of water to dock. Between midnight and noon, during what times can it not dock?
  3. c)
    The wharf floods when the depth reaches 15 m. Use your model to find when that first happens.
Stuck? Peek at a hint

Read the maximum and minimum depths first: the amplitude is half their difference and the midline is their average.

A2 Fall Fair Ferris Wheel

A Ferris wheel turns once every 8 minutes at a constant speed. A rider boards at the bottom, and their height, in metres, after tt minutes is h(t)=22−20cos⁡(π4t)h(t)=22-20\cos\left(\frac{\pi}{4}t\right).

  1. a)
    Determine the average rate of change of height from t=0t=0 to t=1t=1, and from t=1t=1 to t=2t=2. Round to two decimals.
  2. b)
    Estimate the instantaneous rate of change of height at t=2t=2, using the interval 2≤t≤2.0012\le t\le 2.001.
  3. c)
    Estimate the instantaneous rate of change at t=4t=4, using 3.999≤t≤4.0013.999\le t\le 4.001. A rider concludes that the wheel stopped at t=4t=4. Is that right?
Stuck? Peek at a hint

Average rate of change is h(b)−h(a)b−a\dfrac{h(b)-h(a)}{b-a}. Put your calculator in radian mode first.

A3 Georgian Bay Lighthouse

A lighthouse on an island in Georgian Bay is 500 m from a straight shoreline. Let θ\theta be the angle between the beam and the line from the lighthouse straight to the shore. When the beam hits the shore, its length is L(θ)=500sec⁡θL(\theta)=500\sec\theta metres.

  1. a)
    Evaluate L(π3)L\left(\frac{\pi}{3}\right) and L(π4)L\left(\frac{\pi}{4}\right). Give the second to the nearest tenth of a metre.
  2. b)
    The beam turns once every 20 s, so θ=π10t\theta=\frac{\pi}{10}t after tt seconds. For 0≤t<50\le t<5, when is the beam 1000 m long?
  3. c)
    Evaluate LL at t=5t=5 s. What happens to the lit spot on the shore at that moment, and for 5<t<155<t<15?
Stuck? Peek at a hint

Write sec⁡θ=1cos⁡θ\sec\theta=\dfrac{1}{\cos\theta} and use exact values.

Think It Through (T)

Error Analysis · Building Functions

T1 Spot the Flaw

A student solves 2sin⁡xcos⁡x=cos⁡x2\sin x\cos x=\cos x for 0≤x≤2π0\le x\le 2\pi:
"Divide both sides by cos⁡x\cos x: 2sin⁡x=12\sin x=1, so sin⁡x=12\sin x=\frac{1}{2}. The solutions are x=π6x=\frac{\pi}{6} and x=5π6x=\frac{5\pi}{6}."

  1. a)
    Show that x=π2x=\frac{\pi}{2} satisfies the equation.
  2. b)
    Identify the student's error, then solve the equation correctly.
  3. c)
    Without solving again, how many solutions does the equation have for 0≤x≤4π0\le x\le 4\pi? Explain.
Stuck? Peek at a hint

Move every term to one side and look for a common factor.

T2 Build It Backwards

Write an equation for each function. Many answers are possible.

  1. a)
    A sine function with maximum value 77, minimum value −1-1, period π\pi, and a maximum at x=π3x=\frac{\pi}{3}.
  2. b)
    The same graph as (a), written as a cosine function.
  3. c)
    A trig function with period π2\frac{\pi}{2}, vertical asymptotes exactly at x=nπ2x=\frac{n\pi}{2} (n∈Zn\in\mathbb{Z}), whose graph passes through (π8,1)\left(\frac{\pi}{8},1\right).
Stuck? Peek at a hint

Start with the amplitude and midline from the max and min. A sine graph reaches its maximum a quarter period after it crosses the midline going up.

Boss Level

Optional challenge

Squeezed Solutions

Let kk be a positive integer. Consider the equation sin⁡(kx)=12\sin(kx)=\frac{1}{2} for 0≤x≤2π0\le x\le 2\pi.

  1. a)
    For k=3k=3, list every solution.
  2. b)
    For k=3k=3, find the sum of all the solutions.
  3. c)
    For a general positive integer kk, find the number of solutions and their sum, in terms of kk.
Stuck? Peek at a hint

Let u=kxu=kx. If 0≤x≤2π0\le x\le 2\pi, what interval does uu run over?

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Answer key

Final answers only. Try first, then check.

Core Skills

K1
  • a) 7π6\dfrac{7\pi}{6}; 315∘315^\circ
  • b) −12-\dfrac{1}{2}; 22\dfrac{\sqrt{2}}{2}; −3-\sqrt{3}
  • c) Principal angle π4\dfrac{\pi}{4}; e.g. −7π4-\dfrac{7\pi}{4}
K2
  • a) Period π\pi; D={x∈R∣x≠π2+nπ, n∈Z}D=\left\{x\in\mathbb{R}\mid x\ne\frac{\pi}{2}+n\pi,\ n\in\mathbb{Z}\right\}; asymptotes x=π2+nπx=\frac{\pi}{2}+n\pi, n∈Zn\in\mathbb{Z}
  • b) Zeros π2\frac{\pi}{2}, 3π2\frac{3\pi}{2}; maxima (0,1)(0,1) and (2π,1)(2\pi,1); minimum (π,−1)(\pi,-1)
  • c) d=π2d=\dfrac{\pi}{2}
K3
  • a) f(x)=−3sin⁡(2(x−π6))+1f(x)=-3\sin\left(2\left(x-\frac{\pi}{6}\right)\right)+1; amplitude 33, period π\pi, phase shift π6\frac{\pi}{6} right, midline y=1y=1
  • b) R=[−2,4]R=[-2,4]; (11π12,4)\left(\frac{11\pi}{12},4\right)
  • c) y=2cos⁡(π3(x−1))−1y=2\cos\left(\frac{\pi}{3}(x-1)\right)-1
K4
  • a) −2-\sqrt{2}; −2-2; 3\sqrt{3}
  • b) D={x∈R∣x≠π2+nπ, n∈Z}D=\left\{x\in\mathbb{R}\mid x\ne\frac{\pi}{2}+n\pi,\ n\in\mathbb{Z}\right\}; R=(−∞,−1]∪[1,∞)R=(-\infty,-1]\cup[1,\infty); period 2π2\pi
  • c) Period π\pi; asymptotes x=nπ2x=\frac{n\pi}{2}, n∈Zn\in\mathbb{Z}
K5
  • a) x=4π3, 5π3x=\dfrac{4\pi}{3},\ \dfrac{5\pi}{3}
  • b) x=0, 2π3, 4π3, 2πx=0,\ \dfrac{2\pi}{3},\ \dfrac{4\pi}{3},\ 2\pi
  • c) x=7π6+2nπx=\dfrac{7\pi}{6}+2n\pi or x=11π6+2nπx=\dfrac{11\pi}{6}+2n\pi, n∈Zn\in\mathbb{Z}

Real World

A1
  • a) D(t)=6sin⁡(π6t)+8D(t)=6\sin\left(\frac{\pi}{6}t\right)+8
  • b) From 7:00 a.m. to 11:00 a.m. (D(t)<5D(t)<5 for 7<t<117<t<11)
  • c) Never: D(t)=15D(t)=15 needs sin⁡(π6t)=76>1\sin\left(\frac{\pi}{6}t\right)=\frac{7}{6}>1. The model's maximum depth is 14 m, so the wharf does not flood.
A2
  • a) 5.865.86 m/min; 14.1414.14 m/min
  • b) About 15.7115.71 m/min
  • c) 00 m/min. No: the wheel keeps turning at the same speed. At the top (42 m) the rider moves horizontally, so only their height stops changing for an instant.
A3
  • a) 10001000 m; 5002≈707.1500\sqrt{2}\approx 707.1 m
  • b) t=103≈3.33t=\frac{10}{3}\approx 3.33 s
  • c) LL is undefined (sec⁡π2\sec\frac{\pi}{2} is a vertical asymptote): the beam is parallel to the shore and never reaches it. For 5<t<155<t<15 the beam points away from shore (L<0L<0 has no meaning), so there is no lit spot.

Think It Through

T1
  • a) cos⁡π2=0\cos\frac{\pi}{2}=0, so both sides equal 00.
  • b) Dividing by cos⁡x\cos x assumes cos⁡x≠0\cos x\ne0. Factor: cos⁡x(2sin⁡x−1)=0\cos x(2\sin x-1)=0, so x=π6, π2, 5π6, 3π2x=\frac{\pi}{6},\ \frac{\pi}{2},\ \frac{5\pi}{6},\ \frac{3\pi}{2}.
  • c) 88: both factors repeat every 2π2\pi, so each of the 4 solutions has one copy in (2π,4π](2\pi,4\pi], and 00, 2π2\pi, 4π4\pi are not solutions.
T2
  • a) e.g. y=4sin⁡(2(x−π12))+3y=4\sin\left(2\left(x-\frac{\pi}{12}\right)\right)+3
  • b) e.g. y=4cos⁡(2(x−π3))+3y=4\cos\left(2\left(x-\frac{\pi}{3}\right)\right)+3
  • c) e.g. y=cot⁡(2x)y=\cot(2x)

Boss Level

Boss
  • a) x=π18, 5π18, 13π18, 17π18, 25π18, 29π18x=\frac{\pi}{18},\ \frac{5\pi}{18},\ \frac{13\pi}{18},\ \frac{17\pi}{18},\ \frac{25\pi}{18},\ \frac{29\pi}{18}
  • b) 5π5\pi
  • c) 2k2k solutions; sum (2k−1)π(2k-1)\pi

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Free cheat sheet

Trigonometric Functions Cheat Sheet

The whole unit on one page: every rule, a worked example for each idea, and the traps that cost marks.

Unit Summary

One printable page with every rule.

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Most testedSolve 2sin⁡2x−sin⁡x−1=02\sin^2x-\sin x-1=0 for 0≤x≤2π0\le x\le2\pi. (card 6)

1Radians and exact values

π rad=180∘s=rθ\pi\text{ rad}=180^\circ\qquad s=r\theta (θ\theta in radians)

UseFind the related acute angle, then take the sign from CAST. cos⁡(−x)=cos⁡x\cos(-x)=\cos x, sin⁡(−x)=−sin⁡x\sin(-x)=-\sin x. Coterminal: θ±2π\theta\pm2\pi.

Example5π/45\pi/4 is in QIII with related angle π/4\pi/4: sin⁡(5π/4)=−2/2\sin(5\pi/4)=-\sqrt2/2.

Trapcos⁡(−π/4)=+2/2\cos(-\pi/4)=+\sqrt2/2: a negative angle does not make every ratio negative.

2Graphs of sin⁡\sin, cos⁡\cos, tan⁡\tan

Usesin⁡\sin, cos⁡\cos: period 2π2\pi, range [−1,1][-1,1]. tan⁡\tan: period π\pi, zeros at nπn\pi, asymptotes where cos⁡x=0\cos x=0 (x=π/2+nπx=\pi/2+n\pi). Cofunction: sin⁡(π/2−x)=cos⁡x\sin(\pi/2-x)=\cos x.

Traptan⁡x\tan x has period π\pi, and its asymptotes are where cos⁡x=0\cos x=0, not at its zeros.

3Transformations

y=asin⁡[k(x−d)]+cT=2π∣k∣y=a\sin[k(x-d)]+c\qquad T=\dfrac{2\pi}{|k|}

UseAmplitude ∣a∣|a|, midline y=cy=c. Factor kk out first. For a>0a>0, sine reaches its maximum a quarter period after dd.

Trapsin⁡(2x−π/3)=sin⁡[2(x−π/6)]\sin(2x-\pi/3)=\sin[2(x-\pi/6)]: the shift is π/6\pi/6, not π/3\pi/3.

4Modelling

Usea=(max⁡−min⁡)/2a=(\max-\min)/2, c=(max⁡+min⁡)/2c=(\max+\min)/2, k=2π/Tk=2\pi/T. With a maximum at t=dt=d, use cosine.

ExampleHigh tide 55 m at 22 h, low 11 m, every 1212 h: h=2cos⁡[π/6 (t−2)]+3h=2\cos[\pi/6\,(t-2)]+3.

TrapHigh to low tide is half a period: 66 h means T=12T=12 h.

5Reciprocal functions

csc⁡x=1sin⁡xsec⁡x=1cos⁡x\csc x=\dfrac{1}{\sin x}\quad\sec x=\dfrac{1}{\cos x}

cot⁡x=cos⁡xsin⁡x\cot x=\dfrac{\cos x}{\sin x}

UseAsymptotes where the denominator is 00. csc⁡\csc, sec⁡\sec: period 2π2\pi; cot⁡\cot: period π\pi. csc⁡(kx)\csc(kx) has asymptotes at x=nπ/kx=n\pi/k.

Trapcsc⁡x\csc x is 1/sin⁡x1/\sin x, not sin⁡−1x\sin^{-1}x.

6Solving equations

UseRelated angle, then every quadrant CAST allows; general solution +2πn+2\pi n (+πn+\pi n for tan⁡\tan). Factor quadratic forms; never divide by cos⁡x\cos x. Convert csc⁡\csc, sec⁡\sec, cot⁡\cot first.

Example(2sin⁡x+1)(sin⁡x−1)=0(2\sin x+1)(\sin x-1)=0: x=π/2, 7π/6, 11π/6x=\pi/2,\ 7\pi/6,\ 11\pi/6.

Trapsin⁡3x=1/2\sin3x=1/2 on [0,2π][0,2\pi]: 3x3x runs over [0,6π][0,6\pi], so there are 66 solutions, not 22.

7Rates of change

average rate =f(b)−f(a)b−a=\dfrac{f(b)-f(a)}{b-a}

UseInstantaneous: shrink the interval. sin⁡x\sin x changes fastest at its zeros and has rate 00 at a max or min.

Examplesin⁡x\sin x from 00 to π/2\pi/2: (1−0)/(π/2)=2/π≈0.64(1-0)/(\pi/2)=2/\pi\approx0.64.

TrapA rate of 00 at the top of a Ferris wheel means height isn't changing at that instant, not that the wheel stopped.

Which period?

sin⁡\sin, cos⁡\cos, csc⁡\csc, sec⁡\sec 2π/∣k∣2\pi/|k|
tan⁡\tan, cot⁡\cot π/∣k∣\pi/|k|
Data or context max to next max

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