Trigonometric Functions · Lesson 5 of 5
Chapter Summary and Practice
“Connect angle measure, circle coordinates, graphs and identity strategies through a complete chapter review.”
• Connect signed rotations and radian measures to the unit-circle definitions of the six functions. • Use domains, ranges, quadrant signs and periods to interpret and evaluate trigonometric functions. • Recall the identity families through their derivation relationships rather than isolated memorisation. • Select and justify transformations for mixed exact-value and proof problems. • Retain denominator restrictions and choose half-angle signs from the correct interval. • Solve mixed chapter problems and explain the reasoning behind each step.
One chapter, one connected model
The chapter begins with a rotation and ends with methods for combining the values produced by rotations. Radians connect the rotation to arc length; the unit circle connects it to coordinates; graphs show how those coordinates and their quotients repeat. Identities describe what happens when angles are added, subtracted, doubled, tripled or halved.
Keep this chain in mind while reviewing. A sign question can often be answered by a circle diagram, a graph break by a denominator, and a complicated identity by a simpler angle combination. Each approach describes the same underlying functions from a different viewpoint.
| Chapter coverage | Core connection | What to check |
|---|---|---|
| Signed angles and units | Initial ray → rotation → terminal ray | Direction, full turns, degrees/minutes/seconds |
| Radians and arc length | Arc/radius determines angular sweep | Convert degrees; distinguish radius, chord and arc |
| Real radian inputs | A number line can wrap around the unit circle | Different inputs can share a terminal point |
| Unit-circle definitions | P = (cos x, sin x) | Horizontal versus vertical coordinates |
| Six functions and standard values | Reciprocals and quotients of sine and cosine | Nonzero denominators; exact reference values |
| Fundamental identities | Squared coordinates give the unit-radius equation | Division restrictions in tangent/cotangent versions |
| Quadrant signs and reflection | Coordinate signs and symmetry | Choose signs before taking reciprocals |
| Domains, ranges and graphs | Allowed inputs and attainable outputs | Axis values, breaks and unbounded behaviour |
| Periodicity | Repeated terminal points or repeated quotients | 2π for sine/cosine/sec/cosec; π for tan/cot |
| Angle sums and differences | Equal-chord geometry gives cosine addition | Product patterns and signs |
| Related and complementary angles | Substitute axis values into addition identities | Function-name changes under quarter-turn shifts |
| Double and triple angles | Equal angles or 2x + x in addition formulas | Select a useful form and check its domain |
| Sum–product conversions | Add/subtract paired identities | Half-sum, half-difference and term order |
| Mixed and half-angle reasoning | Combine algebra, identities and interval information | Root selection and the actual interval of x/2 |
| Historical applications | Triangles, astronomy, surveying and repeated motion | Connect mathematical representations to real uses |
Angle measurement is part of the calculation
An angle’s number is meaningful only with its unit. A degree angle must be converted before using the arc-length formula. A signed angle and a travelled distance also answer different questions: the first includes direction, while the second is a nonnegative length.
One degree contains 60 angular minutes and 3600 angular seconds. For example, 30′ means half a degree, not thirty hundredths of a degree. Full turns change a rotation measure even when they preserve the terminal ray: 60° and 420° are different rotations with the same endpoint direction.
Problem
A point moves clockwise through 150° on a circle of radius 6 cm, starting on the positive horizontal ray. Find its signed radian rotation, distance travelled, and unit-circle sine and cosine values at that rotation.
- 1.The signed angle is −150° = −150π/180 = −5π/6 radians.
- 2.The travelled arc length is 6|−5π/6| = 5π cm. The negative direction does not make the distance negative.
- 3.The positive angle 5π/6 is π − π/6, so sin(5π/6) = 1/2 and cos(5π/6) = −√3/2.
- 4.Reflection gives sin(−5π/6) = −1/2 and cos(−5π/6) = −√3/2. The terminal point lies in quadrant III, consistent with both coordinates being negative.
Values, signs and graphs are three views of a function
The unit-circle coordinate definition supplies exact values and signs. Quotient and reciprocal definitions supply restrictions. The graph records all allowed input–output pairs, including repeated patterns and branches separated by undefined inputs.
| Function | Definition | Domain exclusions | Range | Fundamental period |
|---|---|---|---|---|
| sin x | Vertical unit-circle coordinate | None | [−1, 1] | 2π |
| cos x | Horizontal unit-circle coordinate | None | [−1, 1] | 2π |
| tan x | sin x/cos x | π/2 + nπ | ℝ | π |
| cot x | cos x/sin x | nπ | ℝ | π |
| sec x | 1/cos x | π/2 + nπ | y ≤ −1 or y ≥ 1 | 2π |
| cosec x | 1/sin x | nπ | y ≤ −1 or y ≥ 1 | 2π |
Here n is any integer. At quadrantal angles, use the coordinates (1, 0), (0, 1), (−1, 0) and (0, −1) before forming quotients. These values show both the zeros and the undefined inputs. They also explain why tangent and cotangent are reciprocal only on the part of their domains where that reciprocal expression is meaningful.
In quadrant I all six values are positive. In quadrant II sine and cosecant are positive; in quadrant III tangent and cotangent are positive; in quadrant IV cosine and secant are positive. The remaining functions are negative within each open quadrant. Axes must be checked separately because zero and undefined are neither positive nor negative values.
Reflection preserves cosine and secant, and reverses sine, cosecant, tangent and cotangent. A full turn preserves every defined value. A half turn changes both sine and cosine signs but leaves their quotient unchanged, explaining tangent’s and cotangent’s shorter period. The graph behaviour follows these facts rather than requiring six unrelated memorised pictures.
Problem
If cos α = −5/13 and π/2 < α < π, find sin α, sin2α and cos2α.
- 1.The fundamental identity gives sin²α = 1 − 25/169 = 144/169. Quadrant II requires sin α = 12/13.
- 2.The double-angle sine is 2(12/13)(−5/13) = −120/169.
- 3.For cosine choose 2cos²α − 1: cos2α = 50/169 − 1 = −119/169.
- 4.The doubled point has both coordinates negative. Its coordinate squares sum to (14400 + 14161)/28561 = 1, providing an exact consistency check.
The graph may grow without bound near an excluded input, but it does not assign that input an output ∞. For example, tan(π/2) is undefined, while values at nearby allowed inputs can have arbitrarily large magnitude.
Reconstructing the identity families
The long formula list has a short logical foundation. Equal-chord geometry gives cosine addition. Reflection gives differences, and complementarity gives sine addition. Setting the two angles equal gives double angles; combining 2x and x gives triple angles; adding or subtracting paired formulas gives sum–product conversions.
Complementary-angle identities are sin(π/2 − x) = cos x and cos(π/2 − x) = sin x. Related-angle substitutions give sin(π − x) = sin x, cos(π − x) = −cos x, sin(π + x) = −sin x, cos(π + x) = −cos x, sin(2π − x) = −sin x and cos(2π − x) = cos x. Quarter-turn addition gives sin(π/2 + x) = cos x and cos(π/2 + x) = −sin x.
An identity involving only sine and cosine usually has all real inputs available. A quotient may lose some of them. This difference is why a formula table must carry its validity conditions, rather than treating the conditions as separate small print.
Choosing a transformation by the expression’s shape
Before expanding, look for a pattern that suggests a useful representation. Symmetric sums suggest half-sum and half-difference, products suggest combined angles, and squares suggest a double-angle expression. The goal is to expose cancellation or a known identity with as little unstructured algebra as possible.
| What you see | A useful first move | Reason |
|---|---|---|
| Two sine/cosine terms with equal spacing | Compute their half-sum and half-difference | The pair often shares a factor with another pair |
| A product of functions at different angles | Product-to-sum | New terms may match and cancel |
| Squared sine or cosine terms | Fundamental or double-angle form | Squares become a constant part and a cosine |
| A doubled or tripled input | Substitute into addition formulas | Larger angles reduce to known values |
| A known value plus a quadrant | Recover its sine/cosine partner | The quadrant selects the square-root sign |
| A half-angle task | Halve the given interval first | The function signs belong to the new angle |
| A quotient proof | Factor numerator and denominator, then check zeros | Cancellation must not restore excluded inputs |
Problem
Prove 2cos(π/13)cos(9π/13) + cos(3π/13) + cos(5π/13) = 0.
- 1.The product-to-sum formula makes the first term cos(10π/13) + cos(−8π/13).
- 2.Reflection gives cos(−8π/13) = cos(8π/13). Since 10π/13 = π − 3π/13, cos(10π/13) = −cos(3π/13).
- 3.Likewise 8π/13 = π − 5π/13, so cos(8π/13) = −cos(5π/13).
- 4.The expression is −cos(3π/13) − cos(5π/13) + cos(3π/13) + cos(5π/13) = 0. No difficult new reference-angle values had to be calculated.
Problem
Prove sin x + sin3x + sin5x + sin7x = 4cos x cos2x sin4x.
- 1.Pair the outside terms sin x + sin7x. Their half-sum is 4x and half-difference is −3x, so the sum is 2sin4x cos3x.
- 2.Pair the inside terms sin3x + sin5x. Their sum is 2sin4x cos x.
- 3.Factor the combined result as 2sin4x(cos3x + cos x). The cosine sum becomes 2cos2x cos x.
- 4.Multiplying gives 4sin4x cos2x cos x, the required result. The chosen grouping made both sine sums share the same factor sin4x.
Problem
Show that [(sin7x + sin5x) + (sin9x + sin3x)]/[(cos7x + cos5x) + (cos9x + cos3x)] equals tan6x where the original denominator is nonzero.
- 1.The first sine pair is 2sin6x cos x and the second is 2sin6x cos3x. Thus the numerator is 2sin6x(cos x + cos3x).
- 2.The cosine pairs similarly give denominator 2cos6x(cos x + cos3x).
- 3.A nonzero original denominator requires both cos6x and cos x + cos3x to be nonzero. Cancel the shared factor to obtain sin6x/cos6x = tan6x.
- 4.The expression is not defined at any additional input merely because tan6x happens to exist there. Keep the original denominator condition.
Problem
Prove (cos x − cos y)² + (sin x − sin y)² = 4sin²((x − y)/2).
- 1.Expand both squares. Group cos²x + sin²x and cos²y + sin²y to obtain 2 − 2(cos x cos y + sin x sin y).
- 2.The expression in parentheses equals cos(x − y), so the result is 2[1 − cos(x − y)].
- 3.Use 1 − cos t = 2sin²(t/2) with t = x − y. The result becomes 4sin²((x − y)/2).
- 4.The left side is the squared distance between two unit-circle endpoints. The identity therefore connects chord distance with half the angular separation.
Problem
Prove (cos x + cos y)² + (sin x − sin y)² = 4cos²((x + y)/2).
- 1.Expanding and grouping the four squared terms gives 2 + 2cos x cos y − 2sin x sin y.
- 2.The two product terms combine as 2cos(x + y), so the expression is 2[1 + cos(x + y)].
- 3.Use 1 + cos t = 2cos²(t/2). This gives 4cos²((x + y)/2).
- 4.The sign choices in the original expression determined whether an angle sum or difference emerged. Carefully tracking them is more reliable than guessing from the final appearance.
Half-angle review: magnitude and sign are separate decisions
The squared half-angle formulas provide magnitudes from cos x. They do not by themselves give signed sine or cosine values. The original interval must be halved first, so that the signs belong to the correct new angle.
Problem
If tan x = −4/3 and π/2 < x < π, find sin(x/2), cos(x/2) and tan(x/2).
- 1.The original angle has positive sine and negative cosine. From sec²x = 1 + 16/9 = 25/9, obtain sec x = −5/3 and cos x = −3/5.
- 2.The halved interval is π/4 < x/2 < π/2, so sine and cosine of x/2 are both positive.
- 3.Their squares are (1 + 3/5)/2 = 4/5 and (1 − 3/5)/2 = 1/5. Hence sin(x/2) = 2/√5 and cos(x/2) = 1/√5.
- 4.The quotient gives tan(x/2) = 2. The original tangent was negative, but its half-angle tangent is positive because its quadrant has changed.
A similar restriction applies to numerical root-finding from tan2x: a quadratic gives candidates that must be filtered using the interval of x. Also retain any denominator condition used in forming the equation. Multiplication by a denominator is legitimate on the common domain, not an excuse to ignore where the original expression was undefined.
Applications and the development of the subject
Trigonometric ideas connect geometry with measuring and predicting change. Triangles support surveying and navigation; the circle model supports repeated behaviour in tides, vibrations and musical tones. Historical development brought together several mathematical traditions and different ways to represent the same relationships.
The historical account highlights contributions from Indian mathematicians and astronomers including Aryabhata, Brahmagupta, Bhaskara I and Bhaskara II. Indian astronomical works developed sine-based methods; mathematical knowledge travelled through the Middle East to Europe alongside the older Greek geometric tradition. Bhaskara I developed methods for sine values beyond 90°. The Yuktibhasa contains a treatment of the sine addition relationship, and Bhaskara II supplied exact sine or cosine expressions at angles such as 18°, 36°, 54° and 72°.
The historical note associates the inverse-function symbols sin⁻¹ and cos⁻¹ with John Herschel in 1813. These symbols are distinct from the reciprocal names cosec and sec. It also recalls accounts of Thales using shadows to find a height and similar triangles to estimate the distance of a ship at sea. Related height and distance problems occur in ancient Indian works. Here the measurement idea, rather than an additional inverse-function topic, connects back to the chapter’s triangle origins.
For a vertical object of height H casting a horizontal shadow of length S, and a smaller vertical staff of height h casting shadow length s under the same sun angle, similar right triangles give H/S = h/s. That common ratio is tan of the sun’s altitude. Equal angles therefore allow a known small height to help determine a much larger one.
Problem
A vertical 2 m staff casts a 3 m shadow. At the same time, a vertical structure casts a 15 m shadow on level ground. Find the structure’s height.
- 1.The two right triangles share the same sun angle, so the height-to-shadow ratios agree.
- 2.Let H be the structure’s height. Then H/15 = 2/3. Multiply both sides by 15 to get H = 10 m.
- 3.The calculation uses similarity, which is the same angle–side relationship behind the tangent ratio. The larger shadow is five times the smaller, so its height is also five times as large.
Mixed chapter checks
These questions combine the models and identities reviewed above. Before calculating, identify the unit, domain or angle interval that matters. In a proof, write the relationship that justifies each transformation so that the chain of reasoning can be followed independently.
Quiz
A point turns clockwise through 210° on a radius-4 cm circle. Which pair gives its signed radian angle and travelled arc length?
Which statement correctly connects a graph break to its definition?
Which first move best simplifies cos5x + cos x?
If π/2 < x < π, which signs apply to sin(x/2) and cos(x/2)?
Which expression equals 1 − cos t for every real t?
A quotient simplifies to tan x after cancelling cos3x. What must remain part of its domain reasoning?
Which observation explains tangent’s half-turn period?
Why are sine and cosine double-angle identities less restricted than their tangent-based versions?
Practice Problems
- Prove 2cos(π/13)cos(9π/13) + cos(3π/13) + cos(5π/13) = 0 using product-to-sum and related-angle identities.
- Prove (sin3x + sin x)sin x + (cos3x − cos x)cos x = 0 by collecting the terms into a cosine difference.
- Prove (cos x + cos y)² + (sin x − sin y)² = 4cos²((x + y)/2), showing the expansion and identity substitutions.
- Prove (cos x − cos y)² + (sin x − sin y)² = 4sin²((x − y)/2), then interpret the left side as a squared chord distance.
- Prove sin x + sin3x + sin5x + sin7x = 4cos x cos2x sin4x by pairing terms with a common half-sum.
- Prove [(sin7x + sin5x) + (sin9x + sin3x)]/[(cos7x + cos5x) + (cos9x + cos3x)] = tan6x. State the original denominator restriction.
- Prove sin3x + sin2x − sin x = 4sin x cos(x/2)cos(3x/2), choosing a useful difference to group first.
- If tan x = −4/3 and π/2 < x < π, derive all three half-angle values and justify their signs.
- If cos x = −1/3 and π < x < 3π/2, derive sin(x/2), cos(x/2) and tan(x/2) exactly.
- If sin x = 1/4 and π/2 < x < π, recover cos x and then derive the three half-angle values.
- Convert −32°45′30″ to radians. Explain how the leading minus sign applies to every subdivision.
- A 9 cm radius sweeps a clockwise angle of 4π/3 radians. Find distance travelled and explain the coordinate signs of its terminal point from the right-facing initial ray.
- Evaluate sin(−11π/3), sec765° and cot(−15π/4), retaining exact values and checking denominators.
- State and explain the domains, ranges and periods of all six functions; sketch their characteristic behaviours over a suitable full repeating interval.
- Starting from the geometric cosine addition proof, explain the derivation chain that yields sine addition, double angles, triple angles and sum-to-product.
- If sin x = 5/13 in quadrant II and cos y = 4/5 in quadrant IV, find sin(x + y) and cos(x + y), showing both sign choices.
- Derive tan(π/8) from a double-angle equation and identify why one algebraic root is rejected.
- Derive cos4x = 1 − 8sin²x cos²x and cos6x = 32cos⁶x − 48cos⁴x + 18cos²x − 1.
- Show with a specific example why reducing an angle by 2π before halving can reverse half-angle sine and cosine signs.
- A vertical 1.5 m staff casts a 2 m shadow while a tower casts an 18 m shadow. Use similar triangles and explain the role of the common sun angle.
Key Takeaways
• Signed rotations, angle units and radians provide the inputs; unit-circle coordinates provide sine and cosine outputs. • The six functions’ signs, domains, ranges and periods follow from coordinates, quotients and reciprocals. • Addition identities generate difference, related-angle, double-angle and triple-angle identities. • Sum–product conversions reorganise an expression so that cancellation or a common factor becomes visible. • Square-root signs come from the appropriate angle interval, especially after halving. • A complete solution explains its transformations, keeps exact values when possible and preserves the original domain.
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Mixed Applications and Half-Angle Reasoning
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