13.16 Compound Angle Formulae
AHL 3.10
This section is for HL students. It builds on 13.15.
- what a compound angle is, and why \(\sin(A + B) \neq \sin A + \sin B\)
- the addition formulae for \(\cos(A \pm B)\), \(\sin(A \pm B)\) and \(\tan(A \pm B)\), and where they come from
- how to use them to find exact values, such as \(\cos 75^\circ\)
- how the double angle formulae come from the addition formulae, and the double angle formula for \(\tan 2A\)
Compound angles
For any two angles \(A\) and \(B\), the angles \(A + B\) and \(A - B\) are called compound angles. In this section we write \(\sin(A + B)\), \(\cos(A - B)\), … in terms of the sines and cosines of \(A\) and \(B\) themselves.
First, a warning. The sine function does not “multiply out” over a bracket:
\[ \sin(30^\circ + 60^\circ) = \sin 90^\circ = 1, \qquad\text{but}\qquad \sin 30^\circ + \sin 60^\circ = \frac{1}{2} + \frac{\sqrt{3}}{2} \neq 1 . \]
So we need proper formulae.
Addition formulae for cosine
We prove the formula for \(\cos(A - B)\) with the unit circle (see 13.6) and the distance formula.
Example 1 In Figure 1, \(P\) and \(Q\) lie on the unit circle. The radius \(OP\) makes an angle \(B\) with the positive \(x\)-axis, and \(OQ\) makes an angle \(A\), where \(A > B\). In Diagram 2 the triangle \(OPQ\) has been rotated clockwise by \(B\) about \(O\), so that \(P\) moves to \((1, 0)\).
(a) Explain why
(i) \(P\) and \(Q\) in Diagram 1 have coordinates \((\cos B, \sin B)\) and \((\cos A, \sin A)\)
(ii) \(Q\) in Diagram 2 has coordinates \(\big(\cos(A - B), \sin(A - B)\big)\).
(b) Use the distance formula \(d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\) to find \(PQ^2\) from each diagram.
(c) Hence show that \(\cos(A - B) = \cos A \cos B + \sin A \sin B\).
(d) By replacing \(B\) with \(-B\), show that \(\cos(A + B) = \cos A \cos B - \sin A \sin B\).
(a)
(i) A point on the unit circle at angle \(\theta\) from the positive \(x\)-axis has coordinates \((\cos\theta, \sin\theta)\). This is the definition of sine and cosine from 13.6.
(ii) After the rotation, \(OP\) lies along the positive \(x\)-axis. The angle \(\angle POQ = A - B\) does not change, so \(OQ\) now makes an angle \(A - B\) with the \(x\)-axis.
(b) Diagram 1:
\[ \begin{aligned} PQ^2 &= (\cos A - \cos B)^2 + (\sin A - \sin B)^2 \\ &= \cos^2 A + \sin^2 A + \cos^2 B + \sin^2 B \\ &\qquad - 2\cos A\cos B - 2\sin A\sin B \\ &= 2 - 2\cos A\cos B - 2\sin A\sin B . \end{aligned} \]
We used \(\cos^2\theta + \sin^2\theta = 1\) twice.
Diagram 2:
\[ \begin{aligned} PQ^2 &= \big(\cos(A - B) - 1\big)^2 + \sin^2(A - B) \\ &= \cos^2(A - B) + \sin^2(A - B) - 2\cos(A - B) + 1 \\ &= 2 - 2\cos(A - B) . \end{aligned} \]
(c) A rotation does not change lengths, so the two expressions are equal:
\[ 2 - 2\cos(A - B) = 2 - 2\cos A\cos B - 2\sin A\sin B , \]
which gives \(\cos(A - B) = \cos A\cos B + \sin A\sin B\).
(d) Replace \(B\) with \(-B\), and use \(\cos(-B) = \cos B\) and \(\sin(-B) = -\sin B\):
\[ \begin{aligned} \cos(A + B) &= \cos A\cos(-B) + \sin A\sin(-B) \\ &= \cos A\cos B - \sin A\sin B . \;\blacksquare \end{aligned} \]
Addition formulae for cosine In the HL formula booklet
\[ \cos(A \pm B) = \cos A\cos B \mp \sin A\sin B \]
- \(A\), \(B\): any two angles (in degrees or radians)
- \(\pm\), \(\mp\): take the top signs together, or the bottom signs together
Notice the sign changes: \(\cos(A + B)\) has a minus in the middle, and \(\cos(A - B)\) has a plus.
Addition formulae for sine
We change sine into cosine with \(\sin\theta = \cos\left(\frac{\pi}{2} - \theta\right)\):
\[ \begin{aligned} \sin(A + B) &= \cos\left(\left(\tfrac{\pi}{2} - A\right) - B\right) \\ &= \cos\left(\tfrac{\pi}{2} - A\right)\cos B + \sin\left(\tfrac{\pi}{2} - A\right)\sin B \\ &= \sin A\cos B + \cos A\sin B . \end{aligned} \]
Replacing \(B\) with \(-B\) again gives \(\sin(A - B)\).
Addition formulae for sine In the HL formula booklet
\[ \sin(A \pm B) = \sin A\cos B \pm \cos A\sin B \]
- \(A\), \(B\): any two angles
- \(\pm\): the sign is the same on both sides
Example 2 Given that \(\cos A = -\dfrac{8}{17}\) and \(\tan B = \dfrac{7}{24}\), where \(\dfrac{\pi}{2} < A < \pi\) and \(\pi < B < \dfrac{3\pi}{2}\), find the exact value of
(a) \(\cos(A - B)\) \(\qquad\) (b) \(\sin(A + B)\)
First find the missing ratios. Draw right-angled triangles, then choose the signs from the quadrant.
\(A\) is in the second quadrant, so \(\sin A > 0\): \(\;\sin A = \sqrt{1 - \dfrac{64}{289}} = \dfrac{15}{17}\).
\(B\) is in the third quadrant, so \(\sin B < 0\) and \(\cos B < 0\). A triangle with sides \(7\) and \(24\) has hypotenuse \(\sqrt{49 + 576} = 25\), so \(\sin B = -\dfrac{7}{25}\) and \(\cos B = -\dfrac{24}{25}\).
(a)
\[ \begin{aligned} \cos(A - B) &= \cos A\cos B + \sin A\sin B \\ &= \left(-\frac{8}{17}\right)\left(-\frac{24}{25}\right) + \frac{15}{17}\left(-\frac{7}{25}\right) \\ &= \frac{192 - 105}{425} \\ &= \frac{87}{425} . \end{aligned} \]
(b)
\[ \begin{aligned} \sin(A + B) &= \sin A\cos B + \cos A\sin B \\ &= \frac{15}{17}\left(-\frac{24}{25}\right) + \left(-\frac{8}{17}\right)\left(-\frac{7}{25}\right) \\ &= \frac{-360 + 56}{425} \\ &= -\frac{304}{425} . \end{aligned} \]
Write an angle as a sum or difference of \(30^\circ\), \(45^\circ\), \(60^\circ\) (or \(\frac{\pi}{6}\), \(\frac{\pi}{4}\), \(\frac{\pi}{3}\)) to find exact values of other angles.
Example 3 Without using a calculator, find the exact value of
(a) \(\cos 75^\circ\) \(\qquad\) (b) \(\sin\dfrac{\pi}{12}\)
(a) \(75^\circ = 45^\circ + 30^\circ\):
\[ \begin{aligned} \cos 75^\circ &= \cos 45^\circ\cos 30^\circ - \sin 45^\circ\sin 30^\circ \\ &= \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4} . \end{aligned} \]
(b) \(\dfrac{\pi}{12} = \dfrac{\pi}{3} - \dfrac{\pi}{4}\):
\[ \begin{aligned} \sin\frac{\pi}{12} &= \sin\frac{\pi}{3}\cos\frac{\pi}{4} - \cos\frac{\pi}{3}\sin\frac{\pi}{4} \\ &= \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} - \frac{1}{2}\cdot\frac{\sqrt{2}}{2} = \frac{\sqrt{6} - \sqrt{2}}{4} . \end{aligned} \]
The answers are equal because \(\cos 75^\circ = \sin 15^\circ\) and \(\dfrac{\pi}{12} = 15^\circ\).
Addition formula for tangent
Example 4 (a) Use \(\tan(A + B) = \dfrac{\sin(A + B)}{\cos(A + B)}\) and the formulae above to show that
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B} . \]
(b) By replacing \(B\) with \(-B\), show that \(\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A\tan B}\).
(a) Divide the numerator and the denominator by \(\cos A\cos B\):
\[ \begin{aligned} \tan(A + B) &= \frac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B} \\[4pt] &= \frac{\dfrac{\sin A\cos B}{\cos A\cos B} + \dfrac{\cos A\sin B}{\cos A\cos B}}{\dfrac{\cos A\cos B}{\cos A\cos B} - \dfrac{\sin A\sin B}{\cos A\cos B}} \\[4pt] &= \frac{\tan A + \tan B}{1 - \tan A\tan B} . \end{aligned} \]
(b) \(\tan(-B) = -\tan B\) (tangent is an odd function), so
\[ \tan(A - B) = \frac{\tan A + \tan(-B)}{1 - \tan A\tan(-B)} = \frac{\tan A - \tan B}{1 + \tan A\tan B} . \;\blacksquare \]
Addition formula for tangent In the HL formula booklet
\[ \tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B} \]
- \(A\), \(B\): angles for which all the tangents are defined
- \(\pm\), \(\mp\): take the top signs together, or the bottom signs together
Example 5 The angles \(A\) and \(B\) are acute, with \(\tan A = 2\) and \(\tan B = 3\).
(a) Find the value of \(\tan(A + B)\).
(b) Hence show that \(A + B = \dfrac{3\pi}{4}\).
(a) \(\tan(A + B) = \dfrac{2 + 3}{1 - 2 \times 3} = \dfrac{5}{-5} = -1\).
(b) \(\tan A > 1\) and \(\tan B > 1\), so both angles are between \(\dfrac{\pi}{4}\) and \(\dfrac{\pi}{2}\). So \(\dfrac{\pi}{2} < A + B < \pi\). The only angle in this interval with tangent \(-1\) is \(\dfrac{3\pi}{4}\). \(\;\blacksquare\)
Double angle formulae from the addition formulae
Put \(B = A\) in the addition formulae. For example, \(\sin(A + A) = \sin A\cos A + \cos A\sin A = 2\sin A\cos A\).
In the same way, \(\cos 2A = \cos^2 A - \sin^2 A\). The other two forms of \(\cos 2A\) are in 13.15.
Put \(B = A\) in the addition formula for tangent to get the double angle formula for tangent.
Double angle formula for tangent In the HL formula booklet
\[ \tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} \]
- \(\theta\): an angle with \(\tan\theta \neq \pm 1\) and \(\tan\theta\) defined
Example 6 Given that \(\cos A = -\dfrac{12}{13}\) and \(\pi < A < \dfrac{3\pi}{2}\), use the double angle formula for tangent to find the exact value of \(\tan 2A\).
\(A\) is in the third quadrant, so \(\sin A < 0\): \(\;\sin A = -\sqrt{1 - \dfrac{144}{169}} = -\dfrac{5}{13}\).
\(\tan A = \dfrac{5}{12}\), so \(\tan 2A = \dfrac{2 \cdot \frac{5}{12}}{1 - \frac{25}{144}} = \dfrac{5}{6} \times \dfrac{144}{119} = \dfrac{120}{119}\).
Example 7 Solve \(\tan 2x = 3\tan x\) for \(0^\circ \leq x \leq 360^\circ\).
Let \(t = \tan x\). Then \(\dfrac{2t}{1 - t^2} = 3t\), so \(2t = 3t(1 - t^2)\) and
\[ t\,(3t^2 - 1) = 0 \;\Longrightarrow\; t = 0 \;\text{ or }\; t = \pm\frac{1}{\sqrt{3}} . \]
- \(\tan x = 0\): \(\;x = 0^\circ, 180^\circ, 360^\circ\).
- \(\tan x = \dfrac{1}{\sqrt{3}}\): \(\;x = 30^\circ, 210^\circ\).
- \(\tan x = -\dfrac{1}{\sqrt{3}}\): \(\;x = 150^\circ, 330^\circ\).
None of these makes \(\tan x\) or \(\tan 2x\) undefined. So \(x = 0^\circ, 30^\circ, 150^\circ, 180^\circ, 210^\circ, 330^\circ, 360^\circ\).
- Check that the angle mode is DEG (shown at the top right of the screen; click on it to change).
- Open a Graphs page.
- Enter \(f1(x) = \tan(2x)\) and \(f2(x) = 3\tan(x)\).
- Menu → Window / Zoom → Window Settings: set \(x\) from \(0\) to \(360\) and \(y\) from \(-10\) to \(10\).
- Menu → Analyze Graph → Intersection, then choose a lower and an upper bound around each crossing point.
The \(x\)-coordinates of the intersection points are the solutions. Count them: there should be seven, as in Example 7.
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \(\sin(A + B) = \sin A + \sin B\) | \(\sin A\cos B + \cos A\sin B\) | Trigonometric functions do not “multiply out” over a bracket. |
| \(\cos(A + B) = \cos A\cos B + \sin A\sin B\) | \(\cos A\cos B - \sin A\sin B\) | For cosine the sign in the middle is the opposite sign. |
Exercises
1 Without using a calculator, find the exact value of
(a) \(\sin 105^\circ\)
(b) \(\tan 105^\circ\), giving your answer in the form \(a + b\sqrt{3}\).
(a) \(\sin(60^\circ + 45^\circ) = \dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2} + \dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2} = \dfrac{\sqrt{6} + \sqrt{2}}{4}\).
(b) \(\tan(60^\circ + 45^\circ) = \dfrac{\sqrt{3} + 1}{1 - \sqrt{3}}\). Multiply the top and the bottom by \(1 + \sqrt{3}\):
\[ \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3} . \]
2 Given that \(\sin A = \dfrac{3}{5}\) and \(\cos B = \dfrac{15}{17}\), where \(90^\circ < A < 180^\circ\) and \(0^\circ < B < 90^\circ\), find the exact value of
(a) \(\sin(A + B)\) \(\qquad\) (b) \(\cos(A - B)\)
\(\cos A = -\dfrac{4}{5}\) (second quadrant) and \(\sin B = \dfrac{8}{17}\) (first quadrant).
(a) \(\sin(A + B) = \dfrac{3}{5}\cdot\dfrac{15}{17} + \left(-\dfrac{4}{5}\right)\dfrac{8}{17} = \dfrac{45 - 32}{85} = \dfrac{13}{85}\).
(b) \(\cos(A - B) = \left(-\dfrac{4}{5}\right)\dfrac{15}{17} + \dfrac{3}{5}\cdot\dfrac{8}{17} = \dfrac{-60 + 24}{85} = -\dfrac{36}{85}\).
3
(a) Show that \(\tan(\theta + 45^\circ) = \dfrac{1 + \tan\theta}{1 - \tan\theta}\).
(b) Hence solve \(\tan(\theta + 45^\circ) = 6\tan\theta\) for \(0^\circ < \theta < 90^\circ\). Give your answers to 1 decimal place.
(a) \(\tan(\theta + 45^\circ) = \dfrac{\tan\theta + \tan 45^\circ}{1 - \tan\theta\tan 45^\circ} = \dfrac{1 + \tan\theta}{1 - \tan\theta}\), because \(\tan 45^\circ = 1\).
(b) Let \(t = \tan\theta\). Then \(1 + t = 6t(1 - t)\), so \(6t^2 - 5t + 1 = 0\), which factorises as \((2t - 1)(3t - 1) = 0\).
\(t = \dfrac{1}{2}\) gives \(\theta = 26.6^\circ\), and \(t = \dfrac{1}{3}\) gives \(\theta = 18.4^\circ\).