Radian measure and why it is the right unit, the six trigonometric functions read off the unit circle, the major angles, the graphs with their periods and asymptotes, amplitude and period of a sinusoid, the Pythagorean and addition identities, and solving trigonometric equations for every solution rather than one.
Subject: Calculus I · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Calculus I · Chapter 1 — Functions and Graphs
Trigonometric Functions
Objectives
Five outcomes. The first looks like a unit conversion and is actually the reason Chapter 3's trigonometric derivatives are as simple as they are.
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 53-64 — the section these objectives are drawn from
Warm-up
Section 1.2 called sine transcendental: no finite formula of arithmetic and roots produces it. That leaves the question of how it is defined at all.
Discussion prompt
You know sine and cosine from right triangles, where they are ratios of sides. But a right triangle cannot have an angle of 200 degrees. How can sine be defined for every real input?
Hint: Think about a point travelling round a circle rather than a triangle with three fixed corners.
Answer:
Put a point on the circle of radius 1 centred at the origin and let it travel counterclockwise. Its coordinates are the cosine and the sine of the angle turned through.
\[ (x, y) = (\cos t, \sin t) \quad \text{on the circle } x^2 + y^2 = 1 \]
Nothing stops the point from going round more than once, or backwards, so the definition works for every real number. The triangle definition is the special case where the angle is between 0 and a quarter turn — and the circle definition agrees with it there.
Concept
Measure an angle by the length of arc it cuts on a circle of radius one. That measure is called the radian, and because it is a length divided by a length it is a pure number. The cosine and sine of that angle are then simply the coordinates of the point the arc ends at.
radian — The measure of a central angle whose subtended arc has the same length as the radius. Since it is an arc length divided by a radius, a radian is a ratio of two lengths and therefore carries no units.
\[ s = r\theta, \qquad 180^\circ = \pi \text{ rad} \]
Being dimensionless is the whole point. Section 3.5 will show that the derivative of sine is cosine only when the input is in radians; in degrees an awkward factor of pi over 180 appears in every derivative and never goes away. Radians are not a convention, they are the measure the calculus works in.
Figure (svg): A circle with an arc equal in length to its radius, subtending an angle of one radian
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 53-55
Section
Section 1
Concept
A circle of radius r has circumference two pi r, so going all the way round covers an arc of two pi radii. That is two pi radians, and it is the same 360 degrees, which fixes the conversion once and for all.
arc length — On a circle of radius r, an angle of theta radians at the centre cuts an arc of length r times theta. This formula is true only in radians, which is one of the simplest signs that radians are the natural measure.
\[ \frac{\theta_{\text{deg}}}{180} = \frac{\theta_{\text{rad}}}{\pi} \]
In practice you rarely need the formula. Remember that pi radians is half a turn, and every common angle follows: a quarter turn is pi over 2, a sixth of a turn is pi over 3, a twelfth is pi over 6.
Figure (svg): A circle with an arc equal in length to its radius, subtending an angle of one radian
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 53-56 — degrees and radians
Picture it
The definition, drawn.
Figure (svg): A circle with an arc equal in length to its radius, subtending an angle of one radian
A radian is a little under 60 degrees, so roughly six of them go round a circle — and the exact number is two pi, about 6.28. That mental picture is a useful check on any conversion.
Worked example
Example 1.21. One proportion serves in both directions.
\[ \text{Express } 225^\circ \text{ in radians, and } \tfrac{5\pi}{3} \text{ radians in degrees.} \]
Multiply degrees by pi over 180
Why: This is the proportion rearranged.
\[ 225 \cdot \frac{\pi}{180} \]
Reduce the fraction
Why: Both 225 and 180 divide by 45.
\[ 5 \pi / 4\text{ radians} \]
For the other direction, multiply radians by 180 over pi
Why: The pi cancels.
\[ (5 \pi / 3) (\frac{180}{\pi}) \]
Simplify
Why: Five times 180 over 3.
\[ 300 ^\circ \]
Figure (svg): The solution to Worked example converting both ways shown as a ladder of expressions, one row per legal move
\[ 225^\circ = \tfrac{5\pi}{4} \text{ rad}, \qquad \tfrac{5\pi}{3} \text{ rad} = 300^\circ \]
Verify: check both against a half turn
Why: Half a turn is 180 degrees, which is pi radians. So 5 pi over 4 should be a bit more than half a turn, and 225 degrees is indeed 45 degrees past 180. Likewise 5 pi over 3 is just under 2 pi, and 300 degrees is indeed 60 short of a full 360. Sanity-checking against the half turn catches an inverted conversion factor instantly, which is the only real error available here.
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 55-56
Matching
Half a turn is pi. Everything else follows.
Match the pairs
Why: Each is the fraction of 180 degrees, written as that fraction of pi. Thirty is a sixth of 180, so pi over 6; 120 is two thirds of 180, so 2 pi over 3. Reading them as fractions of a half turn is faster and far less error-prone than multiplying by pi over 180 each time.
Worked example
Example 1.22. The arc length formula, in a setting where the units matter.
\[ \text{A wheel of radius } 2 \text{ ft turns through } 3\pi \text{ rad. How far does a point on its rim travel?} \]
Use the arc length formula
Why: Arc equals radius times angle, valid only in radians.
\[ s = r \theta \]
Substitute
Why: Two feet times 3 pi.
\[ s = 2(3 \pi) \]
Simplify
Why: Six pi feet.
\[ s = 6 \pi \text{ft} \]
Interpret the angle
Why: Three pi radians is one and a half full turns.
\[ \text{about } 18.85 \text{ft} \]
Figure (svg): The solution to Worked example arc length and angular speed shown as a ladder of expressions, one row per legal move
\[ s = r\theta = 6\pi \text{ ft} \approx 18.85 \text{ ft} \]
Verify: compute it as a fraction of the circumference
Why: The circumference is 2 pi times 2, which is 4 pi feet. One and a half turns is 1.5 times 4 pi, which is 6 pi feet — the same answer by a completely different route. Notice also that the radian, being dimensionless, contributed no unit: feet times radians came out as feet, which is exactly what an arc length should be. Had the angle been in degrees the formula would have given 6 times 2, or 12, which is wrong.
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 56-57
Trap
\[ r = 2 \text{ ft}, \quad \theta = 540^\circ \]
Substitute the degree measure directly
Why: The student uses s equals r theta without converting.
\[ s = 2(540) = 1080 \text{ ft} \quad \text{(wrong)} \]
A wheel of radius 2 feet has a circumference of about 12.6 feet, so one and a half turns cannot be a thousand feet.
\[ 540^\circ = 3\pi \text{ rad} \;\Longrightarrow\; s = 2(3\pi) = 6\pi \approx 18.85 \text{ ft} \]
Convert to radians before any formula that mixes angle with length
Why: The formula s equals r theta is a statement about arc length, and it is only true when the angle is measured by arc length.
This is the first of many places where degrees quietly produce a wrong number rather than an error message. The habit worth forming now is to convert at the door: as soon as an angle enters a calculation, put it in radians and leave it there.
Fill the middle
The angle from the first worked example.
Fill in the blanks
225^\circ \cdot \frac5___ = \frac___\pi}___
Why: Both 225 and 180 are divisible by 45, leaving 5 over 4. So 225 degrees is 5 pi over 4 radians — a quarter turn past a half turn, which the picture confirms.
Prediction
Commit before reasoning.
Predict first
Why does calculus insist on radians rather than degrees?
Correct: Because the arc-length relation holds only in radians, and that is what makes the derivative of sine exactly cosine.
\[ \lim_{t \to 0} \frac{\sin t}{t} = 1 \quad \text{(radians)} \qquad \text{but} \qquad \frac{\pi}{180} \quad \text{(degrees)} \]
Why: In Section 3.5 the derivative of sine is computed from a limit that evaluates to 1 precisely because the angle is measured by arc length. In degrees that limit is pi over 180 instead, so every trigonometric derivative would carry that factor, and every second derivative would carry its square. Degrees can certainly describe angles over 360, and the notation is no harder — the reason is entirely about what the calculus comes out to.
Ranking
Mixed units, smallest first.
Put in order
Why: Converting everything to degrees: pi over 6 is 30, then 45, then 1.2 radians is about 68.8, then 120, then 5 pi over 4 is 225. The one worth pausing on is 1.2 radians — since one radian is about 57.3 degrees, a bare number of radians is always somewhat bigger than it looks.
Section
Section 2
Concept
For a real number t, travel t units of arc counterclockwise from the point one comma zero on the unit circle. The coordinates of where you land are defined to be cosine of t and sine of t. The other four functions are ratios of these two.
the six trigonometric functions — Cosine and sine are the coordinates of the point at arc length t on the unit circle. Tangent is sine over cosine, cotangent its reciprocal, secant is one over cosine and cosecant is one over sine.
\[ \tan t = \frac{\sin t}{\cos t}, \quad \sec t = \frac{1}{\cos t}, \quad \csc t = \frac{1}{\sin t}, \quad \cot t = \frac{\cos t}{\sin t} \]
Because the point is on a circle of radius one, its coordinates satisfy x squared plus y squared equals one. Written in the new names that is cosine squared plus sine squared equals one — the Pythagorean identity, which is therefore not a separate fact at all.
Figure (svg): The unit circle with a point at angle theta, its coordinates labelled as cosine and sine
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 56-59 — the trigonometric functions on the unit circle
Picture it
One point on the unit circle, with both coordinates named.
Figure (svg): The unit circle with a point at angle theta, its coordinates labelled as cosine and sine
Everything else in trigonometry is bookkeeping on this picture. Signs in each quadrant, the values at the major angles, and every identity can be recovered by looking at where the point is.
Worked example
Example 1.23. Locate the point, then read its coordinates.
\[ \text{Evaluate all six functions at } t = \tfrac{2\pi}{3}. \]
Locate the angle
Why: Two thirds of a half turn is 120 degrees, in the second quadrant.
Find the reference angle
Why: It is 60 degrees, or pi over 3, from the horizontal axis.
\[ \text{reference } \frac{\pi}{3} \]
Read the coordinates with the quadrant's signs
Why: In the second quadrant x is negative and y is positive.
\[ (-\frac{1}{2}, \sqrt{3} / 2) \]
Name cosine and sine
Why: The coordinates, in order.
\[ \cos = -\frac{1}{2}, \sin = \sqrt{3} / 2 \]
Build the other four as ratios
Why: Tangent is sine over cosine; the rest are reciprocals.
\[ \tan = -\sqrt{3}, \sec = -2, \csc = 2 / \sqrt{3}, \cot = -1 / \sqrt{3} \]
Figure (svg): The solution to Worked example evaluating at a major angle shown as a ladder of expressions, one row per legal move
\[ \cos\tfrac{2\pi}{3} = -\tfrac{1}{2}, \quad \sin\tfrac{2\pi}{3} = \tfrac{\sqrt{3}}{2}, \quad \tan\tfrac{2\pi}{3} = -\sqrt{3} \]
Verify: check with the Pythagorean identity
Why: Cosine squared is one quarter and sine squared is three quarters, and they sum to 1 exactly as the identity demands. The signs also check against the quadrant: in the second quadrant only sine is positive, so cosine and tangent should both be negative, and both are. Running the identity is a complete check on the magnitudes, and the quadrant is a complete check on the signs.
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 58-59
Sorting
Signs follow from the coordinates' signs in each quadrant.
Sort into buckets
Sort each angle by which of sine and cosine is positive.
Tangent is positive exactly where sine and cosine share a sign, which is the first and third quadrants. Reading the signs off the coordinates is more reliable than any mnemonic, because it also tells you about the other four functions at the same time.
Worked example
Checkpoint 1.23. A ratio is undefined where its denominator vanishes.
\[ \text{Find the domain of } \tan t \text{ and of } \sec t. \]
Write each as a ratio
Why: Both have cosine in the denominator.
\[ \tan = \sin / \cos, \sec = 1 / \cos \]
Find where cosine is zero
Why: On the unit circle the x-coordinate vanishes at the top and bottom.
\[ t = \frac{\pi}{2}\text{ and } t = 3 \pi / 2 \]
Note that these repeat every half turn
Why: The two points are pi apart.
\[ t = \frac{\pi}{2} + \pi n \]
State the domain
Why: All reals except those inputs.
\[ t \ne \frac{\pi}{2} + \pi n \]
Figure (svg): The solution to Worked example the domain of tangent and secant shown as a ladder of expressions, one row per legal move
\[ D = \{t : t \ne \tfrac{\pi}{2} + \pi n, \; n \in \mathbb{Z}\} \]
Verify: check against the graph's asymptotes
Why: The tangent graph has vertical asymptotes exactly at plus and minus pi over 2 and every pi thereafter, which matches the excluded set. That the asymptotes and the domain exclusions coincide is not a coincidence — an asymptote is what a graph does at an input its function cannot accept, and Section 4.6 will make that connection precise.
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 59-60
Error analysis
A student evaluates cosine at two pi over three.
Annotate
On: \( \tfrac{2\pi}{3} \text{ has reference angle } \tfrac{\pi}{3}, \text{ so } \cos\tfrac{2\pi}{3} = \cos\tfrac{\pi}{3} = \tfrac{1}{2} \)
The reference angle gives the size; the quadrant gives the sign. Doing them as two separate steps, in that order, is what keeps them from being confused.
Fill the middle
The angle from the worked example, in the second quadrant.
Fill in the blanks
\left(\cos\tfrac-1/2___, \sin\tfrac______\right) = \left(___, \tfrac___}___\right)
Why: The reference angle pi over 3 gives magnitudes one half and root 3 over 2. The second quadrant makes the x-coordinate negative, so cosine is negative one half while sine stays positive.
Two truths and a lie
All three are about the six functions.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it is the useful distinction. Only sine and cosine have all of the real numbers as their domain. The other four are ratios, and each is undefined wherever its denominator vanishes: tangent and secant fail where cosine is zero, cotangent and cosecant where sine is zero. Those failures are exactly the vertical asymptotes on their graphs.
Notation
Four expressions that look similar and mean different things.
Annotate
On: \( \sin^2 t, \quad \sin t^2, \quad \sin(2t), \quad 2\sin t \)
The last two are the amplitude-versus-period distinction, and they are the same inside-versus-outside rule as Section 1.2's transformations. The first two catch people because the notation for squaring a trigonometric function is genuinely irregular.
Section
Section 3
Concept
Because going once round the circle returns the point to where it started, sine and cosine repeat every two pi. They are bounded between negative one and one. Tangent, being a ratio, repeats twice as often and is not bounded at all.
period and amplitude — A function is periodic with period p if adding p to the input never changes the output, and p is the smallest such positive number. For a sinusoid A times sine of B x, the amplitude is the absolute value of A and the period is two pi divided by the absolute value of B.
\[ y = A\sin(Bx): \quad \text{amplitude } |A|, \quad \text{period } \frac{2\pi}{|B|} \]
The inside constant divides the period rather than multiplying it, which is the same backwards-acting behaviour as a horizontal compression. Doubling B makes the function repeat twice as fast, so its period is halved.
Figure (svg): One sinusoid with amplitude 3 and period pi beside the plain sine, showing what each constant does
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 59-62 — graphs and periods of the trigonometric functions
Picture it
Sine and cosine over one full period.
Figure (svg): The sine and cosine graphs over two periods, with the period and amplitude marked
Cosine is sine shifted left by a quarter turn — which is one identity written as a picture. That relationship is why the derivative of one turns out to be the other, up to a sign.
Worked example
Example 1.25. Two constants, read separately.
\[ \text{Find the amplitude and period of } f(x) = 3\sin(2x) \text{ and sketch one period.} \]
Read the outside constant as the amplitude
Why: It scales the outputs.
\[ | A | = 3 \]
Read the inside constant and divide
Why: Period is 2 pi over the absolute value of B.
\[ \text{period } = 2 \pi / 2 \]
Simplify
Why: The period is pi.
\[ \text{period } = \pi \]
Locate the landmarks within one period
Why: Zero, maximum, zero, minimum, zero at quarter-period spacing.
\[ 0, \frac{\pi}{4}, \frac{\pi}{2}, 3 \pi / 4, \pi \]
Figure (svg): The solution to Worked example amplitude and period from a formula shown as a ladder of expressions, one row per legal move
\[ \text{amplitude } 3, \qquad \text{period } \pi \]
Verify: evaluate at a quarter of the period
Why: One quarter of the period is pi over 4, where the function should reach its maximum. Substituting: 3 sine of 2 times pi over 4 is 3 sine of pi over 2, which is 3 times 1, or 3 — the amplitude, as expected. Checking the value at the quarter-period confirms the period and the amplitude in a single substitution.
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 61-62
Matching
Two pi divided by the inside multiplier.
Match the pairs
Why: Doubling the input halves the period; halving it doubles the period. Tangent is the odd one: it has period pi without any inside multiplier at all, because the ratio of sine to cosine returns to its earlier values after only half a turn.
Worked example
Example 1.26. All four constants at once.
\[ \text{Describe } y = 2\sin\!\left(\tfrac{\pi}{6}(x - 3)\right) + 5. \]
Amplitude from the outside multiplier
Why: It scales the outputs by 2.
\[ \text{amplitude } 2 \]
Period from the inside multiplier
Why: Two pi divided by pi over 6.
\[ \text{period } = 12 \]
Horizontal shift from the inside subtraction
Why: Written in factored form, the shift is read directly.
\[ \text{right } 3 \]
Vertical shift from the outside addition
Why: The midline moves up.
\[ \text{midline } y = 5 \]
State the range
Why: Midline plus or minus the amplitude.
\[ \text{range } [3, 7] \]
Figure (svg): The solution to Worked example a full sinusoidal model shown as a ladder of expressions, one row per legal move
\[ \text{amplitude } 2, \; \text{period } 12, \; \text{shift right } 3, \; \text{midline } y = 5 \]
Verify: check the value at the shifted start
Why: At x equal to 3 the inside is 0, so the sine is 0 and y is 5 — on the midline, which is where a sine starts. One period later, at x equal to 15, the same thing should happen, and it does. Reading the shift required the inside to be in FACTORED form: written as pi x over 6 minus pi over 2 the shift is not 'pi over 2' but 3, and factoring first is what prevents that error.
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 62-63
Trap
\[ y = \sin\!\left(2x - \tfrac{\pi}{2}\right) \]
Read the shift straight off the constant
Why: The student sees minus pi over 2 and reports that as the shift.
\[ \text{shift right by } \tfrac{\pi}{2} \quad \text{(wrong)} \]
Checking where the inside is zero contradicts it: 2x equals pi over 2 gives x equal to pi over 4, not pi over 2.
\[ y = \sin\!\left(2\!\left(x - \tfrac{\pi}{4}\right)\right) \;\Longrightarrow\; \text{shift right } \tfrac{\pi}{4} \]
Factor the inside multiplier out first
Why: The shift is what x itself is reduced by, which only becomes visible once B is factored out.
The safe move, exactly as in Section 1.2, is to solve for where the inside expression equals zero. That single step gives the correct shift without any factoring at all, and it works no matter how the inside is written.
Fill the middle
The sinusoid from the first worked example, with inside multiplier 2.
Fill in the blanks
\text\pi = \frac______ = \frac______ = ___
Why: The period is pi. The function completes a full cycle in half the usual span, because the input runs twice as fast. Note that the amplitude 3 played no part — amplitude and period are set by different constants and do not interact.
Prediction
Commit before reasoning.
Predict first
Going from sin x to sin 2x, what happens to the period?
Correct: It halves, from two pi to pi.
\[ \sin(2x) \text{ completes a cycle when } 2x = 2\pi, \text{ i.e. } x = \pi \]
Why: The input reaches 2 pi when x is only pi, so the whole cycle is finished in half the span — the function repeats twice as often. This is the same inside-acts-backwards rule as horizontal compression in Section 1.2, and it is why the period formula has B in the denominator. The amplitude is set by the outside constant and has no bearing on it.
Sorting
Ask whether the function is a coordinate or a ratio.
Sort into buckets
Sort the six functions.
Secant and cosecant are worth a second look: they are never between negative 1 and 1, which is the opposite of sine and cosine. Being reciprocals of things bounded by 1, their size is always at least 1.
Section
Section 4
Concept
The Pythagorean identity is the unit circle's equation. Dividing it by cosine squared or by sine squared produces the other two Pythagorean forms. The addition formulas are separate facts, and the double-angle formulas follow from them at once.
trigonometric identity — An equation between trigonometric expressions that holds for every input for which both sides are defined. Unlike an equation to be solved, an identity has no particular solutions — it is true throughout.
\[ \cos^2 t + \sin^2 t = 1 \]
The distinction between an identity and an equation is worth being firm about. An identity is a rewriting rule you may apply anywhere; an equation is a question asking which inputs satisfy it. Confusing them leads to solving for x in something true for every x.
Figure (svg): The Pythagorean identity derived from the unit circle, with the two divided forms beside it
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 60-63 — trigonometric identities
Picture it
The circle's equation, and what dividing it gives.
Figure (svg): The Pythagorean identity derived from the unit circle, with the two divided forms beside it
Deriving the second and third on the spot takes one line each and removes two things from memory. That trade is almost always worth making.
Worked example
Example 1.27. Look for the Pythagorean pattern first.
\[ \text{Simplify } \frac{\sec t}{\tan t} \text{ and } \frac{\cos^2 t}{1 - \sin t}. \]
Rewrite everything in sine and cosine
Why: The safest first move on any simplification.
\[ \frac{1 / \cos}{\sin / \cos} \]
Simplify the compound fraction
Why: The cosines cancel.
\[ = 1 / \sin t = \csc t \]
For the second, replace cosine squared using the identity
Why: One minus sine squared.
\[ \frac{1 - \sin ^{2} t}{1 - \sin t} \]
Factor the numerator as a difference of squares
Why: It is one minus sine, times one plus sine.
\[ (1 - \sin) (1 + \sin) / (1 - \sin) \]
Cancel the common factor
Why: Valid wherever sine is not 1.
\[ = 1 + \sin t \]
Figure (svg): The solution to Worked example simplifying with an identity shown as a ladder of expressions, one row per legal move
\[ \frac{\sec t}{\tan t} = \csc t, \qquad \frac{\cos^2 t}{1 - \sin t} = 1 + \sin t \]
Verify: test both at a convenient input
Why: At t equal to pi over 6: sine is one half and cosine is root 3 over 2. The first expression is (2/root 3) divided by (1/root 3), which is 2, and cosecant of pi over 6 is indeed 2. The second is (3/4) divided by (1/2), which is 3/2, and 1 plus one half is 3/2. Both agree. A numerical spot check cannot prove an identity, but it catches a dropped factor immediately.
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 61-62
Matching
Where each one comes from.
Match the pairs
Why: Only two things need remembering: the circle's equation and the addition formulas. Everything else on this list is one line away from those, which makes the memory load far smaller than a table of identities suggests.
Worked example
Checkpoint 1.27. Break an unfamiliar angle into familiar ones.
\[ \text{Evaluate } \cos\tfrac{\pi}{12} \text{ exactly.} \]
Write the angle as a difference of two known angles
Why: A twelfth of pi is 15 degrees, which is 45 minus 30.
\[ \frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4} \]
Apply the cosine subtraction formula
Why: Cosine of a difference is cos cos plus sin sin.
\[ \cos(a - b) = \cos a \cos b + \sin a \sin b \]
Substitute the known values
Why: For pi over 3 and pi over 4.
\[ (\frac{1}{2}) (\sqrt{2} / 2) + (\sqrt{3} / 2) (\sqrt{2} / 2) \]
Combine over a common denominator
Why: Both terms have denominator 4.
\[ \frac{\sqrt{2} + \sqrt{6}}{4} \]
Figure (svg): The solution to Worked example an addition formula shown as a ladder of expressions, one row per legal move
\[ \cos\tfrac{\pi}{12} = \frac{\sqrt{2} + \sqrt{6}}{4} \]
Verify: compare with a decimal
Why: Root 2 is about 1.414 and root 6 about 2.449, so the answer is about 3.863 over 4, or 0.966. A calculator gives cosine of 15 degrees as 0.9659. They agree. The sign check also passes: 15 degrees is in the first quadrant so cosine should be positive and close to 1, which it is — and note the PLUS in the cosine subtraction formula, which is the sign people most often get backwards.
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 62-63
Error analysis
A student simplifies a quotient.
Annotate
On: \( \frac{\cos^2 t}{1 - \sin t} = \frac{1 - \sin^2 t}{1 - \sin t} = 1 - \sin t \)
Test it at t equal to pi over 6: the original is 3/4 over 1/2, which is 3/2, while the student's answer gives 1/2. Only factored expressions may be cancelled, and a one-input numerical check catches the failure immediately.
Fill the middle
Simplifying the quotient from the worked example.
Fill in the blanks
\frac1 + \sin t___ = \frac______ = ___
Why: Factoring the numerator as a difference of squares exposes the common factor, and cancelling it leaves 1 plus sine. Cancelling before factoring would have given 1 minus sine, which fails a numerical check at any input.
Two truths and a lie
All three are about identities.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and the pattern it invents is very tempting. The double-angle formula comes from the addition formula with both angles equal, not from a general rule about multiplying the input. Testing at t equal to pi over 6: sine of pi over 2 is 1, while 3 times one half times root 3 over 2 is about 1.3. They differ, so the claimed formula is simply wrong.
Prediction
Commit before reasoning.
Predict first
You are asked to 'solve' cosine squared plus sine squared equals 1. What is the answer?
Correct: Every real number. It is an identity, not an equation with particular solutions.
\[ \text{true for all } t: \; \cos^2 t + \sin^2 t = 1 \]
\[ \text{true only for some } t: \; \sin t = \tfrac{1}{2} \]
Why: An identity is true for every input in the domain, so the solution set is the whole domain. Being asked to solve one is a sign that the question is really asking you to recognise it. The practical value of the distinction is that an identity may be substituted anywhere, at any time, without justification — whereas an equation constrains the input and may only be used where that constraint holds.
Section
Section 5
Concept
Because the functions repeat, a trigonometric equation with one solution has infinitely many. The method is to find all solutions in one period, then add whole multiples of the period to each.
general solution — The complete set of solutions of a trigonometric equation, written by finding those in a single period and then adding an arbitrary whole-number multiple of the period to each.
\[ \sin t = \tfrac{1}{2} \;\Longrightarrow\; t = \tfrac{\pi}{6} + 2\pi n \;\text{ or }\; t = \tfrac{5\pi}{6} + 2\pi n \]
The commonest error is stopping after the first solution the calculator gives. An inverse trigonometric key returns exactly one value, and it is your job to find the others in the period by symmetry before adding the multiples.
Figure (svg): Solving a trigonometric equation shown on the unit circle and on the graph together
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 62-64 — solving trigonometric equations
Picture it
The same equation on the circle and on the graph.
Figure (svg): Solving a trigonometric equation shown on the unit circle and on the graph together
On the circle, a horizontal line at height one half crosses at two points. On the graph, the same line crosses the wave twice per period. Both pictures say the same thing: two families of solutions, not one.
Worked example
Example 1.28. One period first, then the general answer.
\[ \text{Solve } 1 + \cos t = 0 \text{ for all } t. \]
Isolate the trigonometric function
Why: Subtract 1.
\[ \cos t = -1 \]
Find the solutions in one period
Why: The x-coordinate is negative 1 only at the far left of the circle.
\[ t = \pi \]
Note how many there are per period
Why: Cosine reaches its minimum once per period, so only one.
Add whole multiples of the period
Why: The period of cosine is 2 pi.
\[ t = \pi + 2 \pi n \]
Figure (svg): The solution to Worked example all solutions of a simple equation shown as a ladder of expressions, one row per legal move
\[ t = \pi + 2\pi n, \quad n \in \mathbb{Z} \]
Verify: test two members of the family
Why: At n equal to 0, t is pi and cosine of pi is negative 1, so 1 plus it is 0. At n equal to 1, t is 3 pi, and cosine of 3 pi is also negative 1. Both work. Note this equation is the exception that gives only ONE solution per period, because negative 1 is the extreme value of cosine and the horizontal line is tangent to the wave rather than crossing it.
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 63-64
Sorting
Draw the horizontal line and count crossings.
Sort into buckets
Sort each equation by the number of solutions in one period.
The pattern is worth internalising: strictly inside the range gives two families, exactly at an extreme gives one, outside gives none. Counting crossings before solving tells you when to stop.
Worked example
Checkpoint 1.28. The typical case.
\[ \text{Solve } \sin^2 t = \tfrac{1}{4} \text{ on } [0, 2\pi). \]
Take square roots, keeping both signs
Why: This is where a whole family is usually lost.
\[ \sin t = \frac{1}{2}\text{ or } \sin t = -\frac{1}{2} \]
Solve the positive case in the period
Why: Sine is one half in the first and second quadrants.
\[ t = \frac{\pi}{6}, 5 \pi / 6 \]
Solve the negative case
Why: Sine is negative one half in the third and fourth quadrants.
\[ t = 7 \pi / 6, 11 \pi / 6 \]
Collect all four
Why: Evenly spaced, as the symmetry of the picture predicts.
Figure (svg): The solution to Worked example an equation with two families shown as a ladder of expressions, one row per legal move
\[ t = \tfrac{\pi}{6}, \; \tfrac{5\pi}{6}, \; \tfrac{7\pi}{6}, \; \tfrac{11\pi}{6} \]
Verify: check the symmetry of the answer set
Why: The four values are symmetric about pi, and each is pi over 6 away from a multiple of pi. Squaring means both signs of sine give the same result, so the solutions must be symmetric in both axes — and they are. If your answer set is not symmetric when the equation is, you have lost one. Substituting the third: sine of 7 pi over 6 is negative one half, whose square is one quarter.
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 64-64
Trap
\[ \sin t = \tfrac{1}{2} \]
Take the inverse sine and stop
Why: The calculator returns a single value.
\[ t = \sin^{-1}\!\left(\tfrac{1}{2}\right) = \tfrac{\pi}{6} \quad \text{(incomplete)} \]
Five sixths of pi also satisfies the equation, as does every one of those plus a multiple of two pi. An infinite family has been reduced to one number.
\[ t = \tfrac{\pi}{6} + 2\pi n \quad \text{or} \quad t = \tfrac{5\pi}{6} + 2\pi n \]
Use the inverse to get ONE solution, then symmetry for the rest of the period, then add the period
Why: The inverse function must return a single value to be a function at all, so completing the set is always the solver's job.
Draw the horizontal line on the circle or on the graph and count the crossings in one period. That count is how many families your answer needs, and it is the check that tells you whether you are finished.
Fill the middle
The equation from the visual, solved on the circle.
Fill in the blanks
\sin t = \tfrac5\pi/6___ \;\Longrightarrow\; t = \tfrac______ + 2\pi n \;\text___\; t = ___ + 2\pi n
Why: Sine is positive in the first two quadrants, and the second-quadrant solution is pi minus pi over 6, which is 5 pi over 6. Both families are needed; reporting only the first is the single most common error on this topic.
Ranking
Finding every solution of a trigonometric equation.
Put in order
Why: Step b saves the whole calculation when the value is out of range, and step d is the check that catches the missing family. Both are routinely skipped, and between them they account for most wrong answers on this topic.
Prediction
Commit before reasoning.
Predict first
Why does the inverse sine key return only one of infinitely many solutions?
Correct: Because an inverse has to be a function, so sine's domain is restricted to make it one-to-one first.
\[ \sin^{-1} : [-1, 1] \to \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] \]
Why: Sine is not one-to-one, so it has no inverse until its domain is cut down — conventionally to the span from negative pi over 2 to pi over 2, where it is increasing and hits each value once. The inverse sine therefore returns the unique solution in that window and cannot return more, since a function assigns one output per input. Section 1.4 is entirely about this restriction. Note that it does not return the smallest positive solution: the inverse sine of negative one half is negative pi over 6.
Comparison
Fill the blanks. Domain and period are what you actually reach for.
Comparison matrix
| Function | Period | Domain |
|---|---|---|
| sin t | 2 pi | all real numbers |
| cos t | 2 pi | all real numbers |
| tan t | pi | all reals except pi/2 + pi n |
| sec t | 2 pi | all reals except pi/2 + pi n |
| csc t | 2 pi | all reals except pi n |
Tangent's period of pi, rather than two pi, is the row worth remembering separately: it is the only one of the six that repeats faster than the circle does.
Pattern
Given a trigonometric expression or equation to handle.
Step two split into magnitude then sign is what prevents the commonest evaluation error, and step five's count is what prevents the commonest solving error. Neither costs more than a few seconds.
Stewart, Calculus: Early Transcendentals 8e, Appendix D — Trigonometry Appendix D, pp. A24-A33
Check
Conversion. Half a turn is pi.
Check your understanding
Express 225 degrees in radians.
Answer: A
Why: 225 over 180 reduces to 5 over 4, so the angle is 5 pi over 4 radians.
Check
Evaluation. Magnitude from the reference angle, sign from the quadrant.
Check your understanding
Find cos(2 pi / 3).
Answer: A
Why: The reference angle is pi over 3, giving magnitude one half, and the second quadrant makes cosine negative.
Check
Amplitude and period, read from different constants.
Check your understanding
Find the amplitude and period of f(x) = 3 sin(2x).
Answer: A
Why: The outside constant 3 is the amplitude; the period is 2 pi divided by 2, which is pi.
Real world
The number of hours of daylight in a northern city over a year rises to about 15.3 hours at midsummer and falls to about 9.1 hours at midwinter, repeating annually.
Discussion prompt
Build a sinusoidal model for daylight as a function of the day of the year, and use it to find roughly when daylight passes 14 hours.
Hint: Amplitude is half the peak-to-trough distance; the midline is their average; the period is 365 days.
Answer:
The peak-to-trough swing is 15.3 minus 9.1, or 6.2 hours, so the amplitude is 3.1 and the midline is their average, 12.2 hours.
\[ B = \frac{2\pi}{365}, \qquad D(t) = 3.1\sin\!\left(\tfrac{2\pi}{365}(t - 80)\right) + 12.2 \]
The shift of about 80 days puts the rising crossing of the midline near the spring equinox, which is where a sine starts.
\[ 14 = 3.1\sin\!\left(\tfrac{2\pi}{365}(t-80)\right) + 12.2 \;\Longrightarrow\; \sin(\cdots) = \tfrac{1.8}{3.1} \approx 0.581 \]
\[ \tfrac{2\pi}{365}(t - 80) \approx 0.620 \;\text{ or }\; \pi - 0.620 = 2.522 \]
Solving gives t about 116 and t about 227 — roughly late April and mid August. Two answers, not one, and this is exactly the trap from the last idea: daylight passes 14 hours once on the way up and once on the way down. A solver who stopped at the inverse sine would have reported only April and missed the whole second half of the summer.
Commit first
Answer, then rate your confidence honestly.
Predict first
Going from y = sin x to y = sin(3x), what changes?
Correct: The period becomes two pi over 3 — the wave repeats three times as often.
\[ y = A\sin(Bx): \quad \text{amplitude } |A| = 1, \quad \text{period } \tfrac{2\pi}{3} \]
Why: The constant is inside the function, so it acts on the input and therefore on the period, not the amplitude. The amplitude is set by an outside multiplier and is unchanged at 1 here. Nor is it a shift: a shift would require something added or subtracted inside, not multiplied. The general rule is that the period is two pi divided by the absolute value of the inside multiplier, so a larger multiplier means a shorter period.
Explain it
They can use sine and cosine on right triangles but do not see why anyone would bother with radians.
Discussion prompt
In four sentences or fewer, explain what a radian is and why calculus insists on it.
Hint: Define it as a ratio of lengths first, then say what that buys.
Answer:
A radian is the angle whose arc is exactly as long as the radius. Because it is one length divided by another, it is a pure number with no units — unlike a degree, which is an arbitrary three-hundred-and-sixtieth of a turn.
That matters because arc length then equals radius times angle, with no conversion factor anywhere. When you differentiate sine in Chapter 3, that missing factor is exactly why the answer comes out as plain cosine; in degrees, every trigonometric derivative would drag a factor of pi over 180 along with it forever.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For signs, always do magnitude and sign as two separate steps — reference angle first, quadrant second. For amplitude and period, remember that outside sets height and inside divides the period. For solving, draw the horizontal line and count crossings before you start, so you know how many families to find. For simplifying, rewrite in sine and cosine first, then factor before cancelling anything. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Draw a large unit circle and mark every major angle from 0 to two pi in both degrees and radians, writing the coordinates at each. Beneath it, graph sine and cosine over one full period on the same axes, marking the period, the amplitude, and the quarter-period landmarks. Beside that, graph tangent over two periods with its asymptotes, and write one sentence saying why the asymptotes sit where they do. In the lower half, write the Pythagorean identity and derive the other two from it, showing the division in each case. Then solve sine of t equals one half completely: mark both solutions on your circle, mark them again on your sine graph, and write the general solution with the two families. In a margin, write the arc length formula and one sentence saying what goes wrong if the angle is in degrees.
If your tangent asymptotes are not directly above and below the points where your cosine graph crosses zero, one of the two pictures is wrong — tangent's denominator is cosine, so those inputs must match exactly.
Recap
Five things, and the first is the one that makes Chapter 3's trigonometric derivatives possible.
| If you see | Then |
|---|---|
| An angle in degrees | Convert before any formula mixing angle with length |
| An angle outside quadrant one | Magnitude from the reference angle, sign from the quadrant |
| A multiplier inside | The period is divided by it |
| A multiplier outside | The amplitude is multiplied by it |
| cos^2 or sin^2 | Try the Pythagorean identity |
| A trigonometric equation | Find every solution in one period, then add multiples |
| An inverse trig key | It returns one solution; find the rest yourself |
Section 1.4 takes up the question raised at the end of this one: what exactly is an inverse function, why does sine need its domain cut down before it has one, and how do you find an inverse in general.
OpenStax Calculus Volume 1, §1.3 Trigonometric Functions §1.3, pp. 53-64 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Calculus I — $55/session, free consultation.