Papers 2 and 3. \(\sin(A + B)\) is not \(\sin A + \sin B\). The compound angle formulae give the right expansion, and with them you can find exact values such as \(\sin 75^\circ\), simplify expressions and solve equations with shifted angles.
\(\sin(A \pm B) \equiv \sin A\cos B \pm \cos A\sin B\), \(\cos(A \pm B) \equiv \cos A\cos B \mp \sin A\sin B\), \(\tan(A \pm B) \equiv \dfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B}\). In the cosine formula the sign flips.
Fig. 1: OP = 1 makes angle A + B with the horizontal (here A = 30°, B = 25°, drawn to scale). OQ = cos B and QP = sin B. The height PN = sin(A + B) splits into RN = sin A cos B and PR = cos A sin B, which is the formula for sin(A + B)
Section 01
The six formulae.
Fig. 1 shows where \(\sin(A + B)\) comes from: the height of P is built from two pieces. Replacing \(B\) by \(-B\) gives the minus versions, and dividing sine by cosine gives the tangent formula. All six are in the formula list, but you must know which sign goes where.
Expansion
Sign pattern
\(\sin(A \pm B) \equiv \sin A\cos B \pm \cos A\sin B\)
same sign as the bracket; sin cos, cos sin
\(\cos(A \pm B) \equiv \cos A\cos B \mp \sin A\sin B\)
opposite sign; cos cos, sin sin
\(\tan(A \pm B) \equiv \dfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B}\)
same sign on top, opposite underneath
Worked example Exact value of sin 75°
Find the exact value of \(\sin 75^\circ\).
Split into angles with known exact values: \(75^\circ = 45^\circ + 30^\circ\).
\(\sin(A + B) \ne \sin A + \sin B\). Test it: \(\sin(30^\circ + 60^\circ) = 1\), but \(\sin 30^\circ + \sin 60^\circ = 1.37\). Brackets after sin, cos or tan never multiply out.
Section 02
Using given ratios.
A common question gives one ratio of each angle and asks for a compound ratio as an exact fraction. Draw a right-angled triangle (or use \(\sin^2 + \cos^2 = 1\)) for the missing ratios, then fix each sign from the quadrant.
Worked example Two acute angles
Angles \(A\) and \(B\) are acute, with \(\sin A = \dfrac{3}{5}\) and \(\cos B = \dfrac{5}{13}\). Find the exact values of \(\sin(A + B)\) and \(\cos(A - B)\).
The 3-4-5 triangle gives \(\cos A = \dfrac{4}{5}\); the 5-12-13 triangle gives \(\sin B = \dfrac{12}{13}\). Both positive, since the angles are acute.
\(\cos(A - B) = \cos A\cos B + \sin A\sin B = \dfrac{4}{5}\cdot\dfrac{5}{13} + \dfrac{3}{5}\cdot\dfrac{12}{13} = \dfrac{20 + 36}{65} = \dfrac{56}{65}\).
Answer: \(\sin(A + B) = \dfrac{63}{65}\), \(\cos(A - B) = \dfrac{56}{65}\)
Your turn
Given that \(\tan A = 3\) and \(\tan B = \dfrac{1}{2}\), with both angles acute, find \(\tan(A - B)\) and hence the angle \(A - B\).
\(\tan(A - B) = \dfrac{3 - \frac{1}{2}}{1 + 3 \times \frac{1}{2}} = \dfrac{2.5}{2.5} = 1\), so \(A - B = 45^\circ\).
Examiner trap
Signs come from the quadrant, not from the triangle. If \(A\) is obtuse and \(\sin A = \dfrac{5}{13}\), then \(\cos A = -\dfrac{12}{13}\); forgetting the minus sign wrecks every later line.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Which is the correct expansion of \(\sin(A - B)\)?
Quick check+10 XP
Which is the correct expansion of \(\cos(A + B)\)?
When an equation contains \(\sin(x + 30^\circ)\) or \(\cos(x - 60^\circ)\), expand the bracket with exact values for the known angle, collect the \(\sin x\) and \(\cos x\) terms, and divide to get a single \(\tan x\).
Add: the \(\cos x\) terms cancel, leaving \(2\sin x\cos 60^\circ = 2\sin x \times \dfrac{1}{2}\).
Answer: \(\sin x\)
Worked example Solve an equation
Solve \(\sin(x + 30^\circ) = 2\cos x\) for \(0^\circ \le x \le 360^\circ\).
Expand: \(\dfrac{\sqrt{3}}{2}\sin x + \dfrac{1}{2}\cos x = 2\cos x\).
Collect: \(\dfrac{\sqrt{3}}{2}\sin x = \dfrac{3}{2}\cos x\).
Divide by \(\dfrac{\sqrt{3}}{2}\cos x\) (\(\cos x = 0\) is not a solution, since then \(\sin x = \pm 1\) and the equation fails): \(\tan x = \dfrac{3}{\sqrt{3}} = \sqrt{3}\).
\(x = 60^\circ\) or \(60^\circ + 180^\circ = 240^\circ\).
Answer: \(x = 60^\circ,\ 240^\circ\)
Examiner trap
Do not write \(\sin(x + 30^\circ) = \sin x + \dfrac{1}{2}\). Expand properly: the known angle's sine AND cosine both appear.
Section 04
The tangent formula.
With \(\tan(A \pm B)\), an equation in \(\tan x\) usually turns into a quadratic once you clear the fraction.
Worked example A quadratic in tan x
Solve \(\tan(x + 45^\circ) = 6\tan x\) for \(0^\circ < x < 180^\circ\).
\(\tan(x + 45^\circ) = \dfrac{\tan x + 1}{1 - \tan x}\). Let \(t = \tan x\).
Both positive, so both angles are acute: \(x = \tan^{-1}\dfrac{1}{2} = 26.6^\circ\) and \(x = \tan^{-1}\dfrac{1}{3} = 18.4^\circ\).
Answer: \(x = 18.4^\circ,\ 26.6^\circ\)
Examiner trap
The tangent formula has a 1 in the denominator: \(\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}\), with a MINUS sign for the plus case.
Stage 2 · Exam
Exam-style questions
Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Paper 1 to 6 standard.
Finish the four checks above to unlock the exam questions
Exam style+20 XP
Given that \(\tan A = 2\) and \(\tan B = \dfrac{1}{3}\), find the value of \(\tan(A + B)\).
Exam style+20 XP
Find the exact value of \(\cos 105^\circ\).
Exam style+20 XP
Solve \(2\sin(x - 30^\circ) = \cos x\) for \(0^\circ < x < 180^\circ\).
Exam style+20 XP
Angle \(A\) is obtuse with \(\sin A = \dfrac{5}{13}\), and angle \(B\) is acute with \(\cos B = \dfrac{4}{5}\). Find the exact value of \(\cos(A - B)\).
Skill unlocked
Compound angles, expanded.
This skill is now recorded on your profile. Keep the chain going.