Mathematical Rules of Thumb, illustrated reader · Chapter 3

03Trigonometry

Control Angles, Directions, and Oscillations

4 demonstrations follow the chapter's rules. Choose a value, watch the figure and the numbers change, and check your prediction. Every choice is precomputed from the notebook calculations.

Ask the chapter skill

“Help me use Chapter 3 for my question. Choose a rule, check its assumptions, and show how the result changes when an input changes.”

Use math-thumb-trigonometry from the companion's skill package. The demonstrations below also work on their own.

Examples use constructed inputs or the book's own values, disclosed in each panel. A picture illustrates a rule; its assumptions set its scope.

1Demonstration 1 of 4

Put an error budget on a small angle

Can you substitute 5 directly for a five-degree angle?

The calculation converts degrees to radians before comparing sine with the angle and its cubic bound.

|sin⁡θ−θ|≤|θ|36 |\sin\theta-\theta|\leq\frac{|\theta|^3}{6}

Angle in degrees. Angles in radians inside the formula. The example uses positive angles below one radian.

Predict first. Can you substitute 5 directly for a five-degree angle?

Choose an example

Put an error budget on a small angle. 5° is 0.0872665 radians. Substitute radians into the approximation; the error is bounded by 0.000110762.
Angle in degrees: 5
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Angle in radians
0.0872665
Sine
0.0871557
Approximation
0.0872665
Error bound
0.000110762

5° is 0.0872665 radians. Substitute radians into the approximation; the error is bounded by 0.000110762.

Use the idea

Use rule 3.1.1 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Angles in radians inside the formula. The example uses positive angles below one radian.

Check your understanding: Can you substitute 5 directly for a five-degree angle?
No. Five degrees equals 5π/180 radians. Substituting 5 would change both the angle and the approximation regime.

Book source: Rule 3.1.1: Replace sine by the angle at small scale. Demonstration C03-D01. Worked illustration.

2Demonstration 2 of 4

Keep direction when a ratio loses the quadrant

Why is atan(y/x) wrong for the vector (−1,1)?

A component ratio alone cannot tell opposite vectors apart. The vector picture shows the information atan2 retains.

θ=atan2⁡(y,x) \theta=\operatorname{atan2}(y,x)

Vector direction in degrees. Nonzero unit vectors; the principal angle may be negative for directions above 180 degrees.

Predict first. Why is atan(y/x) wrong for the vector (−1,1)?

Choose an example

Keep direction when a ratio loses the quadrant. The vector and its opposite share the component ratio y/x. Here atan2 gives 135°. Plain atan gives -45°, which points along the opposite vector.
Vector direction in degrees: 135
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

atan2 direction (degrees)
135
Plain atan ratio (degrees)
-45

The vector and its opposite share the component ratio y/x. Here atan2 gives 135°. Plain atan gives -45°, which points along the opposite vector.

Use the idea

Use rule 3.2.1 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Nonzero unit vectors; the principal angle may be negative for directions above 180 degrees.

Check your understanding: Why is atan(y/x) wrong for the vector (−1,1)?
The ratio is −1, whose principal arctangent is −45°. The vector is in quadrant II, so atan2 gives 135°.

Book source: Rule 3.2.1: Use atan2, not a plain arctangent, for direction. Demonstration C03-D02. Worked illustration.

3Demonstration 3 of 4

Avoid subtracting nearly equal floating numbers

If direct subtraction returns zero, must the exact quantity be zero?

The two expressions are mathematically identical. Their floating-point evaluations can disagree at tiny angles.

1−cos⁡x=2sin⁡2(x/2) 1-\cos x=2\sin^2(x/2)

Angle x in radians. Binary64 arithmetic. The stable expression reduces cancellation; it does not remove every rounding error.

Predict first. If direct subtraction returns zero, must the exact quantity be zero?

Choose an example

Avoid subtracting nearly equal floating numbers. At x=1e-08, direct subtraction gives 0, while the equivalent half-angle expression gives 5e-17. Binary64 rounding made cos(x) exactly 1, so the direct result was erased.
Angle x in radians: 1e-08
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Direct 1 − cos(x)
0
Half-angle identity
5e-17
Relative difference
1

At x=1e-08, direct subtraction gives 0, while the equivalent half-angle expression gives 5e-17. Binary64 rounding made cos(x) exactly 1, so the direct result was erased.

Use the idea

Use rule 3.3.4 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Binary64 arithmetic. The stable expression reduces cancellation; it does not remove every rounding error.

Check your understanding: If direct subtraction returns zero, must the exact quantity be zero?
No. Rounding can make cos(x) indistinguishable from one even though 2 sin²(x/2) retains the small positive quantity.

Book source: Rule 3.3.4: Use a half-angle identity for one minus cosine. Demonstration C03-D03. Worked illustration.

4Demonstration 4 of 4

Count the triangles in the sine-law ambiguous case

Why can arcsin give a wrong answer when a=7?

Side a hangs from the top vertex and swings onto the base line. It can miss, hit twice, or hit once on the valid side.

sin⁡B=bsin⁡Aa \sin B=\frac{b\sin A}{a}

Side a opposite A (A=30°, b=10). Given two sides and an acute angle not between them. Height h=b sin A=5.

Predict first. Why can arcsin give a wrong answer when a=7?

Choose an example

Count the triangles in the sine-law ambiguous case. a=7 is between the height 5 and b=10, so it meets the base twice: two valid triangles. Arcsin returns only the acute B; the obtuse angle 180° minus B is also valid.
Side a opposite A (A=30°, b=10): 7
Constructed teaching inputs; calculations executed locally. Supported menu choices are precomputed.

Calculated values

Height b·sin A
5
Side a
7
Number of triangles
2

a=7 is between the height 5 and b=10, so it meets the base twice: two valid triangles. Arcsin returns only the acute B; the obtuse angle 180° minus B is also valid.

Use the idea

Use rule 3.3.2 when its stated conditions fit. Compare the calculation with your own decision threshold; retain the relevant error or uncertainty.

Where the conclusion applies

Given two sides and an acute angle not between them. Height h=b sin A=5.

Check your understanding: Why can arcsin give a wrong answer when a=7?
Arcsin returns only the acute angle B. With h<a<b, the obtuse angle 180° minus B also fits, so there are two triangles.

Book source: Rule 3.3.2: Check both branches in the sine-law ambiguous case. Demonstration C03-D04. Worked illustration.

Bring the idea to a question of your own

Choose the relationship that answers your question, check its conditions, and compare the result with the accuracy or decision threshold you need.

The chapter skill can adapt these calculations to your inputs. It should name the assumptions, explain what the result supports, and say what still needs evidence. The chapter workbook adds a lab and three exercises with answers.