(i) Show that the equation \(\sin \theta + \cos \theta = 2(\sin \theta - \cos \theta)\) can be expressed as \(\tan \theta = 3\).
(ii) Hence solve the equation \(\sin \theta + \cos \theta = 2(\sin \theta - \cos \theta)\), for \(0^\circ \leq \theta \leq 360^\circ\).
Solution
(i) Start with the equation \(\sin \theta + \cos \theta = 2(\sin \theta - \cos \theta)\).
Let \(s = \sin \theta\) and \(c = \cos \theta\).
Then the equation becomes \(s + c = 2s - 2c\).
Rearrange to get \(s + c = 2s - 2c \Rightarrow s + c = 2s - 2c \Rightarrow s = 3c\).
Divide both sides by \(c\) (assuming \(c \neq 0\)) to get \(\frac{s}{c} = 3\).
Thus, \(\tan \theta = 3\).
(ii) From \(\tan \theta = 3\), find \(\theta\) in the range \(0^\circ \leq \theta \leq 360^\circ\).
The principal value is \(\theta = \tan^{-1}(3) \approx 71.6^\circ\).
The other solution in the range is \(\theta = 180^\circ + 71.6^\circ = 251.6^\circ\).
Thus, \(\theta = 71.6^\circ\) or \(251.6^\circ\).
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