13.14 Symmetry Identities and Reciprocal Ratio Equations
AHL 3.9 AHL 3.11
This section is for HL students. It builds on 13.13.
- the symmetry identities such as \(\sin(\pi - \theta) = \sin\theta\)
- how to solve equations with \(\sec^2 x\) and \(\tan x\) (or \(\operatorname{cosec}^2 x\) and \(\cot x\)) together
Symmetry identities
The angles \(\pi - \alpha\), \(\pi + \alpha\) and \(2\pi - \alpha\) in the table of 13.13 give the symmetry properties of the graphs. They come from the unit circle: the point for \(\pi - \theta\) is the reflection of the point for \(\theta\) in the \(y\)-axis.
Symmetry identities Not in the formula booklet
| \(\sin\) | \(\cos\) | \(\tan\) | |
|---|---|---|---|
| \(\pi - \theta\) | \(\sin(\pi - \theta) = \sin\theta\) | \(\cos(\pi - \theta) = -\cos\theta\) | \(\tan(\pi - \theta) = -\tan\theta\) |
| \(\pi + \theta\) | \(\sin(\pi + \theta) = -\sin\theta\) | \(\cos(\pi + \theta) = -\cos\theta\) | \(\tan(\pi + \theta) = \tan\theta\) |
| \(-\theta\) | \(\sin(-\theta) = -\sin\theta\) | \(\cos(-\theta) = \cos\theta\) | \(\tan(-\theta) = -\tan\theta\) |
- \(\theta\): any angle (in radians; in degrees replace \(\pi\) by \(180^\circ\))
Example 1 Given that \(\sin\dfrac{\pi}{5} = k\), write each expression in terms of \(k\).
(a) \(\sin\dfrac{4\pi}{5}\) \(\qquad\) (b) \(\sin\dfrac{6\pi}{5}\) \(\qquad\) (c) \(\sin\left(-\dfrac{\pi}{5}\right)\)
(d) \(\cos\dfrac{4\pi}{5}\) \(\qquad\) (e) \(\tan\dfrac{6\pi}{5}\)
(a) \(\sin\dfrac{4\pi}{5} = \sin\left(\pi - \dfrac{\pi}{5}\right) = \sin\dfrac{\pi}{5} = k\).
(b) \(\sin\dfrac{6\pi}{5} = \sin\left(\pi + \dfrac{\pi}{5}\right) = -\sin\dfrac{\pi}{5} = -k\).
(c) \(\sin\left(-\dfrac{\pi}{5}\right) = -\sin\dfrac{\pi}{5} = -k\).
(d) \(\dfrac{\pi}{5}\) is acute, so \(\cos\dfrac{\pi}{5} = +\sqrt{1 - k^2}\). Then \(\cos\dfrac{4\pi}{5} = -\cos\dfrac{\pi}{5} = -\sqrt{1 - k^2}\).
(e) \(\tan\dfrac{6\pi}{5} = \tan\dfrac{\pi}{5} = \dfrac{k}{\sqrt{1 - k^2}}\).
Equations with reciprocal ratios
When an equation has \(\sec^2 x\) and \(\tan x\) together, replace \(\sec^2 x\) by \(1 + \tan^2 x\) (13.12). In the same way, replace \(\operatorname{cosec}^2 x\) by \(1 + \cot^2 x\). Then solve a quadratic, as in 13.13.
Example 2 Paper 2 Solve \(\sec^2 x - 3\tan x + 1 = 0\), for \(0 \leq x \leq 2\pi\), giving answers to 3 significant figures.
Replace \(\sec^2 x\) by \(1 + \tan^2 x\):
\[ \tan^2 x - 3\tan x + 2 = 0 \;\Longrightarrow\; (\tan x - 1)(\tan x - 2) = 0 . \]
\(\tan x = 1\): \(\;x = \dfrac{\pi}{4}\) or \(\dfrac{5\pi}{4}\). \(\quad\tan x = 2\): \(\;\alpha = \arctan 2 = 1.107\ldots\), quadrants 1 and 3: \(\;x = 1.107\) or \(\pi + 1.107 = 4.249\).
So \(x = 0.785, \; 1.11, \; 3.93, \; 4.25\).
Exercises
1 Solve \(\operatorname{cosec}^2 x - 2\cot x = 0\) for \(0 < x < 2\pi\).
Replace \(\operatorname{cosec}^2 x\) by \(1 + \cot^2 x\):
\[ \cot^2 x - 2\cot x + 1 = 0 \;\Longrightarrow\; (\cot x - 1)^2 = 0 \;\Longrightarrow\; \cot x = 1 . \]
So \(\tan x = 1\): \(\;x = \dfrac{\pi}{4}, \; \dfrac{5\pi}{4}\).
2 Paper 1 Solve \(2\tan^2 x = 3\sec x\) for \(0 \leq x \leq 2\pi\).
Replace \(\tan^2 x\) by \(\sec^2 x - 1\):
\[ \begin{aligned} 2(\sec^2 x - 1) &= 3\sec x \\ 2\sec^2 x - 3\sec x - 2 &= 0 \\ (2\sec x + 1)(\sec x - 2) &= 0 . \end{aligned} \]
\(\sec x = -\dfrac{1}{2}\) means \(\cos x = -2\): no solution (always \(|\sec x| \geq 1\)).
\(\sec x = 2\) means \(\cos x = \dfrac{1}{2}\): \(\;x = \dfrac{\pi}{3}\) or \(\dfrac{5\pi}{3}\).