Review question

# An exact value for $\sin 15^{\circ}$... Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7050

## Solution

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 $CAE$.

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*}