Trigonometry — Complete Notes

What you'll learn

• The special-angle values you must know by heart.
• The compound angle formulae for sine, cosine and tangent.
• The double angle formulae and how they are derived.
• How to write general solutions for trigonometric equations.
• How to prove trigonometric identities step by step.
• How to simplify expressions using reduction formulae.

The big picture

Grade 12 trigonometry extends what you learned in earlier grades. In Grades 10 and 11 you solved simple equations like sinθ = 1/2. In Grade 12 the angles become more complicated: sums and differences of angles, double angles, and equations that need a general solution because they repeat forever.

The whole chapter rests on a handful of formulae. Learn them, then practise proving identities — proving an identity is a favourite CAPS question because it tests both memory and logic.

Special angles

Know these values exactly — no calculator needed:

For signs by quadrant, remember CAST: All ratios positive in the first quadrant, only Sine in the second, only Tangent in the third, only Cosine in the fourth.

Fundamental identities

• sin²θ + cos²θ = 1 (the Pythagorean identity)
• tanθ = sinθ / cosθ

From the first identity:

• sin²θ = 1 − cos²θ
• cos²θ = 1 − sin²θ

Compound angle formulae

These express the sine, cosine and tangent of a sum or difference of two angles:

• 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·sinB
• tan(A + B) = (tanA + tanB) / (1 − tanA·tanB)
• tan(A − B) = (tanA − tanB) / (1 + tanA·tanB)

Notice the sign flip: in the sine and tangent formulae the signs on the right match the sign inside the bracket, but in the cosine formula the sign on the right is opposite to the sign in the bracket.

Double angle formulae

Set B = A in the compound formulae:

• sin2A = 2·sinA·cosA
• cos2A = cos²A − sin²A
• cos2A = 1 − 2sin²A
• cos2A = 2cos²A − 1
• tan2A = 2tanA / (1 − tan²A)

The three forms of cos2A are all the same thing — choose the one that helps you cancel or substitute.

General solutions

Because trigonometric functions repeat, equations usually have infinitely many solutions. We give a general solution using an integer n:

If you must solve within a restricted interval (for example 0° ≤ θ ≤ 360°), generate values by choosing n = 0, ±1, ±2, ... and keep only those inside the interval.

Sine and cosine repeat every 360°, while tangent repeats every 180° — that is exactly why the "+ 360°n" and "+ 180°n" appear.

Worked example 1 — exact value of a compound angle

Find sin 75° without a calculator.

Step 1 — split 75° into known angles. 75° = 45° + 30°.

Step 2 — apply the sine addition formula. sin 75° = sin(45° + 30°) = sin45°·cos30° + cos45°·sin30°

Step 3 — substitute the special values. = (√2/2)(√3/2) + (√2/2)(1/2) = (√6)/4 + (√2)/4 = (√6 + √2) / 4

So sin 75° = (√6 + √2)/4 ≈ 0,966.

Worked example 2 — general solution

Solve cos 2x = 1/2 for the general solution.

Step 1 — take the inverse cosine. arccos(1/2) = 60°.

Step 2 — apply the cosine general solution. 2x = ±60° + 360°n

Step 3 — divide every part by 2. x = ±30° + 180°n

So x = 30° + 180°n or x = −30° + 180°n, which can also be written as x = 150° + 180°n.

Check: for n = 0 we get 30° and 150°; doubling each gives 60° and 300°, and cos 60° = cos 300° = 1/2. Correct.

Worked example 3 — prove an identity

Prove that sin2A / (1 + cos2A) = tanA.

Step 1 — work on the left-hand side only. LHS = sin2A / (1 + cos2A)

Step 2 — replace with double angle formulae. sin2A = 2·sinA·cosA and cos2A = 2cos²A − 1, so: 1 + cos2A = 1 + (2cos²A − 1) = 2cos²A

Step 3 — substitute back and simplify. LHS = (2·sinA·cosA) / (2cos²A) = sinA / cosA

Step 4 — recognise the identity. sinA / cosA = tanA = RHS ∴ the identity is proven.

Reduction formulae

Angles bigger than 90°, and negative angles, can all be reduced to an acute angle between 0° and 90° using these rules:

Two patterns make the table easy to remember. For 180° and 360° the function name never changes; only the sign may change, and that sign follows CAST. For 90° and 270° the function name swaps between sine and cosine, because these are complementary angles.

Example: simplify sin(180° + x) · tan(360° − x). sin(180° + x) = −sinx and tan(360° − x) = −tanx, so the product = (−sinx)(−tanx) = sinx·tanx = sin²x / cosx.

Choosing a method for equations

When an equation contains a double angle together with a single angle (for example sin 2x = sin x), factorise rather than reaching straight for a general-solution formula. Move everything to one side, take out the common ratio as a factor, and solve each factor separately. This avoids losing solutions that would otherwise be divided away.

Common mistakes

• Sign errors in the cosine formula. cos(A + B) has a minus in its expansion — the sign flips.
• Forgetting the "+ 360°n". A general solution without the period term is incomplete and loses marks.
• Working on both sides of an identity. Prove one side until it matches the other; do not move terms across the equals sign as if solving an equation.
• Using the wrong form of cos2A. All three forms are valid, but only one may cancel nicely in a given proof.
• Mixing up degrees and radians. Unless told otherwise, CAPS uses degrees.
• Dropping solutions when dividing an angle. If 2x = ±60° + 360°n, remember to divide the whole right side by 2 to get ±30° + 180°n.

Quick summary / cheat sheet

Practice questions

1. Find the exact value of cos 15°.
2. Give the general solution of tan 2x = 1.
3. Prove that 1 − cos 2A = 2sin²A.
4. Solve sin 2x = sin x for x in [0°; 360°].
5. If sin A = 3/5 and A is acute, find sin 2A and cos 2A.
6. Simplify sin(180° − x) · cos(90° + x).

Answers

1. cos 15° = cos(45° − 30°) = cos45°cos30° + sin45°sin30° = (√6 + √2)/4 ≈ 0,966.
2. 2x = 45° + 180°n, so x = 22,5° + 90°n.
3. LHS = 1 − (1 − 2sin²A) = 1 − 1 + 2sin²A = 2sin²A = RHS ∴ proven.
4. 2sinx·cosx = sinx → sinx(2cosx − 1) = 0. So sinx = 0 → x = 0°, 180°, 360°; or cosx = 1/2 → x = 60°, 300°. Full solution set: x = 0°, 60°, 180°, 300°, 360°.
5. cosA = 4/5 (a 3-4-5 triangle). sin2A = 2(3/5)(4/5) = 24/25. cos2A = 1 − 2sin²A = 1 − 2(9/25) = 1 − 18/25 = 7/25.
6. sin(180° − x) = sinx and cos(90° + x) = −sinx, so the product = −sin²x.

Official EduRank Team notes. Always cross-check with your textbook and teacher.