A running start for a college trigonometry course: the four algebra skills the course leans on hardest — complex fractions, rationalising, the disguised-quadratic pattern and graph transformations — then angle measure in degrees and radians, arc length and sector area, both special triangles rebuilt from scratch, and the unit circle with reference angles and quadrant signs.
Subject: Trigonometry · 72 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Title
Running start before September 8
The algebra the course assumes, and the circle everything is built on
Objectives
Your course starts on September 8. Everything before then is free practice, and the best use of it is not to learn trigonometry early — it is to make the algebra underneath trigonometry automatic, so that when the new ideas arrive you have attention left over for them.
OpenStax, Algebra and Trigonometry 2e chapters 5 and 7
Section
Orientation
Concept
Trigonometry is taught as one course, but it is really two ideas stacked. Students who struggle almost always struggle at the join.
The second idea contains the first. Once you see the ratio as a coordinate, every fact from the triangle chapter is still true and a great deal more becomes available.
OpenStax, Algebra and Trigonometry 2e sections 5.2 and 7.1
Prediction
Picture your syllabus for a second before you answer. The topic list is long, but the time is not evenly spread.
Predict first
Which of these will consume the largest share of a first trigonometry course?
Correct: The unit circle and everything that follows from it
Why: The right-triangle chapter is usually two or three weeks. The unit circle then feeds graphing, identities, inverse functions and equation solving — four or five more chapters that all fail together if the circle is shaky. That is why this deck spends its second half there.
Intuition
Here is the observation the whole subject rests on. Take a right triangle. Scale it up. Every side gets longer, but the ratio of any two sides does not move at all.
Figure (svg): Two nested right triangles sharing the same acute angle, with side lengths marked to show the ratio is unchanged
So the ratio is not a fact about a particular triangle. It is a fact about the angle. That is what makes it worth giving a name and a table of values.
OpenStax, Algebra and Trigonometry 2e section 5.2
Socratic
This is the sentence that makes the rest of the course make sense. Put it in your own words before you read mine.
Discussion prompt
Why can sine be a function of the angle alone, when the triangle you measured it in could have been any size?
Hint: What happens to opposite over hypotenuse if you double both?
Answer:
Because scaling a triangle multiplies every side by the same factor, so the factor cancels in any ratio of two sides.
Similar triangles have equal corresponding ratios. Fix the angle and you have fixed the shape; the size is free.
Concept
Having taught this course a number of times, the failures are remarkably consistent, and none of them are about trigonometry.
Notice that all three are Algebra 2 topics. That is deliberate. The next section is the whole of the algebra you need, and nothing else.
Section
Prerequisites
Concept
Trigonometry produces complex fractions constantly, because tangent is a ratio of two things that are already ratios. You need one reliable move, not four situational ones.
\[ \dfrac{\dfrac{a}{b}}{\dfrac{c}{d}} = \dfrac{a}{b} \cdot \dfrac{d}{c} = \dfrac{ad}{bc} \]
Dividing by a fraction is multiplying by its reciprocal. That single sentence handles every complex fraction you will meet this term.
OpenStax, Algebra and Trigonometry 2e section 1.6
Worked example
This is exactly the arithmetic that appears when you compute tangent of sixty degrees from its sine and cosine.
\[ \tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} \]
Turn the division into a multiplication by the reciprocal
Why: The bottom fraction is one half, and its reciprocal is two.
\[ = \dfrac{\sqrt{3}}{2} \cdot \dfrac{2}{1} \]
Cancel the twos
Why: The two in the denominator and the two in the numerator are the same number, so they divide out.
\[ = \sqrt{3} \]
Verify: the size is believable
Why: Sixty degrees is a steep angle, so the opposite side should be considerably longer than the adjacent one. Root three is about 1.73, which is comfortably bigger than one, so the answer passes the smell test.
OpenStax, Algebra and Trigonometry 2e section 5.2
Picture it
You do not need to memorise the exact values. You need to be able to rebuild them in ten seconds from two pictures. Here is the first.
Figure (svg): A 45-45-90 right triangle with both legs labelled one and the hypotenuse labelled root two
Two equal angles force two equal legs. Set both to one, and the Pythagorean theorem hands you a hypotenuse of root two. Sine and cosine of forty-five are then both one over root two, which we usually write as root two over two.
Concept
Your calculator does not care whether a radical sits on top or on the bottom. Your grader does, and more importantly, matching your answer against the back of the book requires the same convention.
\[ \dfrac{1}{\sqrt{3}} = \dfrac{1}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{\sqrt{3}}{3} \]
You are multiplying by one, so the value does not change. All that changes is where the radical lives.
OpenStax, Algebra and Trigonometry 2e section 1.3
Fill the middle
Same move, different number. Fill in the two blanks.
Fill in the blanks
\dfracroot two2} = \dfrac______} \cdot \dfrac___}___} = \dfrac___}___}
Why: Multiply top and bottom by root two. The bottom becomes root two times root two, which is exactly two, and the top becomes root two. So one over root two equals root two over two — which is why the forty-five degree values are written the way they are.
Pattern
This is the single most useful idea in the prerequisite half of this deck. Almost every trigonometric equation you will be asked to solve is an equation you already know how to solve, with a trig function standing where a variable used to stand.
| what you see | what it really is | the move |
|---|---|---|
| 2 sin x - 1 = 0 | a linear equation in a variable | isolate sin x, then ask which angles |
| 2 cos squared x - cos x - 1 = 0 | a quadratic in a variable | let u be cos x, factor, then unwind |
| sin x cos x = 0 | a product equal to zero | zero product property on each factor |
| tan squared x = 3 | a square equal to a number | take both roots, then unwind |
Substituting a single letter for the trig function is not a trick. It is a way of proving to yourself that you already have the method.
OpenStax, Algebra and Trigonometry 2e section 7.5
Worked example
Solve for the values of cosine, and stop there. Finding the angles themselves is the next chapter; recognising the structure is this one.
\[ 2\cos^2 x - \cos x - 1 = 0 \]
Substitute a single letter
Why: Let u stand for cosine of x. Nothing about the equation changes except how intimidating it looks.
\[ 2u^2 - u - 1 = 0 \]
Figure (svg): The trigonometric equation on the left and the same equation written in u on the right, joined by an arrow labelled let u be cosine of x
Factor the quadratic
Why: Two numbers multiplying to negative two and adding to negative one give the split, and the grouping produces two binomials.
\[ (2u + 1)(u - 1) = 0 \]
Apply the zero product property
Why: A product is zero exactly when one of its factors is zero, so each factor gets its own equation.
\[ u = -\tfrac{1}{2} \quad \text{or} \quad u = 1 \]
Unwind the substitution
Why: Put cosine of x back where u was standing.
\[ \cos x = -\tfrac{1}{2} \quad \text{or} \quad \cos x = 1 \]
Verify: both values are legal
Why: Cosine is a coordinate on a circle of radius one, so it can never leave the interval from negative one to one. Both negative one half and one are inside that interval, so neither solution is impossible on its face.
OpenStax, Algebra and Trigonometry 2e section 7.5
Check
You do not have to solve it. Just name what kind of equation it is once the substitution is made.
Check your understanding
After letting u stand for sine of x, what kind of equation is 3 sin squared x + 5 sin x = 2?
Answer: B
Why: Both trig terms are powers of the same function, sine of x, so a single substitution captures the whole equation. The highest power is two, which makes it a quadratic in u once you move the two across to get 3u squared plus 5u minus 2 equal to zero.
Discrimination
The substitution trick has one requirement: every trig term must be the same function of the same angle. Sort these by whether a single substitution works as written.
Sort into buckets
Does one substitution handle it as it stands?
Concept
Every trigonometric graph you draw this term is the parent sine or cosine curve with four numbers attached. Learning what each number does now means the graphing chapter is bookkeeping rather than discovery.
\[ y = A\sin\big(B(x - C)\big) + D \]
| number | what it controls | what to watch for |
|---|---|---|
| A | amplitude, the height above the midline | a negative A also flips the curve upside down |
| B | how many cycles fit in a full turn | the period is two pi divided by B, not two pi times B |
| C | the horizontal shift | it only reads correctly when B has been factored out first |
| D | the midline, the vertical shift | the curve oscillates about D, not about zero |
OpenStax, Algebra and Trigonometry 2e section 6.1
Picture it
Here is the parent sine curve, dashed, against a version with an amplitude of two, twice the frequency, and a lift of one.
Figure (svg): A dashed parent sine curve plotted against a transformed sine curve with amplitude two, twice the frequency and a vertical shift of one
Read the picture from the outside in. Find the midline first, then the height above it, then count the cycles. That ordering makes the four numbers fall out of the graph almost without algebra.
Error analysis
A very common piece of wrong work. The student read the horizontal shift straight out of the expression without factoring first.
Annotate
On: \( y = \sin(2x - \pi) \;\Rightarrow\; \text{shift right by } \pi \)
Rule of thumb: never read C off an unfactored argument. Factor B out first, every time, even when it feels unnecessary.
Two truths and a lie
Two of these statements are correct. One is a sentence students say all the time and it is wrong.
Eliminate the wrong options
Cross out the false statement.
Survives elimination: c3
Why: Amplitude is measured from the midline, not from zero. Adding D lifts the whole curve, midline included, so the distance from midline to peak is untouched. The maximum output does rise, which is why the sentence sounds plausible, but the amplitude is exactly the same number it was before.
Ranking
You are graphing y equals negative three sine of the quantity two x minus pi, plus one. Put the steps in the order that produces a correct picture.
Put in order
Why: Factoring has to come first or every later number is read wrong. The period follows immediately from B. Then build the frame — midline, then amplitude — before placing anything, shift the frame horizontally, and apply the reflection last so it acts on a curve that is already in the right place.
Explain it to yourself
You will need this one under exam pressure, so bank the reason rather than the rule.
Discussion prompt
Why does the horizontal shift have to be read after factoring B out, and not before?
Hint: Ask what value of x makes the argument equal to zero.
Answer:
Because the shift is defined as the value of x that makes the inside of the function zero, and B scales the x axis before the shift is applied.
Concretely: sine of the quantity two x minus pi is zero when x is pi over two, not when x is pi. Factoring makes that visible: two times the quantity x minus pi over two.
The published shift is always the real shift divided by B when you forget to factor, so the error is systematic rather than random.
Section
Angle measure
Concept
Degrees are a historical accident. Three hundred and sixty is a convenient number with many divisors, and it has nothing to do with circles. Radians are not an alternative unit so much as the natural one.
radian — the angle you get when the arc you have walked around the circle is exactly as long as the radius
Figure (svg): A circle with a radius drawn, and an arc of the same length as the radius highlighted, subtending an angle of one radian
Because the definition compares an arc to a radius, a radian is a ratio of two lengths and therefore has no units at all. That is precisely why the calculus formulas are clean in radians and ugly in degrees.
OpenStax, Algebra and Trigonometry 2e section 7.1
Estimation
Number sense first. If you cannot estimate, an arithmetic slip in a conversion will slide past you unnoticed.
Predict first
Roughly how many degrees is one radian?
Correct: About 57 degrees
Why: A full turn is two pi radians and also 360 degrees, so one radian is 360 divided by two pi, which is about 57.3 degrees. The useful mental anchor is that a radian is a bit less than sixty degrees, so it sits just under the angle in an equilateral triangle.
Concept
Do not learn one rule for each direction. Learn the single fact that both directions come from, and let the units cancel.
\[ 180^\circ = \pi \text{ radians} \]
Everything else is multiplying by a fraction equal to one. To go to radians, multiply by pi over 180. To go to degrees, multiply by 180 over pi. If you forget which way round it is, write the units and see which arrangement cancels the one you started with.
OpenStax, Algebra and Trigonometry 2e section 7.1
Worked example
A degree measure that is not one of the memorised five, done by the general method.
\[ 150^\circ \cdot \dfrac{\pi}{180^\circ} \]
Cancel the degree units
Why: Degrees appear on the top of the first factor and the bottom of the second, so they divide out and leave a pure number times pi.
\[ = \dfrac{150\pi}{180} \]
Reduce the fraction
Why: Thirty divides both 150 and 180, giving five over six.
\[ = \dfrac{5\pi}{6} \]
Verify: the size is right
Why: 150 degrees is between 90 and 180, so the answer should be between pi over two and pi. Five sixths of pi is between one half and one whole pi, so it lands where it should.
OpenStax, Algebra and Trigonometry 2e section 7.1
Picture it
Two formulas that come free once the angle is in radians — and that are simply wrong if it is not.
Figure (svg): A circle with a shaded sector, its central angle theta, radius r and arc marked
Both formulas are just proportions in disguise. The angle theta is the fraction of a full turn, measured in radius-lengths, so multiplying by the radius gives the arc and by half the radius squared gives the area.
Worked example
A central angle of five pi over six on a circle of radius twelve centimetres. Find the arc length.
Figure (svg): A circle with a shaded sector, its central angle theta, radius r and arc marked
\[ s = r\theta = 12 \cdot \dfrac{5\pi}{6} \]
Cancel the six into the twelve
Why: Twelve divided by six is two, which leaves two times five pi.
\[ s = 10\pi \text{ cm} \]
Give a decimal for a sanity check
Why: Ten pi is about 31.4 centimetres.
Verify: against the whole circumference
Why: The full circumference is two pi times twelve, which is about 75.4 centimetres. Our angle is 150 degrees, a little under half a turn, and 31.4 is a little under half of 75.4. The answer is consistent.
OpenStax, Algebra and Trigonometry 2e section 7.1
Check
Solve it on paper before you click.
Check your understanding
You are given a central angle in degrees and asked for the arc length. What must happen first?
Answer: B
Why: The formula says arc length equals radius times theta precisely because a radian is defined as the angle whose arc equals the radius. Feed it degrees and the proportionality constant is wrong by a factor of pi over 180, so the answer comes out about 57 times too large.
Concept
Once you place the angle on a coordinate grid, the restriction to acute angles disappears and the subject opens up.
This is the moment the triangle idea gives way to the circle idea. An angle of 750 degrees makes no sense inside a triangle and perfect sense as two full turns plus thirty degrees.
OpenStax, Algebra and Trigonometry 2e section 7.1
Socratic
No formula needed. Think about what the terminal side is doing.
Discussion prompt
Why must sine of 750 degrees equal sine of 30 degrees exactly, rather than approximately?
Hint: How many full turns is 720 degrees?
Answer:
Because 750 is 30 plus two full turns, and a full turn returns the terminal side to precisely where it started.
Sine is defined by the position of the terminal side, and the position is identical, so the value is identical. There is nothing approximate about it.
This is also the reason sine is periodic with period 360 degrees, which is the same statement written as a graph fact rather than a circle fact.
Picture it
This is the labour-saving device of the whole unit-circle chapter.
Figure (svg): A unit circle with a terminal side at 150 degrees, its reference angle of 30 degrees back to the negative x axis marked
Any angle at all has an acute reference angle back to the horizontal axis. The size of the trig value comes from the reference angle; the sign comes from the quadrant. Memorise the first quadrant once and you have memorised all four.
Matching
Reference angle means the acute angle between the terminal side and the x axis — never the y axis.
Match the pairs
Why: In the second quadrant subtract from 180, in the third subtract 180, in the fourth subtract from 360. So 150 and 210 both reduce to 30, 135 and 225 both reduce to 45, and 300 reduces to 60. Two different angles sharing a reference angle differ only in sign.
Trap
The terminal side is at 150 degrees, sitting in the second quadrant.
Measure the angle up to the y axis
Why: That gap is 60 degrees, so the student writes down a reference angle of 60.
Report sine of 150 as root three over two
Why: Which is the value for 60, and is wrong — the correct value is one half.
The terminal side is at 150 degrees, sitting in the second quadrant.
Measure the angle back to the nearest part of the x axis
Why: The negative x axis is at 180 degrees, so the gap is 30 degrees. Reference angles are always measured to the horizontal.
Take the size from 30 and the sign from the quadrant
Why: Sine of 30 is one half, and sine is positive in the second quadrant, so sine of 150 is one half.
Section
Triangle trigonometry
Concept
Everyone learns the mnemonic. Fewer people notice that it encodes a hierarchy: two of the three ratios are basic, and the third is built from them.
\[ \sin\theta = \dfrac{\text{opp}}{\text{hyp}}, \quad \cos\theta = \dfrac{\text{adj}}{\text{hyp}}, \quad \tan\theta = \dfrac{\text{opp}}{\text{adj}} \]
Divide the first by the second and the hypotenuse cancels, leaving opposite over adjacent. So tangent is not a third independent fact; it is sine over cosine, and that relationship survives into the circle chapter unchanged.
OpenStax, Algebra and Trigonometry 2e section 5.2
Picture it
The second of the two pictures worth rebuilding rather than memorising.
Figure (svg): A 30-60-90 triangle shown as half of an equilateral triangle of side two, with legs one and root three and hypotenuse two
Start with an equilateral triangle of side two and drop a perpendicular. The base is bisected, so the short leg is one; the hypotenuse is still two; the Pythagorean theorem gives the long leg as root three. Every 30 and 60 degree value in the course comes off this picture.
Fill the middle
Read them off the 30-60-90 triangle above rather than from memory.
Fill in the blanks
\sin 30^\circ = one half, \quad \cos 30^\circ = root three over two, \quad \tan 30^\circ = root three over three
Why: Opposite the 30 degree angle is the short leg of length one, over a hypotenuse of two, giving one half. Adjacent is root three over two. Tangent is the ratio of those, which is one over root three, rationalised to root three over three.
Elimination
Before computing anything, some answers can be ruled out on sight. That skill is worth more marks than it looks.
Eliminate the wrong options
One of these could be the sine of some angle. Rule out the other three.
Survives elimination: e2
Why: The range of sine is the closed interval from negative one to one, because sine is a coordinate on the unit circle and no point on that circle is further than one unit from the centre. Negative 0.8 is comfortably inside; the other three all exceed one in size. Checking the range first catches a whole class of arithmetic errors for free.
Worked example
A ladder leans against a wall at 65 degrees to the ground, and its foot is 2.4 metres from the wall. How long is the ladder?
Figure (svg): A ladder leaning against a wall at sixty-five degrees, with the 2.4 metre base labelled adjacent and the ladder labelled hypotenuse
Identify which sides you have and want
Why: The 2.4 metres is adjacent to the 65 degree angle, and the ladder is the hypotenuse. Adjacent and hypotenuse together means cosine.
\[ \cos 65^\circ = \dfrac{2.4}{L} \]
Solve for the length
Why: Multiply both sides by L, then divide both sides by cosine of 65.
\[ L = \dfrac{2.4}{\cos 65^\circ} \]
Evaluate
Why: Cosine of 65 degrees is about 0.4226, and 2.4 divided by that is about 5.68.
\[ L \approx 5.7 \text{ m} \]
Verify: the hypotenuse is the longest side
Why: 5.7 metres is larger than the 2.4 metre base, as any hypotenuse must be. Had the answer come out smaller than 2.4, the ratio was set up upside down.
OpenStax, Algebra and Trigonometry 2e section 5.4
Step zero
Most right-triangle mistakes happen before any arithmetic, in the choice of ratio.
Discussion prompt
You are given an acute angle and the side opposite it, and asked for the hypotenuse. What is step zero?
Hint: The pair of sides involved determines the function, not the other way round.
Answer:
Label the three sides relative to the given angle: opposite, adjacent, hypotenuse. Do it on the diagram, in writing.
Then read off which two you are dealing with — here, opposite and hypotenuse — and that pair names the ratio. Opposite over hypotenuse is sine.
Only then write the equation. Choosing the ratio from the labels rather than from memory is what stops the classic sine-cosine swap.
Real world
A practical warning that costs students real marks every single term.
Discussion prompt
Your calculator returns 0.9939 when you ask for the sine of 80. What has gone wrong, and how would you notice?
Hint: Test the calculator against a value you already know by heart.
Answer:
The calculator is in radian mode. Sine of 80 degrees is about 0.9848; sine of 80 radians is about 0.9939. The two are uncomfortably close and neither is obviously absurd.
That is exactly the danger. Degree and radian answers are often both plausible-looking numbers, so a mode error does not announce itself.
The habit that saves you: check the mode at the start of every session, and sanity check one known value — sine of 30 should return exactly 0.5 in degree mode. If it does not, fix the mode before doing anything else.
Section
The centre of the course
Concept
Draw a circle of radius one centred at the origin. Take any angle in standard position and follow the terminal side out to where it meets the circle. That meeting point has coordinates, and those coordinates are the definition.
\[ (x, y) = (\cos\theta, \; \sin\theta) \]
Figure (svg): A unit circle with the first-quadrant special angles marked, each with its radian measure and its coordinate pair
Cosine is the first coordinate and sine is the second. Every fact in the rest of the course is a consequence of that one sentence.
OpenStax, Algebra and Trigonometry 2e section 7.3
Prediction
The triangle definition simply stops working past a right angle. The circle definition does not.
Predict first
On the unit circle, what is the cosine of 180 degrees?
Correct: Negative one
Why: At 180 degrees the terminal side runs along the negative x axis and meets the circle at the point with coordinates negative one and zero. Cosine is the first coordinate, so it is negative one, and sine is the second, so it is zero. Nothing about the definition needed a triangle.
Picture it
You do not memorise a table of signs. You read them off the coordinate axes.
Figure (svg): A unit circle divided into four quadrants, each labelled with which of sine, cosine and tangent stay positive there
Cosine follows x, so it is positive on the right. Sine follows y, so it is positive on the top. Tangent is their quotient, so it is positive where the two signs agree — the first and third quadrants.
Definition probe
Do this by asking where the terminal side lands, not by computing anything.
Sort into buckets
Is the cosine of this angle positive or negative?
Concept
Every point on the unit circle satisfies the circle's own equation. That is not a new fact you have to learn; it is the definition of the circle.
\[ x^2 + y^2 = 1 \]
Now substitute the definitions of cosine and sine for x and y and you have the most-used identity in the subject.
\[ \cos^2\theta + \sin^2\theta = 1 \]
It is worth pausing on how little work that took. The identity is the equation of the circle, rewritten in trigonometric names. Nothing was proved; something was renamed.
OpenStax, Algebra and Trigonometry 2e section 7.3
Socratic
Your course will hand you three Pythagorean identities and it will feel like three things to remember. It is one.
Discussion prompt
Starting from cosine squared plus sine squared equals one, how do you get an identity involving tangent?
Hint: What do you have to divide by to turn sine into tangent?
Answer:
Divide every term by cosine squared.
The first term becomes one, the second becomes tangent squared, and the right side becomes one over cosine squared, which is secant squared.
So one plus tangent squared equals secant squared. Dividing instead by sine squared gives the third identity, involving cotangent and cosecant. Three identities, one source, two divisions.
Worked example
Cosine of an angle is negative three fifths, and the angle lies in the third quadrant. Find its sine.
Write the Pythagorean identity and substitute
Why: The identity holds for every angle, so it holds for this one.
\[ \left(-\tfrac{3}{5}\right)^2 + \sin^2\theta = 1 \]
Simplify the square
Why: A negative squared is positive, and three fifths squared is nine twenty-fifths.
\[ \tfrac{9}{25} + \sin^2\theta = 1 \]
Isolate the sine squared
Why: Subtract nine twenty-fifths from one, written as twenty-five twenty-fifths.
\[ \sin^2\theta = \tfrac{16}{25} \]
Take the root, then choose the sign from the quadrant
Why: The algebra allows plus or minus four fifths. The third quadrant is below the axis, so the y coordinate is negative, and sine is the y coordinate.
\[ \sin\theta = -\tfrac{4}{5} \]
Verify: the pair sits on the circle
Why: Nine twenty-fifths plus sixteen twenty-fifths is twenty-five twenty-fifths, which is one. The point with coordinates negative three fifths and negative four fifths really does lie on the unit circle, and it lies in the third quadrant as required.
OpenStax, Algebra and Trigonometry 2e section 7.3
Check
Solve it on paper before you click.
Check your understanding
Sine of an angle is 0.6 and the angle is in the second quadrant. What is its cosine?
Answer: B
Why: The identity gives cosine squared equal to one minus 0.36, which is 0.64, so cosine is plus or minus 0.8. In the second quadrant the x coordinate is negative, and cosine is the x coordinate, so the negative root is the right one.
Picture it
Here is the bridge between the circle chapter and the graphing chapter, in one picture.
Figure (svg): A unit circle on the left with a point at sixty degrees and its height marked, next to a sine curve on the right with the same height plotted
Walk around the circle and plot the height against the angle. That is the sine curve. Periodicity, the range from negative one to one, and the zeros at multiples of pi are all visible on the circle before you ever draw the graph.
Comparison
Both definitions are correct. They differ in what they can reach.
Comparison matrix
| question | triangle definition | circle definition |
|---|---|---|
| which angles are allowed | strictly between 0 and 90 degrees | any angle at all, including negative ones |
| what sine means | opposite over hypotenuse | the y coordinate of the point on the circle |
| can it explain periodicity | no, there is no notion of going round again | yes, a full turn returns you to the same point |
| where the identity comes from | the Pythagorean theorem on the sides | the equation of the circle itself |
| what it is good for | surveying, navigation, anything with a physical triangle | graphing, modelling waves, and all of calculus |
The circle definition agrees with the triangle definition everywhere the triangle definition applies. It is a strict extension, which is why the course switches over and never switches back.
Edge cases
Definitions earn their keep at the edges. Push this one.
Discussion prompt
Tangent is sine over cosine. What happens at exactly 90 degrees, and what does the graph do there?
Hint: What is the x coordinate of the point at the top of the circle?
Answer:
At 90 degrees the point on the circle is at coordinates zero and one, so cosine is zero and sine is one. Tangent would be one divided by zero, which is undefined.
Approaching 90 degrees from below, cosine shrinks towards zero while sine stays near one, so the quotient grows without bound. Approaching from above, cosine is a small negative number, so the quotient plunges.
On the graph that shows up as a vertical asymptote at 90 degrees, and at every odd multiple of it. The undefined point is not a flaw in tangent; it is a fact about where the terminal side is vertical.
Analogy
Everything from the triangle chapter survives. It just gets a new name.
Match the pairs
Why: Drop a perpendicular from the point on the circle to the x axis and you have literally drawn the old right triangle, with hypotenuse one. The opposite side is the height, which is y; the adjacent side is the horizontal run, which is x. The triangle never left — it is inscribed in the circle.
Trap
The student memorises all sixteen labelled points on the unit circle the night before the test.
Recall the coordinate at 210 degrees
Why: Under pressure, the root three over two and the one half swap places, because nothing anchors which is which.
Report cosine of 210 as negative one half
Why: The correct value is negative root three over two. The error is invisible because both numbers appear on the circle and both are negative in the third quadrant.
The student memorises the two special triangles and the quadrant sign rule, and rebuilds everything else.
Find the reference angle for 210 degrees
Why: Subtract 180, giving a reference angle of 30 degrees.
Read the size off the 30-60-90 triangle
Why: At 30 degrees the adjacent side is the long leg, so cosine of 30 is root three over two. Sine of 30 is one half.
Attach the sign from the quadrant
Why: The third quadrant has a negative x, so cosine of 210 is negative root three over two. There is nothing left to confuse, because the size came from a picture rather than from recall.
Warm-up
Cover the circle figure. This is retrieval practice, and it is worth about three times what re-reading is worth.
Discussion prompt
Write down, from the two special triangles and the sign rule alone: cosine of 135, sine of 240, tangent of 45, and sine of 270.
Hint: Reference angle for the number, quadrant for the sign.
Answer:
Cosine of 135: reference angle 45, second quadrant so x is negative, giving negative root two over two.
Sine of 240: reference angle 60, third quadrant so y is negative, giving negative root three over two.
Tangent of 45: the 45-45-90 legs are equal, so opposite over adjacent is exactly one, and the first quadrant is positive.
Sine of 270: this is a quadrantal angle, straight down, at coordinates zero and negative one, so sine is negative one.
Section
Consolidation
Missing information
Exam questions occasionally under-specify on purpose, to see whether you notice.
Discussion prompt
Sine of an angle is one half. Find the angle. What is missing?
Hint: How many points on the circle have a height of one half?
Answer:
The interval. Without one, there are infinitely many answers: 30 degrees, 150 degrees, and either of those plus any whole number of full turns.
A well-posed question either restricts the angle to one turn, or asks for a general solution written with an added multiple of 360 degrees.
Noticing this is worth marks in itself. On a question that does supply an interval, the first thing to do is write the interval down and count how many answers it should contain.
Constraint
Same question, now properly posed, and with a restriction that changes the method.
Discussion prompt
Find every angle between 0 and 360 degrees whose sine is negative root two over two, without using a calculator.
Hint: Get the reference angle from the size, then let the sign choose the quadrants.
Answer:
The size root two over two is the 45 degree value, so the reference angle is 45.
Sine is negative in the third and fourth quadrants, so those are the two places to look.
Third quadrant: 180 plus 45 is 225. Fourth quadrant: 360 minus 45 is 315. Those are the two answers, and there are exactly two because a horizontal line cuts the circle in at most two places.
Trade off
You will be asked to work in both. Knowing why each exists stops the conversion from feeling arbitrary.
Comparison matrix
| situation | use degrees | use radians |
|---|---|---|
| a surveying or navigation problem | yes, the field convention is degrees | only if the answer is fed into further mathematics |
| arc length or sector area | no, the formula is simply wrong | yes, both formulas assume radians |
| graphing a trig function | possible but the axis scale is awkward | yes, the period comes out as a clean multiple of pi |
| anything in calculus | no, the derivative formulas gain stray constants | yes, this is the reason radians exist |
The one-line version: degrees for talking to people, radians for talking to mathematics.
Pattern
Every exact-value question this term yields to the same four questions, asked in the same order. Write them on the inside cover of your notebook.
Four questions, in that order, and there is no exact-value problem in a first trigonometry course that survives them.
OpenStax, Algebra and Trigonometry 2e section 7.3
Worked example
Find the exact value of cosine of negative seven pi over three, with no calculator.
Place the angle
Why: It is negative, so it turns clockwise, and it is bigger than a full turn in size. Add two pi, written as six pi over three, to find a coterminal angle.
\[ -\dfrac{7\pi}{3} + \dfrac{6\pi}{3} = -\dfrac{\pi}{3} \]
Reduce once more into the standard range
Why: Adding another full turn, six pi over three, gives five pi over three, which is between zero and two pi.
\[ -\dfrac{\pi}{3} + \dfrac{6\pi}{3} = \dfrac{5\pi}{3} \]
Name the quadrant
Why: Five pi over three is between three pi over two and two pi, which is the fourth quadrant. There x is positive, so the cosine will be positive.
Find the reference angle
Why: Two pi minus five pi over three is pi over three, that is sixty degrees.
Read the size off the 30-60-90 triangle
Why: Cosine of sixty degrees is the short leg over the hypotenuse, which is one half.
\[ \cos\left(-\dfrac{7\pi}{3}\right) = \dfrac{1}{2} \]
Verify: with the even property of cosine
Why: Cosine is an even function, so cosine of negative seven pi over three equals cosine of seven pi over three. Seven pi over three is one full turn plus pi over three, so it is cosine of pi over three, which is one half. Two independent routes, one answer.
OpenStax, Algebra and Trigonometry 2e section 7.3
Picture it
If you keep one image from today, keep this one, and rebuild the rest from it.
Figure (svg): The unit circle with the first-quadrant special angles, their radian measures and their exact coordinate pairs
First quadrant only. Everything else is this picture plus a reference angle and a sign.
Check
Solve it on paper before you click.
Check your understanding
What is the exact value of sine of 225 degrees?
Answer: B
Why: 225 degrees is 180 plus 45, so the terminal side is in the third quadrant and the reference angle is 45. The 45-45-90 triangle gives a size of root two over two, and the third quadrant has a negative y coordinate, so sine is negative.
Explain it
The genuine test of whether today landed. Say this out loud, to a person or to an empty room.
Discussion prompt
In under a minute, explain to someone who has never met trigonometry why sine and cosine can take an angle bigger than ninety degrees.
Hint: Start with the picture, not the formula.
Answer:
Because the modern definition never mentions a triangle. You put the angle at the centre of a circle of radius one and look at where the terminal side crosses the circle.
That point has an x and a y no matter how far round you have gone, so the definition keeps working past ninety degrees, past a full turn, and in the negative direction.
The triangle version is the special case where the point happens to sit in the first quadrant. It is not wrong; it is just short.
Connect it up
No notes. Twenty minutes on this is worth more than an hour of re-reading.
Draw it
Draw a unit circle. Mark the first-quadrant special angles in both degrees and radians. Beside it, draw the two special triangles with every side labelled. Then draw one arrow from each triangle to the circle points it supplies, and write the quadrant sign rule underneath.
Anything you had to look up is your homework between now and September 8.
Exit ticket
Be honest. This decides where the next session opens.
Predict first
Which piece of today would you least want to be tested on tomorrow?
Correct: Whichever one you picked is where session two starts.
Why: There is no wrong answer here. Naming the weak piece is worth more than another pass over the strong ones, and there are almost two weeks before the course starts — enough time to fix one thing thoroughly rather than five things badly.
Recap
Two weeks before the course begins, this is the checklist. If you can do all of it cold on September 8, the first month of the course is revision.
| quantity | the fact to hold | where it came from |
|---|---|---|
| degrees to radians | 180 degrees is pi radians | definition of a radian |
| arc length | radius times the angle in radians | a radian is one radius of arc |
| sector area | one half radius squared times the angle | proportion of the full disc |
| 45 degree values | root two over two, both of them | the equal-leg triangle |
| 30 and 60 degree values | one half and root three over two | half an equilateral triangle |
| Pythagorean identity | cosine squared plus sine squared is one | the equation of the unit circle |
OpenStax, Algebra and Trigonometry 2e — every result above appears there with worked examples and practice sets
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