The two fundamental trigonometric limits and why they require radians, the derivatives of sine and cosine proved from them with the addition formula, the other four derivatives by the quotient rule and the co-function pattern, the four-step cycle of higher derivatives, and simple harmonic motion.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 3 — Derivatives
Derivatives of Trigonometric Functions
Objectives
Five outcomes. The first is the limit Section 2.3 established by squeezing, and it is the foundation everything else here stands on.
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 241-249 — the section these objectives are drawn from
Warm-up
Section 2.3 proved by squeezing that sine of h over h tends to 1 as h tends to zero, using the areas of a triangle, a sector and a triangle on the unit circle.
Discussion prompt
That limit was proved with the sector's area taken as h over 2. What would the limit have been if angles were measured in degrees, and why?
Hint: The sector area formula is the one place radian measure enters the argument.
Answer:
\[ \text{sector area} = \tfrac{1}{2}r^{2}\theta \quad \text{holds only in radians} \]
In degrees the sector of a unit circle with central angle h has area pi h over 360, not h over 2. Running the same squeeze gives pi over 180 rather than 1.
That factor of about 0.01745 would then appear in the derivative of sine, and in every derivative built from it, forever. This section is the payoff for having insisted on radians in Section 1.3 — and the reason a calculator in degree mode gives wrong answers in calculus.
Concept
The derivative of sine is cosine and the derivative of cosine is minus sine. Both are proved from the difference quotient using the addition formula and two fundamental limits, and the other four functions follow by the quotient rule.
the fundamental trigonometric limits — The limit of sine h over h is 1, and the limit of cosine h minus 1 over h is 0, both as h approaches zero and both in radians. Every trigonometric derivative is built from these two.
\[ \lim_{h \to 0}\frac{\sin h}{h} = 1, \qquad \lim_{h \to 0}\frac{\cos h - 1}{h} = 0 \]
The second limit is derived from the first by multiplying by the conjugate, so the entire section rests on a single geometric fact about areas on the unit circle.
Figure (svg): The two fundamental limits, with their values and the geometric source of each
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 241-243
Section
Section 1
Concept
The first limit comes from comparing three areas on the unit circle and squeezing. The second is obtained from it by multiplying the quotient by its conjugate, which converts the cosine difference into a sine squared.
the conjugate derivation — Multiplying cosine h minus 1 over h by cosine h plus 1 over itself turns the numerator into minus sine squared h, which then splits into the first limit times a factor tending to zero.
\[ \frac{\cos h - 1}{h}\cdot\frac{\cos h + 1}{\cos h + 1} = \frac{-\sin^{2}h}{h(\cos h + 1)} \]
Both limits are indeterminate quotients of the kind Section 2.3 catalogued, and both are resolved by techniques from that section: one by squeezing, one by a conjugate.
Figure (svg): The two fundamental limits, with their values and the geometric source of each
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 241-243 — the two fundamental limits
Picture it
Their values and where each comes from.
Figure (svg): The two fundamental limits, with their values and the geometric source of each
Both carry the same warning. In degrees the first becomes pi over 180 and the second stays 0, but the first is the one that matters and its factor propagates into every formula in this section.
Worked example
Example 3.35. The conjugate does the work.
\[ \text{Show that } \lim_{h \to 0}\frac{\cos h - 1}{h} = 0. \]
Note that direct substitution fails
Why: One minus 1 over 0.
\[ \frac{0}{0} \]
Multiply by the conjugate over itself
Why: Cosine h plus 1.
\[ \frac{\cos ^{2} h - 1}{h(\cos h + 1)} \]
Use the Pythagorean identity
Why: Cosine squared minus 1 is minus sine squared.
\[ -\sin ^{2} h / (h(\cos h + 1)) \]
Split into recognisable pieces
Why: One copy of the first limit, times the rest.
\[ -(\sin h / h) (\sin h / (\cos h + 1)) \]
Take the limit of each factor
Why: One times zero over two.
\[ -(1) (\frac{0}{2}) = 0 \]
Figure (svg): The solution to Worked example deriving the second limit shown as a ladder of expressions, one row per legal move
\[ \lim_{h \to 0}\frac{\cos h - 1}{h} = 0 \]
Verify: check numerically and see why the answer is 0 rather than something else
Why: At h equal to 0.01 the quotient is about negative 0.005, and at 0.001 about negative 0.0005 — shrinking by a factor of ten each time, consistent with a limit of 0. The reason it vanishes is that cosine is FLAT at 0: its graph has a horizontal tangent there, so the difference quotient measuring its slope must tend to zero. Indeed this limit is precisely the derivative of cosine at 0, which the next idea will confirm is minus sine of 0, or 0.
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 242-243
Fill the middle
Rewriting a variant so the known limit applies.
Fill in the blanks
\frac5___ = 5\cdot\frac______ \;\longrightarrow\; ___
Why: The bracketed quotient tends to 1 because its argument and denominator now match, leaving the factor 5. In general sine of kx over x tends to k.
Worked example
Checkpoint 3.35. Adjust the expression to match the known form.
\[ \text{Evaluate } \lim_{x \to 0}\frac{\sin 3x}{x}. \]
Note the mismatch
Why: The known limit needs the same expression above and below.
Multiply and divide by 3
Why: So the denominator matches the sine's argument.
\[ 3 \cdot \sin(3 x) / (3 x) \]
Recognise the known limit
Why: As x approaches 0 so does 3x.
\[ \sin(3 x) / (3 x) \to 1 \]
Take the limit
Why: Three times 1.
\[ 3 \]
Figure (svg): The solution to Worked example using the limits on a variant shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 0}\frac{\sin 3x}{x} = 3 \]
Verify: test numerically and note the general pattern
Why: At x equal to 0.001 the quotient is about 2.9999955, closing on 3. The general fact is that sine of kx over x tends to k, since the argument must match the denominator for the known limit to apply. That matching is the whole technique, and forgetting it gives 1 instead of 3 — the commonest error with these limits.
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 243-243
Trap
\[ \lim_{x \to 0}\frac{\sin 3x}{x} \]
Apply the known limit directly
Why: The student sees a sine over a variable and reports 1.
\[ = 1 \quad \text{(wrong)} \]
The known limit requires the SAME expression inside the sine and in the denominator. Here they differ by a factor of 3.
\[ \frac{\sin 3x}{x} = 3\cdot\frac{\sin 3x}{3x} \;\longrightarrow\; 3\cdot 1 = 3 \]
Force the denominator to match the sine's argument
Why: Multiply and divide by whatever factor is needed.
A numerical check settles any doubt in seconds: at x equal to 0.001 the quotient is very close to 3, not to 1. The same matching is needed for every variant, and the general result is that sine of kx over x tends to k.
Matching
Match the arguments before applying anything.
Match the pairs
Why: The last two are mirror images: a factor inside the sine multiplies the answer, a factor in the denominator divides it. Both are handled by the same move of forcing the two arguments to agree.
Two truths and a lie
All three are about the fundamental limits.
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. The limit is k, not 1 — the known result requires the sine's argument and the denominator to be the same expression. Testing at k equal to 3 with x equal to 0.001 gives about 3, not 1. Matching the arguments is the whole technique for every variant of these limits.
Prediction
Commit before reasoning.
Predict first
In the squeeze proof of the first limit, which step requires radians?
Correct: The sector's area, which is h over 2 only in radian measure.
\[ \tfrac12\sin h < \tfrac12 h < \tfrac12\tan h \quad \text{(radians)} \]
Why: The two triangles' areas are unit-independent, and the reciprocal step is pure algebra. Only the sector's area formula depends on how the angle is measured, because that formula comes from the arc-length definition of the radian. In degrees the sector's area is pi h over 360 and the whole squeeze delivers pi over 180 instead of 1. One step in one proof is where the entire convention pays off.
Section
Section 2
Concept
Expanding sine of x plus h with the addition formula splits the difference quotient into two pieces, each containing one of the fundamental limits. Taking those limits gives cosine, and the same argument on cosine gives minus sine.
derivatives of sine and cosine — The derivative of sine is cosine, and the derivative of cosine is negative sine. Both hold only when the input is measured in radians.
\[ \frac{d}{dx}\left[\sin x\right] = \cos x, \qquad \frac{d}{dx}\left[\cos x\right] = -\sin x \]
The graphical evidence comes first and is complete. Sine is flat exactly where cosine is zero, rising where cosine is positive, and steepest where cosine peaks — the slope function of sine simply is cosine.
Figure (svg): The sine curve above the cosine curve, with the slopes of one matching the heights of the other
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 243-245 — derivatives of sine and cosine
Picture it
Sine stacked above cosine with a shared axis.
Figure (svg): The sine curve above the cosine curve, with the slopes of one matching the heights of the other
At every dashed line the upper curve is momentarily flat and the lower one crosses zero. That alignment is the same reading technique as Section 3.2, applied to a pair of curves you already know.
Worked example
Example 3.36. The addition formula, then the two limits.
\[ \text{Prove that } \frac{d}{dx}\left[\sin x\right] = \cos x. \]
Write the difference quotient
Why: The definition.
\[ \frac{\sin(x + h) - \sin x}{h} \]
Expand with the addition formula
Why: Sine of a sum.
\[ \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} \]
Regroup so each limit appears once
Why: Collect the sine x terms.
\[ \sin x(\cos h - 1) / h + \cos x(\sin h) / h \]
Take the limits, noting x is constant here
Why: Zero and one respectively.
\[ \sin x(0) + \cos x(1) \]
Simplify
Why: Only the second term survives.
Figure (svg): The proof of the sine derivative, showing where each of the two limits enters
\[ \frac{d}{dx}\left[\sin x\right] = \cos x \]
Verify: check the result at two landmark inputs
Why: At x equal to 0 the derivative should be cosine of 0, which is 1 — and the sine curve does rise at 45 degrees through the origin. At x equal to pi over 2 it should be cosine of pi over 2, which is 0 — and sine is flat at its peak there. Both match the graph. Note that x was treated as a constant throughout, since the limit is in h — a point worth being explicit about, because it is what allows sine x and cosine x to be pulled outside the limits.
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 243-244
Fill the middle
The regrouped difference quotient for sine, with both limits about to be taken.
Fill in the blanks
\sin x\cdot 0 + \cos x\cdot 1 = \cos x
Why: The first fundamental limit contributes 0 and kills the sine term; the second contributes 1 and leaves cosine. Both limits are used exactly once, which is why both had to be established first.
Worked example
Checkpoint 3.36. The same argument, one sign different.
\[ \text{Prove that } \frac{d}{dx}\left[\cos x\right] = -\sin x. \]
Expand with the cosine addition formula
Why: Note the minus sign it carries.
\[ \frac{\cos x \cos h - \sin x \sin h - \cos x}{h} \]
Regroup
Why: Collect the cosine x terms.
\[ \cos x(\cos h - 1) / h - \sin x(\sin h) / h \]
Take the limits
Why: Zero and one.
\[ \cos x(0) - \sin x(1) \]
Simplify
Why: The surviving term carries a minus.
\[ -\sin x \]
Figure (svg): The solution to Worked example the derivative of cosine shown as a ladder of expressions, one row per legal move
\[ \frac{d}{dx}\left[\cos x\right] = -\sin x \]
Verify: trace the minus sign to its source and check the graph
Why: The minus comes from the cosine addition formula, which has a minus where the sine one has a plus. Graphically it is right: cosine falls on the interval from 0 to pi, and sine is positive there, so the derivative must be negative. At x equal to pi over 2 the derivative is negative 1 and cosine is indeed descending at its steepest. Losing that minus sign is the standard error, and the falling-cosine check catches it instantly.
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 244-245
Trap
\[ \frac{d}{dx}\left[\cos x\right] = \sin x \quad \text{(wrong)} \]
Assume the pattern is symmetric
Why: The student expects sine and cosine to swap cleanly.
On the interval from 0 to pi, cosine falls while sine is positive. A positive derivative for a falling function is impossible.
\[ \frac{d}{dx}\left[\cos x\right] = -\sin x \]
Check the sign against a stretch where the graph's direction is obvious
Why: Cosine descends from 1 to negative 1 across the first half period.
The pattern is not symmetric, and the asymmetry is exactly the co-function rule that governs the whole section: differentiating any co- function produces a minus sign. Cosine, cotangent and cosecant all carry it; sine, tangent and secant do not.
Sorting
Read the derivative's sign at each input.
Sort into buckets
Sort each statement about sine.
The two flat inputs are sine's maximum and minimum, and they sit precisely at cosine's zeros — the alignment the stacked graphs showed. Reading a function's behaviour off its derivative's sign is the technique the whole of Chapter 4 runs on.
Two truths and a lie
All three are about these two derivatives.
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 missing minus sign is the commonest error in the section. Cosine falls on the interval from 0 to pi where sine is positive, so a positive derivative there is impossible. The minus comes from the cosine addition formula and it is the first instance of the co-function pattern that governs three of the six derivatives.
Prediction
Commit before reasoning.
Predict first
What does the addition formula accomplish in the proof?
Correct: It separates the h from the x so the h-dependent parts become recognisable limits.
\[ \sin(x+h) = \sin x\cos h + \cos x\sin h \quad \text{- the whole reason the proof works} \]
Why: Sine of x plus h cannot be split by any algebraic rule — sine does not distribute over addition. The addition formula is the only tool that expresses it in terms of sine and cosine of x and of h separately, and once separated, everything involving x factors out of the limit as a constant while everything involving h forms one of the two fundamental limits. Without that separation the difference quotient cannot be resolved at all.
Section
Section 3
Concept
Tangent, cotangent, secant and cosecant are all quotients of sine and cosine, so the quotient rule gives their derivatives. The results follow a pattern: every co-function's derivative carries a minus sign.
the co-function pattern — For each of the three pairs, the co- function's derivative is the same shape as its partner's with a minus sign and every function replaced by its co-function.
\[ (\tan x)' = \sec^{2}x, \quad (\sec x)' = \sec x\tan x \]
Learning three derivatives plus the pattern is far more reliable than learning six separately, and the pattern also predicts the signs, which is where the errors are.
Figure (svg): The derivatives of all six trigonometric functions, grouped by the co- pattern
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 245-247 — derivatives of the other trigonometric functions
Picture it
All six, grouped by the co- relationship.
Figure (svg): The derivatives of all six trigonometric functions, grouped by the co- pattern
Every entry on the right of a co- function begins with a minus. That single observation halves what has to be remembered and fixes the signs at the same time.
Worked example
Example 3.38. Write it as a quotient and apply the rule.
\[ \text{Prove that } \frac{d}{dx}\left[\tan x\right] = \sec^{2}x. \]
Write tangent as a quotient
Why: Sine over cosine.
\[ \tan x = \sin x / \cos x \]
Apply the quotient rule
Why: Bottom times top prime, minus top times bottom prime.
\[ (\cos x \cos x - \sin x(-\sin x)) / \cos ^{2} x \]
Simplify the numerator
Why: The double negative becomes a plus.
\[ (\cos ^{2} x + \sin ^{2} x) / \cos ^{2} x \]
Use the Pythagorean identity
Why: The numerator is 1.
\[ 1 / \cos ^{2} x \]
Rewrite
Why: The reciprocal of cosine is secant.
\[ \sec ^{2} x \]
Figure (svg): The solution to Worked example the derivative of tangent shown as a ladder of expressions, one row per legal move
\[ \frac{d}{dx}\left[\tan x\right] = \sec^{2}x \]
Verify: check the sign and the behaviour at an asymptote
Why: Secant squared is positive everywhere it is defined, which says tangent is increasing on every branch — and the graph from Section 1.3 confirms that it rises steeply on each. Near an asymptote the secant grows without bound and so does its square, matching tangent's ever-steeper climb. The double negative in step three is where the plus sign comes from, and losing it would give cosine of 2x over cosine squared, which is not positive everywhere and would contradict the graph.
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 245-246
Matching
Watch the signs on the co-functions.
Match the pairs
Why: Each co-function's derivative is its partner's with every function replaced by its co-function and a minus sign attached. That is the entire pattern, and it fixes both the shape and the sign of three of the six.
Worked example
Checkpoint 3.38. A reciprocal, by the same route.
\[ \text{Find } \frac{d}{dx}\left[\sec x\right]. \]
Write secant as a quotient
Why: One over cosine.
\[ \sec x = 1 / \cos x \]
Apply the quotient rule
Why: The numerator's derivative is 0.
\[ (0 - 1(-\sin x)) / \cos ^{2} x \]
Simplify
Why: The double negative again.
\[ \sin x / \cos ^{2} x \]
Split into recognisable factors
Why: One over cosine, times sine over cosine.
\[ (1 / \cos x) (\sin x / \cos x) \]
Rewrite
Why: Secant times tangent.
Figure (svg): The solution to Worked example the derivative of secant shown as a ladder of expressions, one row per legal move
\[ \frac{d}{dx}\left[\sec x\right] = \sec x\tan x \]
Verify: check the sign on a stretch where the graph's direction is clear
Why: On the interval from 0 to pi over 2 both secant and tangent are positive, so the derivative is positive — and secant does rise from 1 toward infinity across that stretch. On the interval from negative pi over 2 to 0, tangent is negative so the derivative is negative, and secant falls from infinity to 1. Both match. Splitting the answer into secant times tangent rather than leaving it as sine over cosine squared is what makes the pattern with cosecant visible.
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 246-247
Error analysis
A student differentiates cotangent by analogy with tangent.
Annotate
On: \( \frac{d}{dx}\left[\cot x\right] = \csc^{2}x \)
The co-function pattern fixes the signs of exactly three of the six derivatives. Checking the answer's sign against whether the graph rises or falls on a branch catches every one of these, and it takes a few seconds.
Fill the middle
The quotient rule applied to sine over cosine, before the identity is used.
Fill in the blanks
\frac1x + \sin^___x}___x} = \frac___}___x} = \sec^___x
Why: The numerator is 1 by the Pythagorean identity, so the quotient is one over cosine squared, which is secant squared. The identity from Section 1.3 is doing the final simplification.
Sorting
Apply the co-function pattern.
Sort into buckets
Sort each of the six functions.
The rule is exactly as mechanical as it looks: the three names beginning with co- are the three derivatives with a minus. Checking against a graph's direction on one branch confirms each, and that check is worth running when the pattern is recalled after a gap.
Prediction
Commit before reasoning.
Predict first
What does secant squared being positive everywhere tell you about tangent?
Correct: Tangent is increasing on every branch.
\[ \sec^{2}x > 0 \;\Longrightarrow\; \tan x \text{ is increasing wherever defined} \]
Why: A positive derivative means the function rises, and secant squared is positive wherever it is defined. So tangent climbs steadily across each branch, from negative infinity to positive infinity — which the graph shows. It is certainly not always positive, since it is negative on half of every branch, and it is emphatically not continuous, having an asymptote every pi. Confusing a statement about the derivative's sign with one about the function's own sign is the error to avoid.
Section
Section 4
Concept
Differentiating sine repeatedly gives cosine, then minus sine, then minus cosine, then sine again. The cycle has length four, so a high-order derivative is found by dividing the order by four and reading the remainder.
the four-step cycle — The derivatives of sine repeat with period four: sine, cosine, minus sine, minus cosine, and back to sine. The same cycle governs cosine, entered at a different point.
\[ \sin x \to \cos x \to -\sin x \to -\cos x \to \sin x \]
The cycle makes an otherwise impossible computation trivial. The hundredth derivative of sine is sine, because 100 leaves remainder 0 on division by 4 — no differentiation required.
Figure (svg): The four-step cycle of derivatives of sine
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 247-248 — higher-order derivatives
Picture it
The four functions the derivatives run through.
Figure (svg): The four-step cycle of derivatives of sine
Cosine enters the same cycle one step along, so its derivatives run cosine, minus sine, minus cosine, sine. Both are the same loop entered at different points.
Worked example
Example 3.40. Divide by four and read the remainder.
\[ \text{Find the } 74\text{th derivative of } \sin x. \]
Recall the cycle
Why: Length four.
\[ \sin, \cos, -\sin, -\cos \]
Divide the order by four
Why: Seventy-four is 4 times 18 plus 2.
\[ \text{remainder } 2 \]
Step that many places from sine
Why: Two steps: cosine, then minus sine.
\[ -\sin x \]
State the answer
Why: The 74th derivative.
\[ -\sin x \]
Figure (svg): The solution to Worked example a high-order derivative shown as a ladder of expressions, one row per legal move
\[ \frac{d^{74}}{dx^{74}}\left[\sin x\right] = -\sin x \]
Verify: check the method on a small order you can do by hand
Why: The 2nd derivative should be minus sine by the same rule, and differentiating twice by hand gives cosine then minus sine — matching. The 4th should be sine, and four differentiations do return to sine. Since 74 and 2 leave the same remainder, they give the same answer. Note that the remainder, not the quotient, is what matters: the 18 complete cycles contribute nothing.
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 247-248
Fill the middle
Finding a high-order derivative of sine by the cycle.
Fill in the blanks
74 = 4(18) + 2 \;\Longrightarrow\; \text___
Why: The remainder is 2, so stepping two places from sine gives minus sine. The eighteen complete cycles return you to the start and contribute nothing.
Worked example
Checkpoint 3.40. Two differentiations with the rules.
\[ \text{Find } y'' \text{ for } y = x\sin x. \]
Differentiate once with the product rule
Why: One factor at a time.
\[ y' = \sin x + x \cos x \]
Differentiate the first term
Why: The derivative of sine.
Differentiate the second term with the product rule again
Why: It is a product too.
\[ \cos x + x(-\sin x) \]
Combine
Why: Add the pieces.
\[ y'' = 2 \cos x - x \sin x \]
Figure (svg): The solution to Worked example a second derivative in context shown as a ladder of expressions, one row per legal move
\[ y'' = 2\cos x - x\sin x \]
Verify: check at a convenient input
Why: At x equal to 0: the formula gives 2 times 1 minus 0, which is 2. Computing directly, y is x sine x, whose first derivative at 0 is sine 0 plus 0, or 0, and the second should measure how fast that is changing. A numerical second difference near 0 gives about 2.000, confirming it. Note that the product rule was needed twice, because differentiating the term x cosine x produces another product.
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 248-248
Trap
\[ 74 \div 4 = 18\text{ remainder }2 \]
Step 18 places round the cycle
Why: The student uses the quotient instead of the remainder.
\[ 18 \bmod 4 = 2 \;\Longrightarrow\; \text{coincidentally right here, wrong in general} \]
For the 30th derivative the quotient is 7 and the remainder 2; stepping 7 places gives the wrong answer while stepping 2 gives the right one.
\[ 74 = 4(18) + 2 \;\Longrightarrow\; \text{step } 2 \text{ places} \;\Longrightarrow\; -\sin x \]
Use the REMAINDER: complete cycles return you to the start
Why: Eighteen full cycles change nothing, so only the leftover steps matter.
The check is to test the method on an order small enough to verify by hand. The 2nd derivative of sine is minus sine, and 2 leaves remainder 2 — consistent. Any method that fails on a small case will fail on a large one.
Matching
Divide by four, read the remainder.
Match the pairs
Why: The last is the striking one: a hundred differentiations return sine unchanged, because 100 is exactly 25 complete cycles. Without the cycle the computation is impossible; with it, it takes one division.
Two truths and a lie
All three are about higher derivatives.
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. Two differentiations give cosine and then minus sine, so the second derivative is MINUS sine. It takes four steps, not two, to return to sine. This also says something worth noticing: sine satisfies the equation that its second derivative is its own negative, which is the defining equation of simple harmonic motion.
Prediction
Commit before reasoning.
Predict first
Since the second derivative of sine is minus sine, what equation does y = sin x satisfy?
Correct: The second derivative equals the negative of the function.
\[ y = \sin x \;\Longrightarrow\; y'' = -\sin x = -y \]
Why: Two differentiations turn sine into minus sine, which is exactly that equation. It is the defining equation of simple harmonic motion, and it says the acceleration always points back toward the centre with magnitude proportional to the displacement — a spring, a pendulum, an oscillating circuit. That sine and cosine are the solutions is why they describe every oscillation in physics, and it is the subject of the next idea.
Section
Section 5
Concept
A mass on a spring has position given by a sinusoid. Differentiating twice returns the negative of the position, which says the acceleration always points back toward the centre and grows with the displacement.
simple harmonic motion — Motion in which the acceleration is proportional to the displacement and directed opposite to it. Its solutions are sinusoids, and it describes springs, pendulums and every small oscillation.
\[ s(t) = A\cos(\omega t) \;\Longrightarrow\; a(t) = -\omega^{2}s(t) \]
The negative sign is the physics: the further the mass is pulled from centre, the harder the spring pulls it back. That single relationship forces the motion to be sinusoidal, which is why oscillations everywhere look the same.
Figure (svg): A mass on a spring with its position, velocity and acceleration as sinusoids
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 248-249 — applications to motion
Picture it
Position, velocity and acceleration for a mass on a spring.
Figure (svg): A mass on a spring with its position, velocity and acceleration as sinusoids
The bottom graph is the top one flipped, which is the equation made visible. Where the mass is furthest from centre, the restoring acceleration is greatest — and at the centre it is zero, which is where the speed peaks.
Worked example
Example 3.42. Differentiate twice and interpret.
\[ \text{A mass has } s(t) = 3\cos t \text{ cm. Find } v \text{ and } a, \text{ and describe the motion at } t = \tfrac{\pi}{2}. \]
Differentiate for the velocity
Why: The derivative of cosine.
\[ v(t) = -3 \sin t \]
Differentiate again for the acceleration
Why: The derivative of minus sine.
\[ a(t) = -3 \cos t \]
Note the relationship
Why: The acceleration is the negative of the position.
\[ a(t) = -s(t) \]
Evaluate at the given time
Why: Sine of pi over 2 is 1, cosine is 0.
\[ s = 0, v = -3, a = 0 \]
Interpret
Why: At the centre, moving fastest, no restoring force.
Figure (svg): The solution to Worked example analysing an oscillation shown as a ladder of expressions, one row per legal move
\[ s = 0, \; v = -3, \; a = 0 \]
Verify: check the physical consistency at both extremes
Why: At t equal to 0 the mass is at 3 centimetres with velocity 0 and acceleration negative 3 — stretched fully, momentarily still, and being pulled hardest back toward centre. At t equal to pi over 2 it is at the centre moving fastest with no restoring force at all. Both are exactly what a spring does, and the pattern that speed peaks where acceleration vanishes is characteristic of every oscillation.
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 249-249
Fill the middle
The spring from the worked example, with the position given.
Fill in the blanks
s = 3\cos t \;\Longrightarrow\; v = -3\sin t \;\Longrightarrow\; a = -3\cos t
Why: The derivative of minus 3 sine t is minus 3 cosine t, which is the negative of the position. That relationship is the defining equation of simple harmonic motion.
Worked example
Checkpoint 3.42. The extremes of position and of speed alternate.
\[ \text{For the same mass, find when the speed is greatest and when it is zero.} \]
Write the speed
Why: The absolute value of the velocity.
\[ | - 3 \sin t | = 3 | \sin t | \]
Find where sine's magnitude is greatest
Why: At the odd multiples of pi over 2.
\[ t = \frac{\pi}{2}, 3 \pi / 2,... \]
Find where it vanishes
Why: At the multiples of pi.
\[ t = 0, \pi, 2 \pi,... \]
Compare with the position
Why: Position is extreme exactly where speed is zero.
Figure (svg): The solution to Worked example where the speed is greatest shown as a ladder of expressions, one row per legal move
\[ \text{fastest at } t = \tfrac{\pi}{2} + \pi n; \quad \text{at rest at } t = \pi n \]
Verify: reason physically rather than only algebraically
Why: At the extremes of displacement the spring is fully stretched or compressed and the mass has momentarily stopped to reverse — zero speed, maximum restoring force. At the centre the spring is relaxed and all the energy is in the motion — maximum speed, zero force. The algebra and the physics agree exactly, and the alternation is a consequence of the derivative of a sinusoid being another sinusoid shifted by a quarter period.
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 249-249
Error analysis
A student computes the derivative of sine numerically to check the rule.
Annotate
On: \( \frac{\sin(30.001^\circ) - \sin(30^\circ)}{0.001} \approx 0.0151, \text{ but } \cos 30^\circ \approx 0.866 \)
The stray factor is not a rounding artefact but pi over 180, appearing exactly as the fundamental limit predicted. This is why a calculator must be in radian mode for any calculus, and the discrepancy's size identifies the cause immediately.
Sorting
For s(t) = 3 cos t, read position, velocity and acceleration.
Sort into buckets
Sort each instant.
The two states alternate every quarter period, which is the same quarter-turn offset that separates sine from cosine. Every oscillation in physics has this structure, which is why one piece of mathematics describes springs, pendulums and circuits alike.
Two truths and a lie
All three are about harmonic motion.
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 conflates two quantities that are exactly out of phase. Where the acceleration is greatest — at the extremes — the mass is momentarily at rest. Where it moves fastest — at the centre — the acceleration is zero. Speed and acceleration peak a quarter period apart, which is the same relationship as between a sinusoid and its derivative.
Prediction
Commit before reasoning.
Predict first
A numerical check of the sine derivative gives 0.0151 where the rule predicts 0.866. What is wrong?
Correct: Degree mode: the factor pi over 180 has appeared.
\[ \frac{0.0151}{0.866} \approx 0.01745 = \frac{\pi}{180} \]
Why: The ratio of the two numbers is 0.01745, which is exactly pi over 180 — the value the fundamental limit takes in degrees. This is not rounding, which would produce a small random discrepancy rather than an exact constant factor, and not a step-size problem, which would shrink as the step shrinks. Recognising the size of the discrepancy identifies the cause immediately, and it is the most practical consequence of everything in this section.
Comparison
Fill the blanks. Three plus the co-pattern is all that needs remembering.
Comparison matrix
| Function | Derivative | Sign |
|---|---|---|
| sin x | cos x | no minus |
| cos x | -sin x | minus: a co-function |
| tan x | sec^2 x | no minus |
| cot x | -csc^2 x | minus: a co-function |
| sec x | sec x tan x | no minus |
The third column is entirely determined by whether the name begins with co-. That pattern fixes half the memory load and all of the sign errors.
Pattern
Given a trigonometric expression to differentiate.
Step five is the one that catches the section's characteristic error. Every missing minus sign shows up as a derivative whose sign contradicts a graph you have known since Section 1.3.
Stewart, Calculus: Early Transcendentals 8e, §3.3 Derivatives of Trigonometric Functions §3.3, pp. 190-196
Check
The fundamental limit. Match the arguments.
Check your understanding
Evaluate the limit of sin(3x)/x as x approaches 0.
Answer: A
Why: Rewrite as 3 times sin(3x)/(3x); the bracket tends to 1, leaving 3.
Check
The co-function pattern. Watch the sign.
Check your understanding
What is the derivative of cot x?
Answer: A
Why: Cotangent is a co-function, so its derivative carries a minus sign.
Check
The cycle. Read the remainder.
Check your understanding
What is the 74th derivative of sin x?
Answer: A
Why: 74 leaves remainder 2 on division by 4, and two steps from sine gives minus sine.
Real world
An alternating current supply delivers a voltage V(t) = 170 sin(120 pi t) volts, where t is in seconds. Engineers care about both the peak voltage and the maximum rate at which it changes, because a fast-changing voltage induces currents in nearby circuits.
Discussion prompt
Find the maximum rate of change of the voltage, say when it occurs, and explain why it is not at the peak voltage.
Hint: The rate of change is the derivative, and its maximum is where the original is steepest.
Answer:
\[ V'(t) = 170\cdot 120\pi\cos(120\pi t) = 20400\pi\cos(120\pi t) \]
The derivative is a cosine of the same frequency, so its maximum magnitude is its amplitude:
\[ |V'|_{\max} = 20400\pi \approx 64{,}088 \text{ volts per second} \]
It occurs where the cosine is at plus or minus 1, which is where the sine is zero — that is, as the voltage passes through zero, not at its peak.
This is the same alternation as the mass on a spring: the quantity is changing fastest where it is momentarily zero, and momentarily unchanging where it is largest. At the 170-volt peak the derivative is zero and the voltage is instantaneously steady.
The engineering consequence is real. Interference is induced by the RATE of change, not the magnitude, so the worst moment for a neighbouring circuit is the zero crossing — which is counterintuitive until you have seen the derivative. Note too where the large factor came from: the 120 pi inside the sine was multiplied out front by the chain rule of Section 3.6, and it is what makes the rate so much larger than the voltage itself.
Commit first
Answer, then rate your confidence honestly.
Predict first
Why does calculus require radians for trigonometric derivatives?
Correct: Because the fundamental limit equals 1 only in radians.
\[ \text{degrees: } \frac{d}{dx}\left[\sin x\right] = \frac{\pi}{180}\cos x \]
Why: The squeeze proof uses the sector-area formula, which holds only in radian measure. In degrees the limit becomes pi over 180, so the derivative of sine would be that factor times cosine, every second derivative would carry it squared, and no formula in this section would be clean. Precision has nothing to do with it — both measures are exact — and degrees describe angles beyond 360 perfectly well. The rules genuinely fail in degrees, which is why calculator mode matters.
Explain it
Their calculator is in degree mode and they cannot see why their numerical derivative of sine is a hundredth of what the rule predicts.
Discussion prompt
In four sentences or fewer, diagnose it from the size of the discrepancy alone.
Hint: Ask them to divide their answer by the predicted one.
Answer:
Have them divide their result by the predicted one. The ratio comes out as 0.01745, which is not a random rounding error but exactly pi divided by 180 — the conversion factor from degrees to radians.
That constant appears because the limit of sine h over h, which the whole derivative rests on, equals 1 in radians and pi over 180 in degrees. So every trigonometric derivative in degree mode is scaled by that factor. Switching to radian mode fixes it, and the exactness of the ratio is what identifies the cause rather than just the symptom.
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 the limits, force the denominator to match the sine's argument before applying anything. For signs, check whether the name begins with co- and confirm against a graph's direction. For the derivations, write the function as sine over cosine or one over cosine first and let the quotient rule do the rest. For the cycle, use the remainder on division by four, never the quotient. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, draw sine above cosine on a shared axis over one full period, dropping vertical guide lines from sine's peaks and troughs to cosine's zeros, and write beneath why this picture already shows the derivative rule. Below, write both fundamental limits with their values, and beside each note where it comes from and what it becomes in degrees. In the middle of the page, prove that the derivative of sine is cosine in full, marking on each line where the addition formula is used and where each fundamental limit enters. Then write all six trigonometric derivatives in a table, circling the three that carry a minus sign and writing the one-word reason. At the bottom, draw the four-step cycle as a loop and use it to write the 74th derivative of sine. Beside it, take the spring position 3 cosine t, differentiate twice, and write the equation relating acceleration to position, with a sentence saying what it means physically. In a margin, write what the derivative of sine becomes in degree mode.
If your six-derivative table has a minus sign on any function whose name does not begin with co-, check it against a graph: tangent and secant both increase on the branch just right of the origin, so neither derivative can be negative there.
Recap
Five things, and the first is where Section 1.3's insistence on radians finally pays.
| If you see | Then |
|---|---|
| sin(kx)/x in a limit | Match the arguments: the answer is k |
| An angle in degrees | Convert before differentiating anything |
| A co-function | Its derivative carries a minus sign |
| tan or sec | Rewrite over cosine and use the quotient rule |
| A high-order derivative of sine | Divide the order by four, use the remainder |
| Acceleration equal to minus the position | Simple harmonic motion |
| A numerical derivative off by 0.01745 | The calculator is in degree mode |
Section 3.6 supplies the one rule still missing: the chain rule, for functions built by composition. It is what lets you differentiate the sine of a polynomial, and it is the rule the alternating-current example above quietly used.
OpenStax Calculus Volume 1, §3.5 Derivatives of Trigonometric Functions §3.5, pp. 241-249 — everything on these slides traces back here
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