1. Prove that \[\cos 3\theta \sin 2\theta - \cos 4\theta \sin \theta = \cos 2\theta \sin \theta.\]

Approach 1

We know \(\sin 2\theta=2\sin\theta\cos\theta\), so the left hand side becomes \[\begin{align*} \cos 3\theta \sin 2\theta - \cos 4\theta \sin \theta &{}= \cos 3\theta (2\sin\theta\cos\theta) - \cos 4\theta \sin \theta\\ &{}= \sin\theta(2\cos 3\theta\cos\theta - \cos 4\theta) \end{align*}\] Now \(\cos3\theta\cos\theta\) appears in the expansion of \(\cos 4\theta=\cos(3\theta+\theta)\), giving \[\begin{align*} &\cos 3\theta \sin 2\theta - \cos 4\theta \sin \theta\\ &\qquad{}=\sin\theta(2\cos 3\theta\cos\theta - \cos 4\theta)\\ &\qquad{}=\sin\theta(2\cos 3\theta \cos\theta - (\cos 3\theta\cos\theta- \sin3\theta\sin\theta))\\ &\qquad{}=\sin\theta(\cos 3\theta \cos\theta + \sin3\theta\sin\theta)\\ &\qquad{}=\sin\theta\cos(3\theta-\theta)\\ &\qquad{}=\sin\theta\cos 2\theta. \end{align*}\]

Approach 2

Using the trig expansions we know for \(\sin(P+Q), \sin(P-Q), \cos(P+Q)\) and \(\cos(P-Q)\), we have

\[\begin{align*} \sin P\cos Q &{}= \dfrac{1}{2}(\sin(P+Q)+\sin(P-Q))\\ \cos P\sin Q &{}= \dfrac{1}{2}(\sin(P+Q)-\sin(P-Q))\\ \cos P\cos Q &{}= \dfrac{1}{2}(\cos(P+Q)+\cos(P-Q))\\ \sin P\sin Q &{}= -\dfrac{1}{2}(\cos(P+Q)-\cos(P-Q)) \end{align*}\]

Now putting \(A = P+Q\) and \(B = P-Q\), we have

\[\begin{align*} \sin A+\sin B &{}= 2\sin \frac{A+B}{2}\cos \frac{A-B}{2} \\ \sin A-\sin B &{}= 2\cos \frac{A+B}{2}\sin \frac{A-B}{2} \\ \cos A+\cos B &{}= 2\cos \frac{A+B}{2}\cos \frac{A-B}{2} \\ \cos A-\cos B &{}= -2\sin \frac{A+B}{2}\sin \frac{A-B}{2} \end{align*}\]
This gives \[\begin{align*} \cos 3\theta \sin 2\theta &{}- \cos 4\theta \sin \theta \\ &\quad{}=\dfrac{1}{2}(\sin 5\theta-\sin\theta-\sin 5\theta+\sin 3\theta)\\ &\quad{}=\dfrac{1}{2}(\sin 3\theta-\sin\theta)\\ &\quad{}=\cos 2\theta \sin \theta. \end{align*}\]
  1. Starting from the relations \[\begin{align*} \sin(A+B) &{}= \sin A \cos B + \cos A \sin B,\\ \cos(A+B) &{}= \cos A \cos B - \sin A \sin B, \end{align*}\] obtain an expression for \(\tan(A+B)\) in terms of \(\tan A\) and \(\tan B\).

We know \(\tan\theta=\dfrac{\sin\theta}{\cos\theta}\), giving

\[\begin{align*} \tan(A+B) &{}= \frac{\sin(A+B)}{\cos(A+B)} \\ &{}= \frac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B}. \end{align*}\]

Now we divide the numerator and denominator of the fraction by \(\cos A\cos B\), giving

\[\begin{align*} \tan(A+B) &{}= \dfrac{\dfrac{\sin A}{\cos A} + \dfrac{\sin B}{\cos B}}{1 - \dfrac{\sin A\sin B}{\cos A\cos B}} \\ &{}= \dfrac{\tan A + \tan B}{1 - \tan A\tan B} \end{align*}\]

Without the use of a calculator or tables, show that the sum of the three acute angles whose tangents are \(\dfrac{2}{9}\), \(\dfrac{1}{4}\) and \(\dfrac{1}{3}\) is \(\dfrac{\pi}{4}\).

Call the three angles \(A\), \(B\) and \(C\), so that \(\tan A = \dfrac{2}{9}\), \(\tan B = \dfrac{1}{4}\) and \(\tan C = \dfrac{1}{3}\).

Then \[\begin{align*} \tan(A+C) &{}= \frac{\tan A + \tan C}{1 - \tan A\tan C}\\ &= \frac{\frac{2}{9}+\frac{1}{3}}{1 - \frac{2}{9}\frac{1}{3}}\\ &= \frac{(\frac{5}{9})}{(\frac{25}{27})}\\ &= \frac{3}{5}. \end{align*}\] Now \[\begin{align*} \tan(A+C+B) &{}= \frac{\tan(A+C) + \tan B}{1 - \tan(A+C)\tan B}\\ &= \frac{\frac{3}{5}+\frac{1}{4}}{1 - \frac{3}{5}\frac{1}{4}}\\ &= \frac{(\frac{17}{20})}{(\frac{17}{20})}\\ &= 1. \end{align*}\]

We are told that \(A\), \(B\) and \(C\) are acute. Also, since \(\tan A\), \(\tan B\) and \(\tan C\) are all less than \(1\), the angles \(A\), \(B\) and \(C\) are all between \(0\) and \(\dfrac{\pi}{4}\). Therefore,\(0 < A + B + C < \dfrac{3\pi}{4}\).

So the only possible solution of \(\tan (A+B+C) = 1\) with \(A+B+C\) in this range is \(A+B+C = \dfrac{\pi}{4}\), as required.