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NotesMath AA HLTopic 3.10Compound angle identities
Back to Math AA HL Topics
3.10.11 min read

Compound angle identities

IB Mathematics: Analysis and Approaches • Unit 3

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Contents

  • sin and cos of A ± B
  • tan(A ± B) and the tan double angle
Picture: open the angle into two known pieces: You can find sin and cos of nice angles like 30°, 45°, 60° exactly. But what about 75°? It is not one of the standard angles — yet 75° = 45° + 30°, both of which you DO know.

The compound-angle formulas let you trade the awkward single angle for two friendly ones. They are in the formula booklet, so you don't memorise them — but you must know how to USE them.

The two cos forms are the tricky ones: cos(A+B) uses a minus in the middle and cos(A−B) uses a plus — the sign FLIPS compared with the bracket.
sin: same sign as the bracket, and the terms 'cross over' (sin·cos + cos·sin).
cos: terms 'match up' (cos·cos, sin·sin) and the middle sign is the OPPOSITE of the bracket.

IB-style question — exact value of sin 75°

A student needs an exact (non-decimal) value of sin 75° for a Paper 1 question.

Find the exact value of sin 75°.

Step by step

  1. Split 75° into two angles you know exactly.
  2. Apply sin(A+B) = sin A cos B + cos A sin B with A = 45°, B = 30°.
  3. Substitute the exact surd values.
  4. Combine over the common denominator 4.

Final answer

sin 75° = (√6 + √2)/4 ≈ 0.966 (and the decimal check confirms it).

IB-style question — cos of a difference

Without a calculator, find the exact value of cos 15°.

Step by step

  1. Write 15° as a difference of known angles.
  2. cos(A−B) = cos A cos B + sin A sin B (the middle sign flips to +).
  3. Substitute the surds.

Final answer

cos 15° = (√6 + √2)/4. (Notice it equals sin 75° — because cos 15° = sin(90°−15°) = sin 75°.)

tan adds 'over' a correction term: Dividing sin(A±B) by cos(A±B) and tidying up gives the tan formula. The top is just the tangents added/subtracted; the bottom is 1 minus/plus the product of the tangents.

Watch the signs: the bottom sign is the opposite of the top sign.

Now comes a free bonus. Put B = A in tan(A+B) and you get the double-angle formula for tan — no new rule to learn, it falls straight out.
Top: tangents added/subtracted. Bottom: 1, then the OPPOSITE sign, then the product.
Set B = A in tan(A+B): tan A + tan A = 2 tan A on top, 1 − tan A·tan A = 1 − tan²A below.

IB-style question — derive tan 2A

Show how the double-angle formula for tan comes from the compound-angle formula.

Starting from tan(A + B), derive an expression for tan 2A.

Step by step

  1. Write 2A as A + A, then use tan(A+B) with B = A.
  2. Add the like terms on top; the product on the bottom is tan²A.

Final answer

tan 2A = 2 tan A / (1 − tan²A), obtained by putting B = A in tan(A+B).

IB-style question — exact value of tan 75°

Find the exact value of tan 75°, giving your answer in the form a + b√3.

Step by step

  1. Use 75° = 45° + 30° in tan(A+B).
  2. Substitute tan 45° = 1 and tan 30° = 1/√3.
  3. Rationalise by multiplying top and bottom by (√3 + 1).
  4. Simplify.

Final answer

tan 75° = 2 + √3 ≈ 3.73.

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Find the exact value of cos 75°. [2 marks]

Related Math AA HL Topics

Continue learning with these related topics from the same unit:

3.1.1Distance & midpoint (3D)
3.1.2Volume & surface area
3.1.3Angles in 3D
3.1.4Solids in 3D coordinates
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