1. Without using tables, show for \(\theta=15^\circ\) that \(\sin\theta=\dfrac{\sqrt{6}-\sqrt{2}}{4}\), and find the values of \(\cos\theta\) and \(\tan\theta\).
We will use the identity \(\sin(x-y) = \sin x \cos y - \sin y \cos x\). We have that \[\begin{align*} \sin 15^\circ=\sin(45-30)^\circ &=\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ \\ &=\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\frac{1}{2} \\ &=\frac{\sqrt{2}\sqrt{3}}{2\times 2}-\frac{\sqrt{2}}{2\times 2} \\ &=\frac{\sqrt{6}-\sqrt{2}}{4}. \end{align*}\] We now have \[\begin{align*} \cos^2 15^\circ=1-\sin^2 15^\circ &=1-\frac{6+2-2\sqrt{6}\sqrt{2}}{16} \\ &=\frac{8+2\sqrt{12}}{16}\\ &=\frac{6+2+2\sqrt{2}\sqrt{6}}{16} \\ &=\frac{(\sqrt{6}+\sqrt{2})^2}{4^2}, \end{align*}\]

and so, since \(\cos \theta\) is positive between \(0^\circ\) and \(90^\circ\), \[\cos 15^\circ=\frac{\sqrt{6}+\sqrt{2}}{4}.\]

Finally, we have \[\begin{align*} \tan 15^\circ=\frac{\sin15^\circ}{\cos15^\circ} &=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \\ &=\frac{(\sqrt{6}-\sqrt{2})^2}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})} \\ &=\frac{6+2-2\sqrt{2}\sqrt{6}}{6-2} \\ &=\frac{8-2\sqrt{12}}{4} \\ &=2-\sqrt{3}. \end{align*}\]
  1. In the triangle \(ABC\), \(B\) is a right angle and the other angles are such that if a point \(E\) on \(BC\) and between \(B\) and \(C\) is chosen so that \(BE=BA\), then there is a point \(D\) on \(AC\) and between \(A\) and \(C\) such that \(AD=DE=EC\). Find the ratio of \(DC\) to \(AB\).

First let’s draw what’s happening, filling in the angles by putting in \(a\) for \(ACB\) and \(b\) for \(CAE\).

Previous diagram with all angles added, all given in terms of alpha

The angle-sum in triangle \(ABC\) tells us that \(a+b = 45^\circ\). The angle at \(D\) tells us that \(a = 2b\). Solving these equations simultaneously gives \(b = 15^\circ, a = 30^\circ\).

We need the ratio of \(DC\) to \(AB\). Without loss of generality we can put \(CE = 1\), so \(CD = 2 \cos 30^\circ\) and \(EA = 2\cos 15^\circ\).

Now, by Pythagoras, \(AB = \dfrac{2\cos 15^\circ}{\sqrt{2}}\).

Thus \[\begin{align*} DC:AB &= 2\cos 30^\circ: \dfrac{2\cos 15^\circ}{\sqrt{2}}\\ &= \sqrt{3} : \sqrt{2}\frac{\sqrt{6}+\sqrt{2}}{4} \\ &= 4\sqrt{3} : \sqrt{12}+2 \\ &=2\sqrt{3} : \sqrt{3}+1. \end{align*}\]