Papers 2 and 3. Set \(B = A\) in the compound angle formulae and you get \(\sin 2A\), \(\cos 2A\) and \(\tan 2A\). The three forms of \(\cos 2A\) are the most useful identities in this topic, for equations now and for integration later.
\(\sin 2A \equiv 2\sin A\cos A\), \(\cos 2A \equiv \cos^2 A - \sin^2 A \equiv 2\cos^2 A - 1 \equiv 1 - 2\sin^2 A\), \(\tan 2A \equiv \dfrac{2\tan A}{1 - \tan^2 A}\). Choose the form of \(\cos 2A\) that matches the other terms in the equation.
Fig. 1: y = cos² x (blue) never goes below 0. It oscillates about the line y = ½ with amplitude ½ and period 180°, exactly the graph of ½ + ½ cos 2x, which is the identity cos 2x = 2cos² x − 1 rearranged
Section 01
From compound to double.
Put \(B = A\) in each compound angle formula. Then use \(\sin^2 A + \cos^2 A = 1\) to get two more versions of \(\cos 2A\). Fig. 1 shows one of them as a graph: \(\cos^2 x\) is a cosine wave of half the size and twice the frequency, lifted up by \(\dfrac{1}{2}\).
Formula
Use it when
\(\sin 2A \equiv 2\sin A\cos A\)
you see \(\sin 2A\), or a product \(\sin A\cos A\)
\(\cos 2A \equiv \cos^2 A - \sin^2 A\)
both \(\sin^2\) and \(\cos^2\) appear
\(\cos 2A \equiv 2\cos^2 A - 1\)
the rest of the equation is in \(\cos A\)
\(\cos 2A \equiv 1 - 2\sin^2 A\)
the rest of the equation is in \(\sin A\)
\(\tan 2A \equiv \dfrac{2\tan A}{1 - \tan^2 A}\)
the equation is in \(\tan A\)
Rearranged: \(\cos^2 A \equiv \dfrac{1}{2}(1 + \cos 2A)\) and \(\sin^2 A \equiv \dfrac{1}{2}(1 - \cos 2A)\). These are how you will integrate \(\sin^2 x\) and \(\cos^2 x\) later.
Any angle doubles: \(\sin 6x = 2\sin 3x\cos 3x\) and \(\cos x = 1 - 2\sin^2\dfrac{x}{2}\).
Worked example All three from one ratio
Given that \(\sin A = \dfrac{3}{5}\) and \(A\) is obtuse, find the exact values of \(\sin 2A\), \(\cos 2A\) and \(\tan 2A\).
\(A\) is in the second quadrant, so \(\cos A = -\dfrac{4}{5}\).
\(\sin 2A \ne 2\sin A\). With \(A = 90^\circ\): \(\sin 180^\circ = 0\), but \(2\sin 90^\circ = 2\). The 2 belongs inside the function, so you need the formula.
Section 02
Solving equations.
Replace the double angle so that only one angle, \(x\), remains. Then either factorise (if every term shares a factor) or make a quadratic in a single function.
Worked example Factorise, do not divide
Solve \(\sin 2x = \cos x\) for \(0^\circ \le x \le 360^\circ\).
\(2\sin x\cos x = \cos x\), so \(2\sin x\cos x - \cos x = 0\).
Factorise: \(\cos x(2\sin x - 1) = 0\).
\(\cos x = 0\): \(x = 90^\circ, 270^\circ\).
\(\sin x = \dfrac{1}{2}\): \(x = 30^\circ, 150^\circ\).
Dividing both sides of \(2\sin x\cos x = \cos x\) by \(\cos x\) throws away the solutions \(\cos x = 0\) (here \(90^\circ\) and \(270^\circ\)). Bring everything to one side and factorise instead.
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 expression is identical to \(\sin 2A\)?
Quick check+10 XP
Which is \(\cos 2A\) written in terms of \(\sin A\) only?
Quick check+10 XP
Find the exact value of \(\sin 15^\circ\cos 15^\circ\).
Quick check+10 XP
If \(\tan A = t\), what is \(\tan 2A\)?
Section 03
Identities and triple angles.
In proofs, choose the version of \(\cos 2x\) that cancels the 1 that is already there: \(1 - \cos 2x = 2\sin^2 x\) and \(1 + \cos 2x = 2\cos^2 x\).
Worked example Prove an identity
Prove that \(\dfrac{1 - \cos 2x}{\sin 2x} \equiv \tan x\).