P3.4 Compound angles, double angles and R-form

Where do the compound angle formulae come from?

It is tempting to think that \(\sin(A + B) = \sin A + \sin B\), but a single test destroys the idea: with \(A = B = 90^\circ\), the left side is \(\sin 180^\circ = 0\) while the right side is 2. The sine of a sum mixes sines and cosines, and a picture shows exactly how.

In the figure, the angle \(A + B\) is built from two right-angled triangles. OP has length 1, so the height of P is \(\sin(A + B)\) and its horizontal distance from O is \(\cos(A + B)\). In triangle OQP, \(OQ = \cos B\) and \(QP = \sin B\). Then \(QM = OQ\sin A = \sin A\cos B\). Because QT is parallel to OM and QP is perpendicular to OQ, the angle TPQ is also \(A\), so \(PT = QP\cos A = \cos A\sin B\). Adding the two heights gives \(\sin(A + B) = \sin A\cos B + \cos A\sin B\). The horizontal distances give \(ON = OM - TQ\), which is \(\cos(A + B) = \cos A\cos B - \sin A\sin B\).

Replacing \(B\) by \(-B\), with \(\cos(-B) = \cos B\) and \(\sin(-B) = -\sin B\), gives the "minus" versions, and dividing the sine formula by the cosine formula (then dividing top and bottom by \(\cos A\cos B\)) gives the tan formula.

How do you find \(\sin(A + B)\) from given ratios?

A typical question gives one ratio of each angle and says which quadrant it lies in. To use the formulae you need both the sine and the cosine of each angle. Draw a right-angled triangle (or use \(\sin^2 + \cos^2 = 1\)) to find the size of the missing ratio, then use the quadrant to fix its sign. Acute angles have every ratio positive; obtuse angles (90° to 180°) have sin positive and cos and tan negative. Never find the angles on a calculator and round them: the question asks for exact values.

How do you solve equations involving compound angles?

Expand every compound angle, using the exact values of the fixed angle (for example \(\cos 30^\circ = \frac{\sqrt3}{2}\)). Collect the \(\sin\theta\) terms on one side and the \(\cos\theta\) terms on the other, then divide by \(\cos\theta\) to reach \(\tan\theta = k\). Dividing is allowed provided \(\cos\theta = 0\) is not a solution, so check that by substituting. tan has period 180°, so in \(0^\circ \le \theta < 360^\circ\) there are two solutions: the principal value and that value plus 180°.

What are the double-angle formulae?

Put \(B = A\) in the addition formulae. \(\sin(A + A) = \sin A\cos A + \cos A\sin A = 2\sin A\cos A\), and \(\cos(A + A) = \cos^2 A - \sin^2 A\). Replacing \(\sin^2 A\) by \(1 - \cos^2 A\), or \(\cos^2 A\) by \(1 - \sin^2 A\), gives two more versions of \(\cos 2A\). The formulae work for any angle and its double: \(\sin 6x = 2\sin 3x\cos 3x\), and \(\cos x = 1 - 2\sin^2\frac{x}{2}\).

The rearranged forms are worth seeing as graphs. \(y = \cos^2 x\) is a cosine wave of half the period, squeezed between 0 and 1 and centred on \(y = \frac12\), which is exactly what \(\frac12(1 + \cos 2x)\) describes.

How do you solve equations containing \(\sin 2\theta\) or \(\cos 2\theta\)?

The aim is always one trigonometric function of one angle. If the equation also contains \(\sin\theta\) or \(\cos\theta\), replace the double angle using the formulae, and choose the form that matches:

• \(\cos 2\theta\) with \(\sin\theta\) elsewhere: use \(\cos 2\theta = 1 - 2\sin^2\theta\). With \(\cos\theta\): use \(2\cos^2\theta - 1\). The result is a quadratic.
• \(\sin 2\theta\) with \(\sin\theta\) or \(\cos\theta\): write \(\sin 2\theta = 2\sin\theta\cos\theta\) and factorise. Never divide by \(\sin\theta\) or \(\cos\theta\), which loses the solutions where it is zero.
• If every term already has the same double angle (for example \(3\sin 2\theta = \cos 2\theta\)), keep it: solve for \(2\theta\) over the doubled interval, then halve.
• Reject any root of the quadratic outside \(-1 \le \sin\theta \le 1\), and say why.

How do you prove identities and use half angles?

Start from the more complicated side and work towards the other. Do not treat the identity as an equation: never do the same thing to both sides. A double angle on one side and single angles on the other tells you which formula to use, and you should choose the form of \(\cos 2A\) that cancels something: \(1 + \cos 2A = 2\cos^2 A\) and \(1 - \cos 2A = 2\sin^2 A\) are the most useful. Finish with "= RHS" or "\(\equiv \tan x\) as required".

Putting \(A = \frac{\theta}{2}\) in the cosine formulae gives \(\cos\theta = 2\cos^2\frac{\theta}{2} - 1 = 1 - 2\sin^2\frac{\theta}{2}\), and so the half-angle results below.

How do you write \(a\cos\theta + b\sin\theta\) as a single sine or cosine?

Add a cosine wave and a sine wave of the same period and something remarkable happens: the result is a single wave of the same period, with a bigger amplitude \(R\) and a horizontal shift \(\alpha\). The figure shows \(3\cos\theta + 4\sin\theta\), which is exactly \(5\cos(\theta - 53.1^\circ)\). Writing the sum this way is called the R-form (or harmonic form).

The method is always the same three lines:

• expand the required form, for example \(R\cos(\theta + \alpha) = R\cos\theta\cos\alpha - R\sin\theta\sin\alpha\);
• compare coefficients of \(\cos\theta\) and \(\sin\theta\) to get \(R\cos\alpha = \ldots\) and \(R\sin\alpha = \ldots\) (both positive when \(0 < \alpha < 90^\circ\) and \(R > 0\));
• square and add for \(R = \sqrt{a^2 + b^2}\); divide for \(\tan\alpha\).

Use the form the question gives. As a guide, these choices keep \(\alpha\) acute: \(a\cos + b\sin \to R\cos(\theta - \alpha)\); \(a\cos - b\sin \to R\cos(\theta + \alpha)\); \(b\sin + a\cos \to R\sin(\theta + \alpha)\); \(b\sin - a\cos \to R\sin(\theta - \alpha)\).

To solve \(a\cos\theta + b\sin\theta = c\), write it as \(R\cos(\theta - \alpha) = c\), so \(\cos(\theta - \alpha) = \frac{c}{R}\). Shift the interval by \(\alpha\) before finding the values of the bracket, then add \(\alpha\) back (with a "+" form, subtract it). There are no solutions if \(|c| > R\).

How does the R-form give maximum and minimum values?

Since \(-1 \le \cos(\ldots) \le 1\), the expression \(R\cos(\theta \pm \alpha)\) lies between \(-R\) and \(R\), and the same holds for \(R\sin(\theta \pm \alpha)\). The value of \(\theta\) at the maximum comes from setting the bracket equal to 0° for cos (or 90° for sin); at the minimum, 180° for cos (or 270° for sin). Then solve for \(\theta\).

Linked expressions follow from this. For \(k + R\sin(\ldots)\) the range is \(k - R\) to \(k + R\). For a reciprocal such as \(\dfrac{m}{k + R\sin(\ldots)}\) with \(k > R\), the greatest value occurs when the denominator is least. For a squared term such as \((R\sin(\ldots))^2\), the least value is 0, not \(-R^2\). Contexts such as tides, temperatures and the heights of rotating objects use exactly this: the "greatest height" is \(k + R\).

Exam tips for P3.4

Common misconceptions

How this topic connects

P3.3 Reciprocal and inverse trigonometric functions

Exact values, CAST and solving \(\sin\theta = k\) come from earlier units; sec, cosec, cot and \(\sec^2\theta = 1 + \tan^2\theta\) from P3.3 are often mixed into the same identity or equation as the formulae here.

P3.6 Differentiation and P4 integration

The derivatives of \(\sin x\) and \(\cos x\) rest on the compound angle formulae: a first-principles argument expands \(\sin(x + h)\). In P4, \(\sin^2 x\) and \(\cos^2 x\) are integrated by first writing them as \(\frac12(1 \mp \cos 2x)\).

Mechanics and Physics

Resolving forces gives terms like \(P\cos\theta + Q\sin\theta\), and the R-form finds the angle that makes such a force greatest or least; projectile ranges use \(2\sin\theta\cos\theta = \sin 2\theta\). Adding two oscillations of the same frequency gives a single oscillation: the R-form is the algebra of superposition, with \(R\) the resultant amplitude and \(\alpha\) the phase.

Key formulae

Check your understanding

Specification checklist

• (P3 2.3) I can quote and use the compound angle formulae for \(\sin(A \pm B)\), \(\cos(A \pm B)\) and \(\tan(A \pm B)\), and explain a geometric proof of them.
• (P3 2.3) I can find exact values of compound angles, and of sin, cos or tan of \(A \pm B\) when ratios of \(A\) and \(B\) and their quadrants are given.
• (P3 2.3) I can derive and use the double-angle formulae for \(\sin 2A\), \(\cos 2A\) (all three forms) and \(\tan 2A\).
• (P3 2.3) I can use half-angle results such as \(\cos^2\frac{\theta}{2} = \frac12(1 + \cos\theta)\) to find exact values.
• (P3 2.3) I can solve equations involving compound or double angles in a given interval, in degrees or radians, without losing solutions.
• (P3 2.3) I can prove trigonometric identities using these formulae, writing a full chain of reasoning and a conclusion.
• (P3 2.3) I can express \(a\cos\theta + b\sin\theta\) as \(R\cos(\theta \pm \alpha)\) or \(R\sin(\theta \pm \alpha)\), and use it to solve \(a\cos\theta + b\sin\theta = c\).
• (P3 2.3) I can find the maximum and minimum values of expressions built from the R-form, including reciprocals and contexts, and the angles where they occur.