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
- Usually work on one side only.
- Convert tan into sin/cos when useful.
- Use sin²θ + cos²θ = 1.
- Look for common denominators.
- Factor where possible.
- 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
- Rewrite the equation as a polynomial in one trig function.
- Factorise or use the quadratic formula.
- Solve each resulting trigonometric equation.
- 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?
- A known side and its opposite angle are available.
- You need another side when its opposite angle is known.
- You need an angle when its opposite side is known.
- It is particularly important in the ambiguous SSA case.
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
- Two sides and the included angle are known.
- All three sides are known and an angle is required.
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
- Amplitude = 3
- Period = π
- Midline = y = −1
- Maximum = 2
- Minimum = −4
- Range: −4 ≤ y ≤ 2
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
- Amplitude = 2
- Period = 2π/3
- Shift right by π/6
- Midline = y = 1
- Range = −1 to 3
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
-
Check the interval.
Write it down before calculating.
-
Check the calculator mode.
Determine whether the question uses degrees or radians.
-
Simplify first.
Factorise or use identities before pressing buttons.
-
Find the reference angle.
Use the positive principal value.
-
Use signs.
Determine the relevant quadrants using CAST or the unit circle.
-
Generate every solution.
Do not stop after inverse trig gives one answer.
-
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:
- Is my calculator in the correct mode?
- Did I use degrees or radians correctly?
- Did I identify the correct trigonometric relationship?
- Did I match sides with their opposite angles?
- Did I use the included angle for the area formula?
- Did I consider every quadrant?
- Did I check the entire interval?
- Did I preserve exact values when required?
- Did I avoid premature rounding?
- Does my answer make geometric sense?
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:
- Use SOHCAHTOA in right-angled triangles.
- Convert between degrees and radians.
- Recall exact standard-angle values.
- Use the unit-circle interpretation of sine and cosine.
- Determine signs in all four quadrants.
- Use reciprocal trigonometric functions.
- Manipulate fundamental trigonometric identities.
- Solve sine, cosine and tangent equations.
- Solve quadratic trigonometric equations.
- Use compound-angle identities.
- Use double-angle identities.
- Use the Sine Rule.
- Handle the ambiguous Sine Rule case.
- Use the Cosine Rule.
- Find the area of a triangle using 1/2ab sinC.
- Use radians for arc length and sector area.
- Understand sine, cosine and tangent graphs.
- Determine amplitude, period, range and midline.
- Handle graph transformations.
- Prove trigonometric identities.
- Convert a sinx + b cosx into R-form.
- Find maximum and minimum values.
- Solve parameter-based trigonometric problems.
- Handle multi-step examination problems.
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.