Here is the truth about Grade 12 trigonometry. Most students walk into the NSC exam and throw away 20 to 30 marks on this topic. Not because it is impossible. Because nobody ever showed them the method properly. Trig in Paper 2 carries around 40 marks total, and compound angles plus general solutions make up a big chunk of that. If you learn the patterns in this post, you will pick up marks that most of your classmates are leaving on the table.
In This Post You Will LearnPaper 2
✓ What compound angle formulas are and how to use them without memorising blindly
✓ How to prove trig identities step by step without getting lost
✓ The exact method for solving general solutions (and why your calculator alone will not save you)
✓ How to handle restricted values and choose the right quadrant
✓ What double angle formulas are and when to use each version
✓ How the NSC exam tests this topic year after year
The Compound Angle Formulas You Must Know
Stop trying to memorise these by staring at them. Write them out by hand 10 times. Then use them in problems. That is how they stick.
Here are the four you need:
sin(A + B) = sinA.cosB + cosA.sinB
sin(A - B) = sinA.cosB - cosA.sinB
cos(A + B) = cosA.cosB - sinA.sinB
cos(A - B) = cosA.cosB + sinA.sinBNotice the pattern. For sin, the sign in the middle matches the sign in the bracket. For cos, it flips. That one observation saves you from mixing them up.
Quick memory trick: "Sin keeps the Sign, Cos Changes." Write that on a sticky note. Put it on your desk. Look at it every day until the exam.
Worked Example: Using Compound Angles to Find an Exact Value
Calculate the exact value of cos15° without a calculator.
Think: 15° = 45° - 30°. Both of those are special angles you know.
cos15° = cos(45° - 30°)
cos15° = cos45°.cos30° + sin45°.sin30°
cos15° = (√2/2)(√3/2) + (√2/2)(1/2)
cos15° = √6/4 + √2/4
cos15° = (√6 + √2) / 4
That is your exact answer. No calculator needed. The NSC exam loves asking for exact values like this.
The Special Angles You Must Know Cold
You cannot do compound angles if you do not know your special angles. Here they are. No excuses.
| Angle | sin | cos | tan |
|-------|--------|--------|--------|
| 0° | 0 | 1 | 0 |
| 30° | 1/2 | √3/2 | 1/√3 |
| 45° | √2/2 | √2/2 | 1 |
| 60° | √3/2 | 1/2 | √3 |
| 90° | 1 | 0 | undef |If you do not know these from memory, stop reading and go learn them right now. Everything else in trig builds on top of these.
Double Angle Formulas in Grade 12 Trigonometry
Double angle formulas come from compound angles. Just set A = B in the compound angle formula and you get:
sin2A = 2sinA.cosA
cos2A = cos²A - sin²A
= 2cos²A - 1
= 1 - 2sin²AYes, there are three versions of cos2A. Each one is useful in different situations.
When to use which version of cos2A:
Use cos²A - sin²A when you have both sin and cos in the problem.
Use 2cos²A - 1 when you only have cos values.
Use 1 - 2sin²A when you only have sin values.
Worked Example: Simplify Using Double Angles
Simplify: sin80°.cos80°
Look at this. It is half of sin2A where A = 80°.
sin80°.cos80° = ½ sin(2 x 80°) = ½ sin160°
But wait. sin160° = sin(180° - 160°) = sin20° (using reduction formulas).
So sin80°.cos80° = ½ sin20°
That kind of simplification is worth 3 to 4 marks in the exam.
How to Prove Trigonometric Identities
Identity proofs scare students. But there is a method. Follow it every time and you will get through them.
The Golden Rule of Trig Identities:
Pick the MORE complicated side. Work on that side only. Simplify it until it looks like the other side. Never work on both sides at the same time.
Step-by-Step Method for Proving Identities
Step 1: Choose the more complicated side (usually the left).
Step 2: Look for compound angle or double angle formulas you can expand.
Step 3: Convert everything to sin and cos if there are tan, sec, or cosec terms.
Step 4: Factorise or find common denominators if needed.
Step 5: Simplify until you reach the other side.
Worked Example: Prove an Identity
Prove that: cos2x / (cosx + sinx) = cosx - sinx
Start with the left hand side (LHS).
LHS = cos2x / (cosx + sinx)
Use cos2x = cos²x - sin²x (choose this version because it factorises nicely).
LHS = (cos²x - sin²x) / (cosx + sinx)
Factorise the numerator. It is a difference of two squares.
LHS = (cosx + sinx)(cosx - sinx) / (cosx + sinx)
Cancel (cosx + sinx) from top and bottom.
LHS = cosx - sinx
LHS = RHS. Done.
See? Not scary. You just need to pick the right version of cos2x and spot the factorisation.
For full live lessons on this topic, see our Grade 12 Maths tuition page.
General Solutions: The Method That Always Works
A general solution gives you ALL the angles that satisfy a trig equation, not just the ones between 0° and 360°.
Here is the method. Burn it into your brain.
For sin equations:
If sinθ = sinα, then:
θ = α + k.360° OR θ = (180° - α) + k.360°where k is any integer (k ∈ Z).
For cos equations:
If cosθ = cosα, then:
θ = α + k.360° OR θ = -α + k.360°For tan equations:
If tanθ = tanα, then:
θ = α + k.180°Tan only has one general solution because its period is 180°, not 360°.
Worked Example: Finding the General Solution
Solve for θ: 2sinθ - 1 = 0
Step 1: Isolate sinθ.
sinθ = 1/2
Step 2: Find the reference angle.
α = 30° (because sin30° = 1/2)
Step 3: Write the general solution.
θ = 30° + k.360° OR θ = (180° - 30°) + k.360°
θ = 30° + k.360° OR θ = 150° + k.360°, where k ∈ Z
Worked Example: General Solution With a Restricted Range
Now find all values of θ between -360° and 360°.
From θ = 30° + k.360°:
k = 0: θ = 30°
k = -1: θ = 30° - 360° = -330°
k = 1: θ = 30° + 360° = 390° (too big, stop)
From θ = 150° + k.360°:
k = 0: θ = 150°
k = -1: θ = 150° - 360° = -210°
k = 1: θ = 150° + 360° = 510° (too big, stop)
Solutions: θ ∈ {-330°, -210°, 30°, 150°}
Pro tip from Mr Sawaya: Students lose marks here because they forget to test negative values of k. The question says -360° to 360°. That means you must try k = -1 (and sometimes k = -2) as well. Do not stop at k = 0 and k = 1.
Solving Harder Trig Equations
Some NSC questions combine compound angles with general solutions. Here is how to handle them.
Worked Example: Equation With Double Angles
Solve for x: cos2x = cosx, for x ∈ [0°, 360°]
Step 1: Replace cos2x with 2cos²x - 1 (choose this version because the RHS is cosx).
2cos²x - 1 = cosx
Step 2: Rearrange into standard form.
2cos²x - cosx - 1 = 0
Step 3: This is a quadratic in cosx. Factorise.
(2cosx + 1)(cosx - 1) = 0
Step 4: Solve each factor.
2cosx + 1 = 0, so cosx = -1/2
Reference angle = 60°. Cos is negative in quadrant 2 and 3.
x = 120° or x = 240°
cosx - 1 = 0, so cosx = 1
x = 0° or x = 360°
Solutions: x ∈ {0°, 120°, 240°, 360°}
If you are still building your foundation in functions, check out our guide on Grade 12 Functions and Graphs - Everything You Need to Know.
Common Mistakes Students Make
- Getting the sign wrong in compound angle formulas
Students write cos(A + B) = cosA.cosB + sinA.sinB. That plus sign is wrong. It should be minus. Remember: "Cos Changes" the sign. Write the formulas out by hand until they are automatic.
- Forgetting to give the general solution when asked
If the question says "determine the general solution," you MUST include the k.360° or k.180° part. Writing just θ = 30° and θ = 150° without the k.360° will cost you marks. The k part is what makes it general.
- Not finding all solutions in a restricted range
When the question says "for x ∈ [-180°, 360°]," students only try k = 0 and miss solutions where k = -1 or k = 1. Substitute multiple values of k and check each one against the range.
- Choosing the wrong version of cos2A
If you are solving an equation and the other terms involve sinx only, use cos2x = 1 - 2sin²x. If the other terms involve cosx only, use cos2x = 2cos²x - 1. Choosing the wrong version means you cannot simplify or factorise. This is a decision you make before you start calculating.
- Working on both sides of an identity simultaneously
In a "prove that" question, you must pick one side and simplify it to look like the other. Working on both sides and meeting in the middle is not accepted by the examiners. It does not prove anything because you are assuming the thing you are trying to prove.
How This Topic Appears in the NSC Exam
Grade 12 trigonometry appears in Paper 2 of the NSC Maths exam.
Trig as a whole carries around 40 marks in Paper 2. Compound angles, double angles, and general solutions make up approximately 20 to 25 of those marks. The rest goes to trig graphs and 2D/3D trig problems.
This section typically appears as Question 5 or Question 6 in Paper 2. The DBE structures it in three parts most years.
Part (a) asks you to prove an identity using compound or double angle formulas. This is usually worth 4 to 6 marks.
Part (b) gives you a trig equation and asks for the general solution. This is worth 6 to 8 marks.
Part (c) may ask for specific solutions in a restricted range, or it may give you a harder equation that combines quadratic factorisation with trig.
In the 2023 NSC exam, this section included a double angle identity proof and a general solution involving cos2x and sinx. The general solution required students to use the identity 1 - 2sin²x for cos2x, form a quadratic in sinx, and factorise.
The DBE is predictable here. If you practise 5 past paper questions on compound angle identities and 5 on general solutions, you will have seen every variation they throw at you.
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