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Trigonometry

Edexcel International A Level Mathematics • Deep Theory • Identities • Equations • Graphs • Radians • Triangle Geometry

1. Trigonometry — The Core Idea

Trigonometry is the study of relationships between angles and lengths. At its simplest level, it connects the sides of a right-angled triangle. At A Level, the subject develops into a much broader system involving functions, identities, equations, graphs, radians and non-right-angled triangles.

Important distinction: A trigonometric ratio is a relationship between sides of a triangle, whereas a trigonometric function such as sin x is a function defined for angles measured in degrees or radians.

Angle notation

The usual convention is to use capital letters for angles and lowercase letters for their opposite sides:

a is opposite A
b is opposite B
c is opposite C

Degrees and radians

Degrees divide a complete revolution into 360 equal parts. Radians measure an angle using the arc length of a circle.

180° = π radians
360° = 2π radians
θ radians = θ × 180/π degrees
θ degrees = θ × π/180 radians
Calculator warning: If a question uses π, radians or an interval such as 0 ≤ x ≤ 2π, your calculator must normally be in RADIAN mode. Degree questions require DEGREE mode.

2. Basic Trigonometric Ratios

In a right-angled triangle, the three fundamental ratios are sine, cosine and tangent.

sin θ = opposite / hypotenuse
cos θ = adjacent / hypotenuse
tan θ = opposite / adjacent
SOH — CAH — TOA

Finding a side

If the angle and another side are known, choose the ratio that contains the required side and the known side.

Worked Example

A right-angled triangle has an angle of 37° and a hypotenuse of 12 cm. Find the side opposite the angle.

sin 37° = opposite / 12
opposite = 12 sin 37°
opposite ≈ 7.223 cm

Finding an angle

Use an inverse trigonometric function when a side ratio is known.

θ = sin⁻¹(value)
θ = cos⁻¹(value)
θ = tan⁻¹(value)
The notation sin⁻¹θ means the inverse function arcsin θ. It does not mean 1/sin θ. The reciprocal of sin θ is cosec θ.

3. Exact Trigonometric Values

The standard angles 0°, 30°, 45°, 60° and 90° produce exact values that are fundamental to A Level trigonometry.

Sine

sin 0° = 0
sin 30° = 1/2
sin 45° = √2/2
sin 60° = √3/2
sin 90° = 1
Cosine

cos 0° = 1
cos 30° = √3/2
cos 45° = √2/2
cos 60° = 1/2
cos 90° = 0
Tangent

tan 0° = 0
tan 30° = √3/3
tan 45° = 1
tan 60° = √3

Why do these values occur?

The values can be derived from special triangles rather than memorised as isolated facts.

45°–45°–90° triangle

If the equal perpendicular sides have length 1, Pythagoras gives a hypotenuse of √2.

sin 45° = 1/√2 = √2/2
cos 45° = 1/√2 = √2/2
tan 45° = 1

30°–60°–90° triangle

The side lengths can be represented as 1, √3 and 2.

sin 30° = 1/2
cos 30° = √3/2
tan 30° = 1/√3 = √3/3

4. The Unit Circle and Trigonometric Signs

The unit circle gives a powerful way to understand trigonometric functions for angles beyond the first quadrant.

A point on the unit circle corresponding to angle θ has coordinates:

(cos θ, sin θ)

Therefore the x-coordinate represents cosine and the y-coordinate represents sine.

Signs in the four quadrants

Quadrant I: sin +, cos +, tan +

Quadrant II: sin +, cos −, tan −

Quadrant III: sin −, cos −, tan +

Quadrant IV: sin −, cos +, tan −

CAST

A → All positive
S → Sine positive
T → Tangent positive
C → Cosine positive

CAST is a memory aid. The actual mathematics comes from the signs of the coordinates and tan θ = sin θ / cos θ.

5. Reciprocal Trigonometric Functions

The reciprocal functions are cosecant, secant and cotangent.

cosec θ = 1/sin θ
sec θ = 1/cos θ
cot θ = 1/tan θ = cos θ/sin θ

Important identities

sin²θ + cos²θ = 1
1 + tan²θ = sec²θ
1 + cot²θ = cosec²θ

Domain restrictions

Since a reciprocal cannot have zero as its denominator:

sec θ is undefined when cos θ = 0
cosec θ is undefined when sin θ = 0
cot θ is undefined when sin θ = 0

6. Trigonometric Identities

An identity is an equation that is true for every value for which both sides are defined.

An equation is something you solve for particular values. An identity is a relationship that remains true throughout its entire valid domain.

The fundamental identity

sin²θ + cos²θ = 1

Deriving the tangent identity

Divide the fundamental identity by cos²θ:

sin²θ/cos²θ + cos²θ/cos²θ = 1/cos²θ
tan²θ + 1 = sec²θ

Deriving the cotangent identity

Dividing the fundamental identity by sin²θ gives:

cot²θ + 1 = cosec²θ

Identity manipulation strategy

  1. Usually work on one side only.
  2. Convert tan into sin/cos when useful.
  3. Use sin²θ + cos²θ = 1.
  4. Look for common denominators.
  5. Factor where possible.
  6. Do not divide by an expression without considering where it may be zero.

Worked Example

Show that:

(1 − cos²θ)/sin θ = sin θ
1 − cos²θ = sin²θ
sin²θ/sin θ = sin θ
Therefore the identity is proved.

7. Solving Trigonometric Equations

A trigonometric equation may have several solutions within a specified interval. The principal inverse-trigonometric answer is usually only the starting point.

Example 1 — sine

sin θ = 1/2
0° ≤ θ ≤ 360°

The reference angle is 30°. Sine is positive in Quadrants I and II.

θ = 30°, 150°

Example 2 — cosine

cos θ = −√3/2
0° ≤ θ ≤ 360°

The reference angle is 30°. Cosine is negative in Quadrants II and III.

θ = 150°, 210°

Example 3 — tangent

tan θ = 1
0° ≤ θ ≤ 360°

The reference angle is 45°. Tangent is positive in Quadrants I and III.

θ = 45°, 225°

Radian form

sin x = 1/2
0 ≤ x ≤ 2π

x = π/6, 5π/6
Always read the interval carefully. A correct trigonometric solution outside the specified interval must not be included.

8. Quadratic Trigonometric Equations

Many harder trigonometric equations are actually algebraic equations in disguise. Treat sin x, cos x or tan x as an algebraic variable.

Worked Example

Solve:

2sin²x − 3sin x + 1 = 0
0° ≤ x ≤ 360°

Factorise:

(2sin x − 1)(sin x − 1) = 0

Therefore:

sin x = 1/2
or
sin x = 1

The first equation gives x = 30°, 150°. The second gives x = 90°.

x = 30°, 90°, 150°

Key method

  1. Rewrite the equation as a polynomial in one trig function.
  2. Factorise or use the quadratic formula.
  3. Solve each resulting trigonometric equation.
  4. Check the complete interval.

9. Transforming Trigonometric Equations

Some equations cannot be solved immediately because they contain different trigonometric functions. Identities allow them to be transformed into a solvable form.

Example

1 − 2sin²x = 0

Recognise the double-angle identity:

cos 2x = 1 − 2sin²x

Therefore:

cos 2x = 0

The important point is that solving for 2x requires careful treatment of the interval. If:

0° ≤ x ≤ 180°

then:

0° ≤ 2x ≤ 360°

Hence:

2x = 90°, 270°
x = 45°, 135°
Exam habit: Whenever you substitute u = 2x, 3x or another multiple, transform the interval too.

10. The Sine Rule

The Sine Rule connects each side of a triangle with the sine of its opposite angle.

a/sin A = b/sin B = c/sin C

When is it useful?

Worked Example

A = 30°, B = 60°, a = 5. Find b.

5/sin 30° = b/sin 60°
b = 5sin60°/sin30°
b = 5(√3/2)/(1/2)
b = 5√3

11. The Ambiguous Case of the Sine Rule

When two sides and a non-included angle are known, the Sine Rule can produce two possible angles.

Suppose:

a/sin A = b/sin B

If you calculate B using sin⁻¹, remember:

sin B = sin(180° − B)

Therefore both B and 180° − B may need to be checked.

Example

a = 10, b = 12, A = 30°
10/sin30° = 12/sin B
sin B = 12sin30°/10
sin B = 0.6
B = 36.87°

But:

B = 180° − 36.87°
B = 143.13°

Both must be checked against the triangle condition A + B + C = 180°.

Never automatically reject the second inverse-sine solution. Check whether it produces a valid triangle.

12. The Cosine Rule

The Cosine Rule generalises Pythagoras to any triangle.

a² = b² + c² − 2bc cos A

Cyclic versions are:

b² = a² + c² − 2ac cos B
c² = a² + b² − 2ab cos C

When to use it

Worked Example

Let b = 5, c = 7 and A = 60°. Find a.

a² = 5² + 7² − 2(5)(7)cos60°
a² = 25 + 49 − 70(1/2)
a² = 39
a = √39

Finding an angle

Rearrange:

cos A = (b² + c² − a²)/(2bc)
A = cos⁻¹[(b² + c² − a²)/(2bc)]

13. Area of a Triangle

If two sides and the included angle are known, the area is:

Area = 1/2 ab sin C

The same relationship can be written cyclically:

Area = 1/2 bc sin A
Area = 1/2 ca sin B

Why does this work?

If side b is used as the base and side a is the sloping side, the perpendicular height is a sin C.

Area = 1/2 × base × height
= 1/2 × b × a sin C
= 1/2 ab sin C

Worked Example

Find the area when a = 8, b = 5 and C = 30°.

Area = 1/2(8)(5)sin30°
= 20 × 1/2
= 10 square units
The angle must be the angle between the two sides used in the formula.

14. Compound-Angle Identities

Compound-angle identities express trigonometric functions of sums and differences in terms of the individual angles.

sin(A + B) = sin A cos B + cos A sin B
sin(A − B) = sin A cos B − cos A sin B
cos(A + B) = cos A cos B − sin A sin B
cos(A − B) = cos A cos B + sin A sin B
tan(A + B) = (tan A + tan B)/(1 − tan A tan B)
tan(A − B) = (tan A − tan B)/(1 + tan A tan B)

Memory pattern

For cosine, the sign between the two products is opposite to the sign inside the original angle.

cos(A + B) → minus
cos(A − B) → plus

Worked Example

Find the exact value of sin 75°.

Write:

75° = 45° + 30°
sin75° = sin45°cos30° + cos45°sin30°
= (√2/2)(√3/2) + (√2/2)(1/2)
sin75° = (√6 + √2)/4

15. Double-Angle Identities

Set A = B in the compound-angle identities.

sin 2A = 2sin A cos A
cos 2A = cos²A − sin²A
cos 2A = 2cos²A − 1
cos 2A = 1 − 2sin²A
tan 2A = 2tan A/(1 − tan²A)

Choosing the useful form

If the expression contains cos²A, use:

cos 2A = 2cos²A − 1

If it contains sin²A, use:

cos 2A = 1 − 2sin²A

If both sin²A and cos²A appear, the form cos²A − sin²A is often most convenient.

Worked Example

Given sin A = 3/5 and cos A = 4/5, find sin 2A and cos 2A.

sin2A = 2(3/5)(4/5)
sin2A = 24/25
cos2A = (4/5)² − (3/5)²
= 16/25 − 9/25
cos2A = 7/25

16. The Graphs of sin x, cos x and tan x

y = sin x

Domain: all real x
Range: −1 ≤ y ≤ 1
Period: 2π
Maximum: 1
Minimum: −1

Important points over one cycle:

x = 0 → sin x = 0
x = π/2 → sin x = 1
x = π → sin x = 0
x = 3π/2 → sin x = −1
x = 2π → sin x = 0

y = cos x

Domain: all real x
Range: −1 ≤ y ≤ 1
Period: 2π
x = 0 → cos x = 1
x = π/2 → cos x = 0
x = π → cos x = −1
x = 3π/2 → cos x = 0
x = 2π → cos x = 1

y = tan x

Domain: x ≠ π/2 + nπ
Range: all real numbers
Period: π

Tangent has vertical asymptotes wherever cos x = 0.

x = π/2 + nπ
n ∈ ℤ

17. Transformations of Trigonometric Graphs

The general sinusoidal form is:

y = a sin(bx + c) + d

Effect of a

The magnitude |a| controls the amplitude.

Amplitude = |a|

If a is negative, the graph is reflected in the x-axis.

Effect of b

The period becomes:

Period = 2π/|b|

In degrees:

Period = 360°/|b|

Effect of d

d moves the graph vertically.

Midline: y = d

Range

d − |a| ≤ y ≤ d + |a|

Worked Example

y = 3sin(2x) − 1

18. Horizontal Translation and Phase

Expressions of the form:

y = a sin[b(x − h)] + d

have a horizontal shift of h.

y = a sin[b(x − h)] + d
→ shift right by h
y = a sin[b(x + h)] + d
→ shift left by h
The sign inside the bracket appears opposite to the direction of the horizontal translation.

Example

y = 2cos[3(x − π/6)] + 1

19. Proving Trigonometric Identities

Proof questions test whether you can manipulate expressions systematically rather than simply calculate numerical values.

Worked Proof

Prove:

tan θ + cot θ = sec θ cosec θ

Start with the left-hand side:

tan θ + cot θ = sin θ/cos θ + cos θ/sin θ

Use a common denominator:

= (sin²θ + cos²θ)/(sin θ cos θ)

Use the fundamental identity:

= 1/(sin θ cos θ)

Split the reciprocal:

= 1/sin θ × 1/cos θ
= cosec θ sec θ
Proof principle: You do not need to make both sides look complicated. Choose the side that can be transformed naturally into the other.

20. Radians

A radian is defined using the relationship between arc length and radius.

θ = s/r

where θ is measured in radians, s is the arc length and r is the radius.

Arc length

s = rθ

Sector area

Area = 1/2 r²θ

Worked Example — Arc Length

A circle has radius 8 cm and central angle π/3 radians.

s = rθ
= 8(π/3)
s = 8π/3 cm

Worked Example — Sector Area

Area = 1/2(8²)(π/3)
= 32π/3 cm²
The formulas s = rθ and Area = 1/2r²θ require θ to be in radians.

21. Why Radians Matter

Radians are not merely an alternative angle unit. They make many trigonometric relationships take their simplest mathematical form.

For example, when θ is small and measured in radians:

sin θ ≈ θ

This approximation is not generally valid when θ is measured in degrees.

Similarly:

tan θ ≈ θ

for sufficiently small θ in radians.

This is an extension/advanced idea. The key Edexcel skill is to understand that radians are the natural angle measure used in calculus and circular geometry.

22. Periodicity and General Solutions

Trigonometric functions repeat their values periodically.

sin(x + 2π) = sin x
cos(x + 2π) = cos x
tan(x + π) = tan x

General sine solution

If sin x = sin α, the general solutions can be written:

x = α + 2nπ
or
x = π − α + 2nπ
n ∈ ℤ

General cosine solution

x = ±α + 2nπ
n ∈ ℤ

General tangent solution

x = α + nπ
n ∈ ℤ

In examination questions, the interval usually determines which members of these infinite families are required.

23. Expressing a sin x + b cos x in R Form

Extension / advanced technique: This method is extremely useful for combining a sine term and a cosine term into one sinusoidal function.

Consider:

a sin x + b cos x = R sin(x + α)

Expand the right-hand side:

R sin(x + α) = R sin x cos α + R cos x sin α

Compare coefficients:

R cos α = a
R sin α = b

Squaring and adding gives:

R = √(a² + b²)

Then:

tan α = b/a

Worked Example

Express:

3sin x + 4cos x

in the form R sin(x + α).

R = √(3² + 4²)
R = 5
cos α = 3/5
sin α = 4/5
α = tan⁻¹(4/3)
3sin x + 4cos x = 5sin(x + α)

Why R-form matters

Once an expression is written as R sin(x + α), its maximum and minimum become immediately visible:

−R ≤ R sin(x + α) ≤ R

24. Maximum and Minimum Values

Since:

−1 ≤ sin x ≤ 1
−1 ≤ cos x ≤ 1

it follows that:

−R ≤ R sin(x + α) ≤ R

Example

y = 7 − 3cos x

Since −1 ≤ cos x ≤ 1:

−3 ≤ −3cos x ≤ 3
4 ≤ y ≤ 10

Therefore the minimum value is 4 and the maximum value is 10.

25. Parameter Problems

Trigonometric questions often introduce an unknown constant and ask you to determine its possible values.

Example — range

Suppose:

y = k + 4sin x

The range is:

k − 4 ≤ y ≤ k + 4

If the question says the minimum value is 7:

k − 4 = 7
k = 11

Parameter + number of roots

Questions may ask how many solutions an equation has for a particular interval. A graph or range analysis can be more efficient than solving every equation directly.

For parameter questions, first identify the possible range of the trigonometric expression. Then compare that range with the required value.

26. Choosing the Correct Triangle Method

Use SOHCAHTOA when:

The triangle is right-angled and the relevant sides/angle are known.

Use the Sine Rule when:

You know an opposite side-angle pair and need another side or angle.

Use the Cosine Rule when:

You know two sides and their included angle, or all three sides.

Use the area formula when:

Two sides and the included angle are known and the area is needed.

Right triangle → SOHCAHTOA
Opposite side-angle pair → Sine Rule
SAS or SSS → Cosine Rule
Two sides + included angle + area → 1/2ab sin C

27. Master Strategy for Trigonometric Equations

  1. Check the interval. Write it down before calculating.
  2. Check the calculator mode. Determine whether the question uses degrees or radians.
  3. Simplify first. Factorise or use identities before pressing buttons.
  4. Find the reference angle. Use the positive principal value.
  5. Use signs. Determine the relevant quadrants using CAST or the unit circle.
  6. Generate every solution. Do not stop after inverse trig gives one answer.
  7. Check the interval. Remove values outside the required range.
A common exam error is mathematically correct working followed by an incomplete solution set.

28. Common Trigonometry Mistakes

1. Degree/radian confusion

The calculator may return a completely different answer if it is in the wrong mode.

2. Confusing sin⁻¹ with cosec

sin⁻¹x means inverse sine, whereas cosec x = 1/sin x.

3. Missing the second solution

Inverse trig functions return only a principal value. The complete interval must still be considered.

4. Wrong CAST quadrant

Remember that tangent is positive in Quadrants I and III, not Quadrants I and IV.

5. Wrong angle in the Sine Rule

Each side must correspond to its opposite angle.

6. Wrong angle in the area formula

1/2ab sin C requires C to be the included angle between a and b.

7. Using the Cosine Rule incorrectly

The angle must be opposite the side being squared in:

a² = b² + c² − 2bc cos A

8. Treating an identity as an equation

An identity is true for all valid values; it is not solved for a small collection of answers.

9. Rounding too early

Keep exact values or calculator precision until the final step.

10. Forgetting transformed intervals

If you replace x with 2x, the interval for 2x changes as well.

29. Trigonometry Exam Checklist

Before submitting a trigonometry answer, ask:

30. Trigonometry Formula Atlas

Basic ratios

sin θ = opposite/hypotenuse
cos θ = adjacent/hypotenuse
tan θ = opposite/adjacent

Fundamental identities

sin²θ + cos²θ = 1
1 + tan²θ = sec²θ
1 + cot²θ = cosec²θ

Reciprocals

sec θ = 1/cos θ
cosec θ = 1/sin θ
cot θ = cos θ/sin θ

Compound angles

sin(A ± B) = sin A cos B ± cos A sin B

cos(A ± B) = cos A cos B ∓ sin A sin B

tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B)

Double angles

sin2A = 2sinAcosA
cos2A = cos²A − sin²A
cos2A = 2cos²A − 1
cos2A = 1 − 2sin²A
tan2A = 2tanA/(1 − tan²A)

Triangle geometry

a/sinA = b/sinB = c/sinC

a² = b² + c² − 2bc cosA

Area = 1/2 ab sinC

Radians

s = rθ
Area of sector = 1/2r²θ
180° = π radians

Sinusoidal graphs

y = a sin(bx + c) + d

Amplitude = |a|
Period = 2π/|b|
Midline = y = d
Range = d − |a| to d + |a|

R-form

a sinx + b cosx = R sin(x + α)

R = √(a² + b²)

31. Practice Questions — Foundation

Attempt each question before opening the solution.

Question 1 — Basic ratio

In a right-angled triangle, the opposite side is 8 cm and the hypotenuse is 10 cm. Find θ.

Show Full Solution
sin θ = 8/10 = 0.8
θ = sin⁻¹(0.8)
θ ≈ 53.13°

Question 2 — Exact identity

Given sin θ = 5/13 and θ is acute, find cos θ.

Show Full Solution
sin²θ + cos²θ = 1
25/169 + cos²θ = 1
cos²θ = 144/169
cos θ = 12/13

Question 3 — Equation

Solve sin θ = √3/2 for 0° ≤ θ ≤ 360°.

Show Full Solution
Reference angle = 60°
Sine positive in QI and QII
θ = 60°, 120°

Question 4 — Cosine Rule

Two sides are 6 and 8 and the included angle is 60°. Find the third side exactly.

Show Full Solution
a² = 6² + 8² − 2(6)(8)cos60°
= 36 + 64 − 48
= 52
a = 2√13

Question 5 — Area

Find the area of a triangle with sides 10 cm and 6 cm and included angle 30°.

Show Full Solution
Area = 1/2(10)(6)sin30°
= 30(1/2)
= 15 cm²

32. Practice Questions — Intermediate

Question 6 — Quadratic trig equation

Solve: 2sin²x − 3sinx + 1 = 0, for 0° ≤ x ≤ 360°.

Show Full Solution
(2sinx − 1)(sinx − 1) = 0
sinx = 1/2 → x = 30°,150°
sinx = 1 → x = 90°
x = 30°, 90°, 150°

Question 7 — Compound angle

Find the exact value of cos 75°.

Show Full Solution
cos75° = cos(45° + 30°)
= cos45°cos30° − sin45°sin30°
= (√2/2)(√3/2) − (√2/2)(1/2)
cos75° = (√6 − √2)/4

Question 8 — Double angle

Given cos θ = 5/13 and θ is acute, find cos 2θ.

Show Full Solution
cos2θ = 2cos²θ − 1
= 2(25/169) − 1
= 50/169 − 169/169
= −119/169

Question 9 — Radians

A sector has radius 9 cm and angle 2π/3 radians. Find its area.

Show Full Solution
Area = 1/2r²θ
= 1/2(81)(2π/3)
= 27π cm²

Question 10 — Graph range

Find the range of y = 4 − 3sin x.

Show Full Solution
−1 ≤ sinx ≤ 1
−3 ≤ −3sinx ≤ 3
1 ≤ y ≤ 7

33. Practice Questions — Hard / Multi-Step

Question 11 — Identity + equation

Solve: 1 − 2sin²x = 0, for 0° ≤ x ≤ 180°.

Show Full Solution
1 − 2sin²x = cos2x
cos2x = 0
0° ≤ 2x ≤ 360°
2x = 90°,270°
x = 45°,135°

Question 12 — R-form

Express 5sinx + 12cosx in the form R sin(x + α), where R > 0 and 0 < α < 90°.

Show Full Solution
R = √(5² + 12²)
= 13
cosα = 5/13
sinα = 12/13
α = tan⁻¹(12/5)
5sinx + 12cosx = 13sin(x + α)

Question 13 — Maximum value

Find the maximum value of 8 + 6sin x − 8cos x.

Show Full Solution
6sinx − 8cosx = R sin(x + α)
R = √(6² + 8²) = 10
−10 ≤ 6sinx − 8cosx ≤ 10
Maximum = 8 + 10 = 18

Question 14 — Identity proof

Prove: (1 − cosx)/sinx = sinx/(1 + cosx).

Show Full Solution

Start with the left-hand side and multiply numerator and denominator by 1 + cosx.

(1 − cosx)/sinx × (1 + cosx)/(1 + cosx)
= (1 − cos²x)/(sinx(1 + cosx))
= sin²x/(sinx(1 + cosx))
= sinx/(1 + cosx)

Question 15 — Mixed triangle reasoning

In triangle ABC, a = 7, b = 9 and C = 60°. Find the area exactly.

Show Full Solution
Area = 1/2ab sinC
= 1/2(7)(9)(√3/2)
Area = 63√3/4

34. Deep Reasoning — A Level Challenge

Challenge 1

Given:

sin θ + cos θ = √2

Find θ for 0° ≤ θ ≤ 360°.

Show Full Solution

Square both sides:

sin²θ + 2sinθcosθ + cos²θ = 2
1 + sin2θ = 2
sin2θ = 1
2θ = 90°,450°

Since 0° ≤ θ ≤ 360°, we have 0° ≤ 2θ ≤ 720°. Therefore:

2θ = 90°,450°
θ = 45°,225°

Challenge 2

Show that the maximum value of a sinx + b cosx is √(a² + b²).

Show Full Solution

Write:

a sinx + b cosx = R sin(x + α)
R = √(a² + b²)

Since:

−1 ≤ sin(x + α) ≤ 1

we have:

−R ≤ Rsin(x + α) ≤ R
Maximum = R = √(a² + b²)

Challenge 3

Solve: 3cos²x − 2cosx − 1 = 0 for 0° ≤ x ≤ 360°.

Show Full Solution
(3cosx + 1)(cosx − 1) = 0
cosx = −1/3
or
cosx = 1

For cosx = 1:

x = 0°,360°

For cosx = −1/3:

x ≈ 109.47°,250.53°
x = 0°,109.47°,250.53°,360°

35. Trigonometry Master Checklist

You should be able to confidently:

Final rule: Do not treat trigonometry as a collection of unrelated formulas. The major ideas are connected: the unit circle explains the functions, identities connect the functions, compound angles generate double angles, graphs show periodic behaviour, and triangle rules apply the same relationships geometrically.