MCR3U · Unit 4 test review

Trigonometric Functions: Test Review

Everything for the Unit 4 test on one page: the key facts from all 10 lessons, the mistakes that cost the most marks, and a practice test that points you back to the lessons you need.

Key facts

Right Triangle Trigonometry

$sin θ = \frac{opposite}{hypotenuse}$ , $cos θ = \frac{adjacent}{hypotenuse}$ , $tan θ = \frac{opposite}{adjacent}$ .
Reciprocal ratios: $csc θ = \frac{1}{sin θ} = \frac{hyp}{opp}$ , $sec θ = \frac{1}{cos θ} = \frac{hyp}{adj}$ , $cot θ = \frac{1}{tan θ} = \frac{adj}{opp}$ .
To find a missing SIDE, set up the correct ratio and isolate the unknown.
To find a missing ANGLE, use the inverse function: $θ = sin^{- 1} (\dots)$ , $cos^{- 1} (\dots)$ , or $tan^{- 1} (\dots)$ .
Angle of elevation (looking up) = angle of depression (looking back down) between the same two points, since horizontal lines are parallel.
Two-triangle problems: solve for the shared side first, then add or subtract, rounding only at the end.

Special Triangles and Exact Values

45-45-90 triangle side ratio: $1 : 1 : 2$ .
30-60-90 triangle side ratio: $1 : 3 : 2$ (shortest side opposite $30°$ , longest leg opposite $60°$ , hypotenuse opposite $90°$ ).
$sin 30° = cos 60° = \frac{1}{2}$ ; $sin 60° = cos 30° = \frac{3}{2}$ ; $sin 45° = cos 45° = \frac{2}{2}$ .
$tan 30° = \frac{3}{3}$ , $tan 45° = 1$ , $tan 60° = 3$ ; $tan 90°$ is undefined.
On the unit circle, the point at angle $θ$ has coordinates $(cos θ, sin θ)$ .
$sin^{2} θ$ means $(sin θ)^{2}$ ; reciprocal exact values come from flipping, e.g. $csc 30° = 2$ , $sec 45° = 2$ .

Angles from 0° to 360° and the CAST Rule

Reference angle: the acute angle between the terminal arm and the $x$ -axis; its trig ratios share the SAME magnitude as the original angle's.
For a point $P (x, y)$ on the terminal arm: $r = x^{2} + y^{2}$ , $sin θ = \frac{y}{r}$ , $cos θ = \frac{x}{r}$ , $tan θ = \frac{y}{x}$ .
CAST rule: QI all positive, QII sine positive, QIII tangent positive, QIV cosine positive (everything else in that quadrant is negative).
A negative angle $- θ$ is coterminal with $360° - θ$ , and has identical trig ratios.
Reciprocal ratios: $csc θ = \frac{1}{sin θ}$ , $sec θ = \frac{1}{cos θ}$ , $cot θ = \frac{1}{tan θ}$ — same sign as the original ratio.
Solving $sin θ$ , $cos θ$ or $tan θ = k$ (with $0 < ∣ k ∣ < 1$ for sine or cosine, or any $k \neq = 0$ for tangent) on $0° \leq θ < 360°$ gives TWO angles: same reference angle, two quadrants chosen by CAST.

The Sine Law

Sine law: $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ , or flipped for angles: $\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}$ .
Each side is paired with the angle directly opposite it.
Use the sine law when you know an angle and its opposite side (ASA, AAS, or SSA).
Angles in any triangle always sum to $180°$ .
It comes from writing the same altitude two ways: $h = b sin A = a sin B$ .
3D problems: split into flat triangles that share a side; solve the one with enough information first.

The Ambiguous Case of the Sine Law (SSA)

SSA (two sides and a non-included angle) can produce 0, 1, or 2 triangles — it's the only ambiguous case.
The cause: $sin θ = sin (180° - θ)$ , so one sine value fits an acute AND an obtuse angle.
Test height: $h = b sin A$ (with $A$ acute, $b$ adjacent to $A$ , $a$ opposite $A$ ).
$a < h$ : no triangle. $a = h$ : one (right) triangle. $h < a < b$ : two triangles. $a \geq b$ : one triangle.
Given angle $90°$ or more: one triangle if $a > b$ , none if $a \leq b$ .
In the two-triangle case, the second angle is $B_{2} = 180° - B_{1}$ ; check it still leaves $A + B_{2} < 180°$ .

The Cosine Law

Cosine law: $c^{2} = a^{2} + b^{2} - 2 ab cos C$ , where $C$ is included between sides $a$ and $b$ .
Use it for SAS (find the missing side) or SSS (find any angle).
Rearranged to solve for an angle: $cos C = \frac{a ^{2} + b ^{2} - c ^{2}}{2 ab}$ .
A negative value of $cos C$ means $C$ is obtuse — this is expected, not an error.
It comes from Pythagoras plus $sin^{2} C + cos^{2} C = 1$ ; at $C = 90°$ it reduces to $c^{2} = a^{2} + b^{2}$ .
Solving a whole triangle: cosine law first (largest angle in SSS), then the sine law for the smaller angles, then the angle sum.

Trigonometric Identities

Pythagorean identity: $sin^{2} θ + cos^{2} θ = 1$ for every angle $θ$ .
Quotient identity: $tan θ = \frac{sin θ}{cos θ}$ .
Rearranged Pythagorean forms: $sin^{2} θ = 1 - cos^{2} θ$ and $cos^{2} θ = 1 - sin^{2} θ$ .
To prove an identity, simplify LS and RS independently until they match — never operate across the equals sign.
One angle where LS and RS differ disproves a statement; checking a few angles never proves one.
Useful moves: write $tan$ as $\frac{s i n}{c o s}$ , combine fractions over a common denominator, factor, and replace $sin^{2} θ + cos^{2} θ$ with $1$ .

Periodic Functions: Period, Amplitude, and Axis

Period: the horizontal distance for one complete repeating cycle.
Axis (midline): $\frac{max + min}{2}$ — the horizontal centre line of the graph.
Amplitude: $\frac{max - min}{2}$ — always reported as a positive distance.
A function is periodic if its values repeat in a fixed cycle forever.
Because $f (x + p) = f (x)$ , subtracting whole periods predicts future values: $f (x + n p) = f (x)$ .
Checks: $max = axis + amplitude$ and $min = axis - amplitude$ .

Graphs of y = sin x and y = cos x — Transformations

Parent curves: $y = sin x$ and $y = cos x$ both have amplitude $1$ , period $360°$ , and axis $y = 0$ .
In $y = a sin (k (x - d)) + c$ : $∣ a ∣$ = amplitude, $\frac{360°}{∣ k ∣}$ = period, $d$ = phase shift (right if positive), $c$ = axis.
A negative $a$ reflects the graph vertically (upside down) but leaves the amplitude, period, and axis unchanged.
$sin x$ starts at the axis and rises; $cos x$ starts at its maximum — they're the same shape, just $90°$ apart.
Order: stretches and reflections first, then translations. Mapping rule: $(x, y) \to (\frac{x}{k} + d, a y + c)$ .
Factor $k$ out before reading $d$ : $sin (2 x - 60°) = sin (2 (x - 30°))$ has $d = 30°$ .

Modelling with Sinusoidal Functions

Axis: $c = \frac{max + min}{2}$ . Amplitude: $a = \frac{max - min}{2}$ . Both always come first.
$k = \frac{360°}{period}$ .
If a scenario starts at a maximum, model with $+ cos$ ; if it starts at a minimum, model with $- cos$ ; if it starts at the axis rising, use $+ sin$ .
Once $a, k, c$ are set from the max/min/period, only the starting behaviour decides which parent function and sign to use — often avoiding the need for a $d$ term entirely.
In context: $c$ is the average level (a Ferris wheel's centre height), $∣ a ∣$ is half the range (the radius), the period is the time for one cycle, and $d$ is the time of the chosen starting feature.
To find WHEN a model reaches a value: set it equal, isolate the trig ratio, use the related angle and CAST (or read the graph), then add whole periods for later cycles. Each value between max and min occurs twice per cycle.

Where marks get lost

Leaving the calculator in RADIAN mode — MCR3U works entirely in degrees, so always confirm the calculator says DEG before evaluating any trig function. (Right Triangle Trigonometry)
Mislabelling opposite vs. adjacent relative to the wrong angle (they swap depending on which angle you're using). (Right Triangle Trigonometry)
Forgetting to apply the INVERSE trig function when solving for an angle — writing $θ = sin (0.7)$ instead of $θ = sin^{- 1} (0.7)$ . (Right Triangle Trigonometry)
Mixing up angle of elevation and angle of depression — both are measured from the HORIZONTAL, never from the vertical. (Right Triangle Trigonometry)
Rounding each distance in a two-triangle problem before subtracting — for example using $84.6$ and $147.2$ gives $62.6$ by luck here, but in general early rounding shifts the final difference. Keep full values until the last step. (Right Triangle Trigonometry)
Mixing up which special value goes with $30°$ vs $60°$ — remember sine INCREASES from $30°$ to $60°$ while cosine DECREASES. (Special Triangles and Exact Values)
Forgetting that $tan 90°$ is undefined (division by $cos 90° = 0$ ), not simply a very large number. (Special Triangles and Exact Values)
Rationalizing incorrectly — leaving $tan 30°$ as $\frac{1}{3}$ is fine, but if asked to rationalize, multiply top and bottom by $3$ to get $\frac{3}{3}$ . (Special Triangles and Exact Values)
Assuming the side labelled ' $1$ ' in the 30-60-90 ratio is always the hypotenuse — it's actually the SHORTEST side, opposite the $30°$ angle. (Special Triangles and Exact Values)
Treating $sin^{2} 30°$ as $sin (30 °^{2})$ or as $2 sin 30°$ — it means $(sin 30°)^{2} = (\frac{1}{2})^{2} = \frac{1}{4}$ . (Special Triangles and Exact Values)
Forgetting to check the SIGN after finding the reference angle — the reference angle is always positive, but the final answer might need a minus sign. (Angles from 0° to 360° and the CAST Rule)
Using the wrong reference-angle formula for the quadrant (e.g. using $180° - θ$ in Quadrant III instead of $θ - 180°$ ). (Angles from 0° to 360° and the CAST Rule)
Thinking a negative angle gives a negative trig ratio automatically — the SIGN still depends on which quadrant the negative angle lands in, not on the angle being negative. (Angles from 0° to 360° and the CAST Rule)
Confusing $csc θ$ (reciprocal of sine) with $sin^{- 1} θ$ (inverse sine, a completely different operation) — $csc θ = \frac{1}{sin θ}$ , not the angle whose sine is $θ$ . (Angles from 0° to 360° and the CAST Rule)
Stopping at the calculator's single answer when solving — $cos^{- 1} (- 0.4) \approx 113.6°$ is only one solution; cosine is also negative in QIII, giving $360° - 113.6° = 246.4°$ . (Angles from 0° to 360° and the CAST Rule)
Pairing a side with the WRONG angle — always match each side to the angle directly opposite it, never an adjacent one. (The Sine Law)
Trying to apply the sine law to an SSS or SAS triangle where no full angle/side pair is known yet. (The Sine Law)
Forgetting that a triangle's angles must sum to $180°$ when a third angle is needed. (The Sine Law)
Rounding intermediate angle or side values too early, which can shift later answers noticeably. (The Sine Law)
In 3D problems, mixing a horizontal angle and an angle of elevation in the same triangle — each flat triangle only uses angles that lie in its own plane. (The Sine Law)
Skipping the $h = b sin A$ test and assuming there's always exactly one triangle, like every other case. (The Ambiguous Case of the Sine Law (SSA))
Forgetting to check whether $A + B_{2} < 180°$ before accepting the second (obtuse) triangle — sometimes only one of the two candidate triangles is actually valid. (The Ambiguous Case of the Sine Law (SSA))
Applying the ambiguous-case test when the given angle is obtuse — the simpler $a > b$ rule applies instead, not the height comparison. (The Ambiguous Case of the Sine Law (SSA))
Mixing up which side is $a$ (opposite the given angle) and which is $b$ (adjacent to it) when computing $h = b sin A$ . (The Ambiguous Case of the Sine Law (SSA))
Finding $B_{2} = 180° - B_{1}$ but reusing triangle 1's $C$ and $c$ — each triangle needs its own third angle and its own third side. (The Ambiguous Case of the Sine Law (SSA))
Using the cosine law with an angle that ISN'T included between the two known sides — for SAS, the angle must sit between the two given sides. (The Cosine Law)
Forgetting to take the square root at the end when solving for a side (leaving the answer as $c^{2} = \dots$ instead of $c = \dots$ ). (The Cosine Law)
Panicking when $cos C$ comes out negative — that's a valid, expected result for an obtuse angle, not a mistake. (The Cosine Law)
Mixing up which side is squared alone vs. which two are multiplied together in the rearranged angle formula. (The Cosine Law)
Evaluating $a^{2} + b^{2} - 2 ab cos C$ in the wrong order: computing $(a^{2} + b^{2} - 2 ab) cos C$ or subtracting before multiplying. Work out $2 ab cos C$ as one number first, then subtract it. (The Cosine Law)
Working across the equals sign while 'proving' an identity (adding the same thing to both sides) — this can make false statements look proven. (Trigonometric Identities)
Forgetting the $\pm$ when taking a square root of $sin^{2} θ$ or $cos^{2} θ$ , and then forgetting to pick the correct sign using the quadrant. (Trigonometric Identities)
Cancelling a $sin θ$ or $cos θ$ term across a sum instead of a product (you can only cancel matching FACTORS, not terms added together). (Trigonometric Identities)
Trying to prove an identity by plugging in just one specific angle — that only checks one case, not all angles; algebraic simplification is required. (Trigonometric Identities)
Misreading the notation: $sin^{2} θ$ means $(sin θ)^{2}$ , not $sin (θ^{2})$ , and $\frac{s i n θ}{θ}$ does not simplify to $sin$ , because sine is a function, not a number multiplied by $θ$ . (Trigonometric Identities)
Reporting the amplitude as the FULL vertical distance (max minus min) instead of HALF of it. (Periodic Functions: Period, Amplitude, and Axis)
Confusing the axis (midline) with the minimum value — the axis is the average, not the bottom of the graph. (Periodic Functions: Period, Amplitude, and Axis)
Reading the period as the distance to the halfway point of a cycle instead of a FULL repeat. (Periodic Functions: Period, Amplitude, and Axis)
Assuming a negative amplitude is possible — amplitude is a distance, so it's always non-negative. (Periodic Functions: Period, Amplitude, and Axis)
Measuring the period between two points at the same HEIGHT where one is rising and the other falling; the matching point must be doing the same thing, so measure peak to peak. (Periodic Functions: Period, Amplitude, and Axis)
Reading the phase shift's sign backwards — $(x - d)$ shifts RIGHT for positive $d$ , so $y = sin (x + 30°)$ actually means $d = - 30°$ (a shift LEFT). (Graphs of y = sin x and y = cos x — Transformations)
Forgetting to take $∣ k ∣$ (absolute value) when computing the period — a negative $k$ still gives a positive period. (Graphs of y = sin x and y = cos x — Transformations)
Confusing $c$ (vertical shift / new axis) with $d$ (horizontal shift) — one moves the graph up-down, the other left-right. (Graphs of y = sin x and y = cos x — Transformations)
Assuming a negative amplitude means the curve goes lower — really $∣ a ∣$ is still the amplitude, and the negative sign only flips the graph vertically. (Graphs of y = sin x and y = cos x — Transformations)
Reading $d$ before factoring out $k$ : in $y = sin (3 x - 90°)$ the shift is $30°$ right, not $90°$ , because $3 x - 90° = 3 (x - 30°)$ . (Graphs of y = sin x and y = cos x — Transformations)
Jumping straight to a phase-shifted sine formula instead of first checking whether the scenario naturally starts at a max or min — using $\pm cos$ is almost always simpler than forcing $sin$ with an awkward $d$ . (Modelling with Sinusoidal Functions)
Computing the axis or amplitude from the WRONG pair of numbers (e.g. using the radius as the axis instead of the centre height, in a Ferris wheel problem). (Modelling with Sinusoidal Functions)
Forgetting to convert the given cycle length into $k = \frac{360°}{period}$ , and instead leaving the raw period number inside the trig function. (Modelling with Sinusoidal Functions)
Mixing units (e.g. writing $t$ in minutes in one part of the formula and seconds in another) — keep the time unit for $t$ consistent throughout. (Modelling with Sinusoidal Functions)
Stopping at one time when asked when the model reaches a value: on a Ferris wheel with $h (t) = - 10 cos (9 t) + 12$ , the rider is at $17$ m both rising ( $t \approx 13.3$ s) and falling ( $t \approx 26.7$ s), and again every $40$ s after that. (Modelling with Sinusoidal Functions)

Practice test

20 questions from across the unit. Each one shows the lesson it comes from.

Right Triangle Trigonometry
1
A right triangle has hypotenuse $12$ and an angle of $25°$ . Find the length of the side opposite that angle (round to 2 decimals).
Right Triangle Trigonometry
2
Put the steps for finding a building's height from an angle of elevation in order.
1. 1Evaluate in degree mode and round at the end
2. 2Write the equation and isolate the unknown
3. 3Choose the ratio linking the known side to the unknown side
4. 4Sketch the right triangle and mark the given angle
5. 5Label opposite, adjacent and hypotenuse from that angle
Special Triangles and Exact Values
3
State the exact value of $sin 45°$ .
Special Triangles and Exact Values
4
Match each expression to its exact value.
1. A. $\frac{2}{2}$
2. B. $\frac{1}{2}$
3. C. $3$
4. D. $0$
5. E. $2$
$sin 30°$
$cos 45°$
$tan 60°$
$csc 30°$
$cos 90°$
Angles from 0° to 360° and the CAST Rule
5
Find the reference angle for $θ = 205°$ .
°
Angles from 0° to 360° and the CAST Rule
6
Put the steps for solving $tan θ = - 1.2$ on $0° \leq θ < 360°$ in order, taking the quadrants counterclockwise.
1. 1Check both answers on the calculator
2. 2QIV: $θ = 360° - 50.2° = 309.8°$
3. 3QII: $θ = 180° - 50.2° = 129.8°$
4. 4Find the reference angle $tan^{- 1} (1.2) \approx 50.2°$
5. 5Use CAST to place $50.2°$ : tangent is negative in QII and QIV
The Sine Law
7
In $△ A BC$ , $A = 50°, B = 60°, a = 10$ . Find $b$ (round to 2 decimals).
The Sine Law
8
Put the steps for a 3D tower problem (two observers on the ground, one angle of elevation) in order.
1. 1Check that the height is reasonable
2. 2Use tangent in the vertical triangle to find the height
3. 3Use the sine law for the ground distance to the tower's base
4. 4Sketch the ground triangle and the vertical triangle separately
5. 5Find the third angle of the ground triangle
The Ambiguous Case of the Sine Law (SSA)
9
In $△ A BC$ , $A = 40°, a = 12, b = 8$ . How many triangles are possible?
The Ambiguous Case of the Sine Law (SSA)
10
Put the steps for handling an SSA problem in order.
1. 1If two triangles exist, set $B_{2} = 180° - B_{1}$ and solve each triangle
2. 2Find $B_{1}$ with the inverse sine
3. 3Compute $h = b sin A$ and compare it with $a$ and $b$
4. 4Check that the angle is not between the two given sides
5. 5Decide whether the given angle is acute or obtuse
The Cosine Law
11
In $△ A BC$ , $a = 6, b = 8$ , included angle $C = 60°$ . Find $c$ (round to 2 decimals).
The Cosine Law
12
Put the steps for solving an SSS triangle in the safest order.
1. 1Check that larger angles sit opposite longer sides
2. 2Find the third angle from the $180°$ angle sum
3. 3Use the sine law for a second angle
4. 4Identify the longest side
5. 5Use the cosine law for the angle opposite it
Trigonometric Identities
13
State the Pythagorean identity.
Trigonometric Identities
14
Put the lines of the proof of $cos^{4} θ - sin^{4} θ = cos^{2} θ - sin^{2} θ$ in order.
1. 1LS $=$ RS, so the identity is true
2. 2LS $= cos^{2} θ - sin^{2} θ$
3. 3LS $= (cos^{2} θ - sin^{2} θ) (1)$
4. 4LS $= cos^{4} θ - sin^{4} θ$
5. 5LS $= (cos^{2} θ - sin^{2} θ) \cdot (cos^{2} θ + sin^{2} θ)$
Periodic Functions: Period, Amplitude, and Axis
15
A graph has maximum $50$ and minimum $10$ . Find the amplitude.
Periodic Functions: Period, Amplitude, and Axis
16
Put the steps for predicting a far-future value from a table of periodic data in order.
1. 1State the prediction with its units
2. 2Read the value at the reduced time from the table
3. 3Subtract whole periods from the future time
4. 4Find two successive maximums in the table
5. 5Subtract their times to get the period
Graphs of y = sin x and y = cos x — Transformations
17
Find the period of $y = sin (4 x)$ .
°
Graphs of y = sin x and y = cos x — Transformations
18
Put the steps for graphing $y = 2 sin (3 (x - 20°)) + 1$ in order.
1. 1Repeat the cycle every $120°$ in both directions
2. 2Plot the five new points and join them with a smooth wave
3. 3Add $d = 20°$ to each $x$ and $c = 1$ to each $y$
4. 4List the five key points of $y = sin x$
5. 5Divide each $x$ by $k = 3$ and multiply each $y$ by $a = 2$
Modelling with Sinusoidal Functions
19
A Ferris wheel has radius 8 m, centre height 10 m, period 30 s, and a rider starts at the bottom. Find the amplitude $∣ a ∣$ .
Modelling with Sinusoidal Functions
20
Put the steps of this lesson's translation checklist for modelling a situation with a sinusoid in order.
1. 1Check the model at a known point
2. 2Choose $sin$ , $cos$ or $- cos$ and $d$ from the starting behaviour
3. 3Find the period and compute $k = \frac{360°}{period}$
4. 4Identify the maximum and minimum values
5. 5Compute the axis $c$ and amplitude $∣ a ∣$

Endless drill

Fresh numbers every time, for the calculation questions in this unit.

Endless drillnew numbers every time

Generating a problem…

Trigonometric Functions: Test Review

Key facts

Where marks get lost

Practice test

Endless drill

Keep going