Additional Mathematics Synthesis Guide 19: Trigonometric Calculus — Differentiation, Integration, Stationary Points and Exact Areas

Radians → derivative → sign → stationary point → antiderivative → accumulation → exact area.

1. Why the standard calculus formulas use radians

The familiar formulas

• d(sinx)/dx=cosx
• d(cosx)/dx=−sinx
• d(tanx)/dx=sec²x

assume x is measured in radians. Radian measure is built from arc length and makes the limiting ratio behind these derivatives equal to 1.

If x were measured numerically in degrees, an extra conversion factor would appear. For school calculus, use radians whenever applying the standard trigonometric derivative and integral formulas unless a question explicitly defines another convention.

2. The basic derivative set

For radian x:

• d(sinx)/dx=cosx
• d(cosx)/dx=−sinx
• d(tanx)/dx=sec²x, where tanx is defined

The sign in the cosine derivative is especially important. Differentiating cosx to sinx loses the direction of change: near x=π/2, cosine is decreasing through zero, so its derivative should be negative.

3. The chain rule supplies the inner-angle factor

Differentiate

y=sin(3x−1).

The outer derivative is cosine and the inner derivative is 3, so

y′=3cos(3x−1).

Similarly,

d[cos(5x+2)]/dx=−5sin(5x+2).

The inner coefficient is not an optional extra. It is the rate at which the angle itself changes with x.

4. Tangents to trigonometric curves

Find the tangent to y=2sinx+cosx at x=0.

The point is (0,1). The derivative is

y′=2cosx−sinx.

At x=0, the gradient is 2. Therefore the tangent is

y−1=2x,

or y=2x+1.

The point came from the original function; the gradient came from the derivative. Keeping those roles separate avoids a common substitution error.

5. Product-rule synthesis with a trigonometric factor

Differentiate

f(x)=xsinx.

Using the product rule:

f′(x)=sinx+xcosx.

This derivative is not obtained by multiplying the separate derivatives. The product rule preserves one factor while differentiating the other, then reverses the roles.

Stationary points of xsinx require solving sinx+xcosx=0, which generally does not reduce to elementary exact angles. This is a useful reminder that a derivative can be exact even when the resulting equation needs numerical methods outside a basic exact-solution exercise.

6. Stationary points of a simple trigonometric combination

Let

f(x)=sin2x+2cosx

for 0≤x≤2π. Then

f′(x)=2cos2x−2sinx.

Set f′=0:

cos2x=sinx.

Use cos2x=1−2sin²x. Let u=sinx:

1−2u²=u.

Thus 2u²+u−1=0, so (2u−1)(u+1)=0. Therefore

sinx=1/2 or sinx=−1.

Hence the stationary x-values are π/6, 5π/6 and 3π/2.

7. Completing the stationary-point coordinates

For f(x)=sin2x+2cosx:

• f(π/6)=sin(π/3)+2cos(π/6)=√3/2+√3=3√3/2.
• f(5π/6)=sin(5π/3)+2cos(5π/6)=−√3/2−√3=−3√3/2.
• f(3π/2)=sin3π+2cos(3π/2)=0.

The stationary points are therefore the full ordered pairs, not just the three x-values.

8. Classify by derivative signs

For the previous example, test the sign of f′ on the intervals separated by π/6, 5π/6 and 3π/2. A positive-to-negative change gives a local maximum; negative-to-positive gives a local minimum. If the sign does not change, the point is stationary without being a local turning extremum.

This method is often safer than assuming every solution of f′=0 is a maximum or minimum. Periodic functions can contain repeated patterns and stationary inflexion-like behaviour in more complicated examples.

9. Second derivative as a classification shortcut

For f(x)=sin2x+2cosx,

f″(x)=−4sin2x−2cosx.

At a stationary point, a negative second derivative indicates a local maximum and a positive second derivative a local minimum, provided the standard smoothness conditions hold.

If f″=0, the test is inconclusive. Do not automatically rename the point as an inflexion point without further evidence.

10. R-form can simplify derivative equations

Suppose

f(x)=3sinx−4cosx.

Then

f′(x)=3cosx+4sinx.

Guide 17 rewrites this as 5cos(x−α), where cosα=3/5 and sinα=4/5. Therefore stationary points satisfy

cos(x−α)=0.

This may be cleaner than manipulating the original sine-cosine equation term by term.

R-form and calculus are different chapters on paper, but the derivative can create exactly the structure R-form is designed to simplify.

11. Basic trigonometric antiderivatives

Again with radian x:

• ∫cosx dx=sinx+C
• ∫sinx dx=−cosx+C
• ∫sec²x dx=tanx+C on intervals where tanx is defined

The negative sign in ∫sinx dx is required because d(−cosx)/dx=sinx.

Differentiating a proposed antiderivative is the quickest structural check.

12. Reverse chain rule for linear angles

Evaluate

∫cos(3x) dx.

An antiderivative is

(1/3)sin(3x)+C.

Similarly,

∫sin(4x−1) dx=−(1/4)cos(4x−1)+C.

The reciprocal inner coefficient compensates for the chain-rule multiplier that differentiation would produce.

13. Definite integral with exact special angles

Evaluate

∫₀^π/32cosx dx.

An antiderivative is 2sinx. Therefore

2sin(π/3)−2sin0=√3.

The exact answer √3 is preferable to an early decimal because the endpoint is a standard exact angle.

14. Signed integral versus geometric area

Evaluate the signed integral of sinx from 0 to 2π:

∫₀^2πsinx dx=[−cosx]₀^2π=0.

The geometric area between the curve and x-axis is not zero. The positive area from 0 to π and the negative signed contribution from π to 2π cancel in the integral.

Total geometric area is

∫₀^πsinx dx − ∫_π^2πsinx dx = 2+2=4.

15. Find sign changes before an area calculation

For y=2sinx−1 on 0≤x≤π, solve 2sinx−1=0. The roots are x=π/6 and 5π/6.

The graph is negative near 0, positive between the roots, and negative near π. A total-area calculation must therefore split into three regions.

An antiderivative is

F(x)=−2cosx−x.

The sign analysis determines which definite contributions should be negated when converting signed integrals into geometric area.

16. Worked exact area for 2sinx−1

On [0,π/6], the signed integral is

F(π/6)−F(0)=2−√3−π/6

which is negative, so the geometric area is √3+π/6−2.

By symmetry, the right outer region has the same area.

The central signed area is

F(5π/6)−F(π/6)=2√3−2π/3.

Therefore total area is

2(√3+π/6−2)+(2√3−2π/3)=4√3−4−π/3.

A sketch provides an immediate check that the total area should be positive.

17. Area between a sine curve and a horizontal line

Suppose the region is bounded by y=sinx and y=1/2 between their intersections in [0,π]. The intersections occur at x=π/6 and 5π/6.

On that interval sinx≥1/2, so the area is

∫_π/6^5π/6(sinx−1/2) dx.

Evaluate:

[−cosx−x/2]_π/6^5π/6=√3−π/3.

Upper-minus-lower is the geometric relationship. Integrating sinx alone would answer a different area question.

18. Periodicity can simplify a definite integral

Over a complete period, the signed integral of sinx or cosx is zero:

∫_a^a+2πsinx dx=0,

∫_a^a+2πcosx dx=0.

For a shifted sinusoid M+Rcos(x−α), the integral over one complete period is 2πM because the oscillatory cosine contribution cancels.

This gives a useful interpretation: the mean value over a full period is the midline M.

19. Worked mean-value connection

Let

f(x)=7+3cosx+4sinx.

Over 0≤x≤2π,

∫f(x)dx=∫7dx+3∫cosx dx+4∫sinx dx=14π.

Divide by the interval length 2π to obtain average value 7.

R-form also gives f(x)=7+5cos(x−α), making the same mean value visually obvious from the midline.

20. Optimisation on a restricted trigonometric interval

Let f(x)=2sinx+cosx on 0≤x≤π/2. Find the greatest value.

Differentiate:

f′(x)=2cosx−sinx.

Set f′=0: tanx=2. The stationary point lies inside the interval. If tanx=2 in Quadrant I, take a right triangle with opposite 2 and adjacent 1, giving sinx=2/√5 and cosx=1/√5. Hence

f=5/√5=√5.

Endpoints give f(0)=1 and f(π/2)=2, so the global maximum on the interval is √5.

R-form would also give amplitude √5, but the endpoint check remains essential whenever the allowed interval may not contain an amplitude-achieving phase.

21. Restricted intervals can block the global sinusoidal maximum

Suppose f(x)=3cosx+4sinx but x is restricted to 0≤x≤π/6. The unrestricted amplitude is 5, attained at x=α=tan⁻¹(4/3)≈0.927, which is greater than π/6≈0.524.

The amplitude value 5 is therefore not achievable on this restricted interval. Differentiate or inspect the R-form phase to see that f is increasing over the interval, so the maximum occurs at x=π/6:

f(π/6)=3√3/2+2.

A global formula must still respect the problem’s feasible domain.

22. Common failure patterns

• Using degree mode with standard radian derivative formulas.
• Omitting the inner-angle factor in the chain rule.
• Differentiating cosx to +sinx.
• Solving only one trigonometric branch for f′=0.
• Reporting stationary x-values without y-coordinates when points are requested.
• Assuming every stationary point is a maximum or minimum.
• Forgetting the reciprocal inner factor during integration.
• Taking the absolute value of one net definite integral and calling it total area after sign cancellation has already occurred.

23. A reliable trigonometric-calculus routine

1. Confirm radian measure for standard calculus formulas.
2. Differentiate with the correct chain, product or quotient structure.
3. For stationary points, solve the derivative equation over the complete stated interval.
4. Recover y-coordinates from the original function.
5. Classify with sign change or a valid second-derivative test.
6. For integration, verify the antiderivative by differentiation when uncertain.
7. For geometric area, locate every sign or boundary change first.
8. Keep exact special-angle values until the requested approximation stage.

24. Practice set

1. Differentiate 4sinx−3cosx.
2. Differentiate sin(5x).
3. Differentiate cos(2x−1).
4. Differentiate xcosx.
5. Find the tangent to y=sinx at x=π/3.
6. Find stationary x-values of f(x)=sinx+cosx on 0≤x<2π.
7. Find the corresponding stationary values in Question 6.
8. Use R-form to state the maximum of sinx+cosx.
9. Find the greatest value of sinx+cosx on 0≤x≤π/6.
10. Integrate cos(4x).
11. Integrate sin(3x+2).
12. Evaluate ∫₀^π/2cosx dx.
13. Evaluate ∫₀^πsinx dx.
14. Find the signed integral of sinx on [0,2π].
15. Find the total area between y=sinx and the x-axis on [0,2π].
16. Find the area between y=cosx and the x-axis on [−π/2,π/2].
17. Find the area between y=sinx and y=1/2 between their intersections in [0,π].
18. Find the average value of 5+2cosx on [0,2π].
19. Explain why an amplitude found by R-form may fail to be the maximum on a restricted interval.
20. Explain why ∫|f(x)|dx and |∫f(x)dx| answer different questions.

Answers

1. 4cosx+3sinx.
2. 5cos5x.
3. −2sin(2x−1).
4. cosx−xsinx.
5. Point (π/3,√3/2), gradient 1/2, so y−√3/2=(1/2)(x−π/3).
6. cosx−sinx=0, so tanx=1; x=π/4,5π/4.
7. √2 and −√2 respectively.
8. √2.
9. The unrestricted maximum occurs at π/4, outside the interval; f is increasing there, so maximum at π/6 is 1/2+√3/2.
10. (1/4)sin4x+C.
11. −(1/3)cos(3x+2)+C.
12. 1.
13. 2.
14. 0.
15. 4.
16. 2.
17. √3−π/3.
18. 5.
19. Because the phase that achieves amplitude R may lie outside the allowed interval; endpoints and feasible stationary points must be checked.
20. The first accumulates magnitudes and therefore geometric/total variation; the second takes the magnitude only after positive and negative contributions may have cancelled.

25. What mastery looks like

Mastery means the learner can differentiate and integrate trigonometric expressions in radians, preserve chain-rule factors, solve derivative equations completely over intervals, use R-form when it simplifies the calculus, distinguish local extrema from restricted-domain global extrema, and convert signed integrals into geometric area only after checking the graph’s sign structure.

The transfer test is to mix these roles. Give a derivative that becomes an R-form equation, or an R-form model whose exact area over a period is required. If the learner can choose the representation that makes the next mathematical job simpler, the chapters are beginning to operate as one system.

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• Guide 17: R-Form Trigonometry, Amplitude, Phase, Maxima and Modelling
• Guide 18: Linear Law Transformations, Straight-Line Graphs, Parameter Recovery and Model Checking
• Guide 20: Coordinate Circles, Lines, Chords, Tangents and Intersection Conditions
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