13.15 Double Angle Formulae
SL 3.6
- the double angle formulae for \(\sin 2A\) and \(\cos 2A\)
- how to use double angle formulae to find exact values, prove identities and solve equations
Double angle formulae
The double angle formulae give \(\sin 2A\) and \(\cos 2A\) in terms of \(\sin A\) and \(\cos A\). One form is \(\cos 2A = \cos^2 A - \sin^2 A\). Using \(\sin^2 A + \cos^2 A = 1\) we can write \(\cos 2A\) in two more ways:
\[ \begin{aligned} \cos 2A &= \cos^2 A - (1 - \cos^2 A) = 2\cos^2 A - 1 \\ \cos 2A &= (1 - \sin^2 A) - \sin^2 A = 1 - 2\sin^2 A \end{aligned} \]
Double angle formulae for sine and cosine In the formula booklet
\[ \sin 2\theta = 2\sin\theta\cos\theta \]
\[ \begin{aligned} \cos 2\theta &= \cos^2\theta - \sin^2\theta \\ &= 2\cos^2\theta - 1 \\ &= 1 - 2\sin^2\theta \end{aligned} \]
- \(\theta\): any angle; \(2\theta\) is double the angle
SL students can still find \(\tan 2\theta\) as \(\dfrac{\sin 2\theta}{\cos 2\theta}\). Choose the form of \(\cos 2\theta\) that matches the question: use \(2\cos^2\theta - 1\) when you know \(\cos\theta\), and \(1 - 2\sin^2\theta\) when you know \(\sin\theta\).
Example 1 Given that \(\cos A = -\dfrac{12}{13}\) and \(\pi < A < \dfrac{3\pi}{2}\), find, without using a calculator, the exact value of
(a) \(\sin 2A\) \(\qquad\) (b) \(\cos 2A\) \(\qquad\) (c) \(\tan 2A\)
\(A\) is in the third quadrant, so \(\sin A < 0\): \(\;\sin A = -\sqrt{1 - \dfrac{144}{169}} = -\dfrac{5}{13}\).
(a) \(\sin 2A = 2\left(-\dfrac{5}{13}\right)\left(-\dfrac{12}{13}\right) = \dfrac{120}{169}\).
(b) \(\cos 2A = 2\left(-\dfrac{12}{13}\right)^2 - 1 = \dfrac{288}{169} - 1 = \dfrac{119}{169}\).
(c) \(\tan 2A = \dfrac{\sin 2A}{\cos 2A} = \dfrac{120}{119}\).
The formula \(\cos 2A = 1 - 2\sin^2 A\) can be rearranged to give \(\sin^2 A\) in terms of \(\cos 2A\). This lets us find exact values of half of a known angle.
Example 2 (a) Show that \(\sin^2 A = \dfrac{1 - \cos 2A}{2}\).
(b) Hence show that \(\sin 112.5^\circ = \dfrac{\sqrt{2 + \sqrt{2}}}{2}\).
(a) From \(\cos 2A = 1 - 2\sin^2 A\) we get \(2\sin^2 A = 1 - \cos 2A\), so \(\sin^2 A = \dfrac{1 - \cos 2A}{2}\).
(b) Put \(A = 112.5^\circ\), so \(2A = 225^\circ\) and \(\cos 225^\circ = -\dfrac{\sqrt{2}}{2}\):
\[ \sin^2 112.5^\circ = \frac{1 + \frac{\sqrt{2}}{2}}{2} = \frac{2 + \sqrt{2}}{4} . \]
\(112.5^\circ\) is in the second quadrant, where sine is positive, so take the positive root: \(\sin 112.5^\circ = \dfrac{\sqrt{2 + \sqrt{2}}}{2}\). \(\;\blacksquare\)
Solving equations with double angles
When an equation contains both \(2x\) and \(x\), use a double angle formula so that everything is in terms of one angle \(x\). Then factorise, as in 13.13.
- \(\sin 2x\) with \(\sin x\) or \(\cos x\): replace \(\sin 2x\) by \(2\sin x\cos x\) and factorise (never divide by \(\sin x\) or \(\cos x\), or you lose solutions).
- \(\cos 2x\) with \(\sin x\): use \(1 - 2\sin^2 x\) to get a quadratic in \(\sin x\).
- \(\cos 2x\) with \(\cos x\): use \(2\cos^2 x - 1\) to get a quadratic in \(\cos x\).
Example 3 Solve each equation for \(0^\circ \leq x \leq 360^\circ\).
(a) \(\sin 2x = \sin x\)
(b) \(\cos 2x - 3\sin x + 1 = 0\)
(a) \(2\sin x\cos x - \sin x = 0\), so \(\sin x\,(2\cos x - 1) = 0\).
- \(\sin x = 0\): \(\;x = 0^\circ, 180^\circ, 360^\circ\).
- \(\cos x = \dfrac{1}{2}\): \(\;x = 60^\circ, 300^\circ\).
So \(x = 0^\circ, 60^\circ, 180^\circ, 300^\circ, 360^\circ\).
(b) \((1 - 2\sin^2 x) - 3\sin x + 1 = 0\), so \(2\sin^2 x + 3\sin x - 2 = 0\), which factorises as
\[ (2\sin x - 1)(\sin x + 2) = 0 . \]
\(\sin x = -2\) has no solutions. \(\sin x = \dfrac{1}{2}\) gives \(x = 30^\circ, 150^\circ\).
HL students: continue with 13.16 Compound Angle Formulae.
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \(\sin 2A = 2\sin A\) | \(\sin 2A = 2\sin A\cos A\) | Doubling the angle does not double the sine. |
| \(\cos 2A = 2\cos A - 1\) | \(\cos 2A = 2\cos^2 A - 1\) | The cosine is squared. |
| Dividing \(\sin x\cos x = \frac{1}{2}\sin x\) by \(\sin x\) | Factorise: \(\sin x\left(\cos x - \frac{1}{2}\right) = 0\) | Dividing loses the solutions of \(\sin x = 0\). |
| \(\sin A = +\frac{15}{17}\) for \(A\) in the third quadrant | Check the quadrant: \(\sin A < 0\) there | The square root gives the size; the quadrant gives the sign. |
Exercises
1 Given that \(\cos\theta = -\dfrac{1}{4}\) and \(\dfrac{\pi}{2} < \theta < \pi\), find the exact value of
(a) \(\sin\theta\)
(b) \(\sin 2\theta\)
(c) \(\cos 2\theta\)
(a) \(\sin\theta > 0\) in the second quadrant: \(\;\sin\theta = \sqrt{1 - \dfrac{1}{16}} = \dfrac{\sqrt{15}}{4}\).
(b) \(\sin 2\theta = 2 \cdot \dfrac{\sqrt{15}}{4} \cdot \left(-\dfrac{1}{4}\right) = -\dfrac{\sqrt{15}}{8}\).
(c) \(\cos 2\theta = 2\left(\dfrac{1}{16}\right) - 1 = -\dfrac{7}{8}\).
2 Solve \(\cos 2x = \cos x\) for \(0 \leq x \leq 2\pi\).
\(2\cos^2 x - 1 = \cos x\), so \(2\cos^2 x - \cos x - 1 = 0\), which factorises as \((2\cos x + 1)(\cos x - 1) = 0\).
- \(\cos x = 1\): \(\;x = 0, 2\pi\).
- \(\cos x = -\dfrac{1}{2}\): \(\;x = \dfrac{2\pi}{3}, \dfrac{4\pi}{3}\).
So \(x = 0, \dfrac{2\pi}{3}, \dfrac{4\pi}{3}, 2\pi\).
3 Solve \(2\sin 2x = 3\sin x\) for \(0^\circ \leq x \leq 360^\circ\). Give non-exact answers to 1 decimal place.
\(4\sin x\cos x - 3\sin x = 0\), so \(\sin x\,(4\cos x - 3) = 0\).
- \(\sin x = 0\): \(\;x = 0^\circ, 180^\circ, 360^\circ\).
- \(\cos x = \dfrac{3}{4}\): \(\;x = \cos^{-1}(0.75) = 41.4^\circ\) or \(x = 360^\circ - 41.4^\circ = 318.6^\circ\).
So \(x = 0^\circ, 41.4^\circ, 180^\circ, 318.6^\circ, 360^\circ\).