Derivation of $\sin(15^\circ)$ geometrically

Question

This is my attempt:-

Let us consider a right $\triangle ABC$ such that angle $A$ is $15^\circ$ and $C$ is $75^\circ$.

On the line $AB$, let us assume a point $D$ such that $\frac{BC}{BD} =\frac{ 1}{\sqrt{3}}$ (Without Loss Of Generality). So $\angle BDC$ becomes $30^\circ$ and $\angle BCD$ becomes $60^\circ$. Then $\angle DCA$ becomes equal to $\angle BAC$, that is $15^\circ$; so $CD = DA$.

$CD$ will be $2$ times $BC$ (angle $BDC = 30^\circ$; $\sin 30^\circ$). On adding $BD$ and $AD$ we get $$AB = BC(2+\sqrt{3})$$

$$BC^2 + AB^2 = AC^2$$ $$\therefore AC = 2BC\sqrt{2 + \sqrt{3}}$$ $$\sin(15^\circ) = \frac{BC}{AC}$$ $$\sin(15^\circ) = \frac{1}{\sqrt{2 + \sqrt{3}}}$$

Rationalising the denominator 2 times we get:- $$\sin(15^\circ) = \frac{(4-2\sqrt{3})(\sqrt{2+\sqrt{3}})}{4}$$

Further simplifying:- $$\sin(15^\circ) = \frac{(2-\sqrt{3})(\sqrt{2+\sqrt{3}})}{2}$$ $$\sin(15^\circ) = \frac{(\sqrt{2-\sqrt{3}})(\sqrt{2-\sqrt{3}}) (\sqrt{2+\sqrt{3}})}{2}$$

$$\sin(15^\circ) = \frac{\sqrt{2-\sqrt{3}}}{2}$$

Is my answer correct? And is there any other method or way to get the value of $\sin(15^\circ)$ geometrically?

Answer 1

You obtained the correct answer. However, not every step you wrote down is correct.

You found that $$AC = 2BC\sqrt{2 + \sqrt{3}}$$ from which it follows that $$\sin(15^\circ) = \frac{BC}{AC} = \frac{1}{2\sqrt{2 + \sqrt{3}}}$$ Multiplying the numerator and denominator by $\sqrt{2 - \sqrt{3}}$ yields $$\sin(15^\circ) = \frac{\sqrt{2 - \sqrt{3}}}{2\sqrt{4 - 3}} = \frac{\sqrt{2 - \sqrt{3}}}{2}$$

As for an alternative method, consider the diagram below.

Since $\angle AEC$ is a right angle, $\triangle ACE$ is a right triangle. Since the acute angles of a right triangle are complementary and $m\angle CAE = 45^\circ$, $m\angle ACE = 45^\circ$. Since $CE = 1$ and the ratio of the side lengths of a $45^\circ, 45^\circ, 90^\circ$ right triangle is $1: 1: \sqrt{2}$, $AE = 1$ and $AC = \sqrt{2}$.

Since $\angle AEC$ is a right angle, $\overline{AD} \perp \overline{CE}$. Thus, $\angle CED$ is also a right angle. Therefore, $\triangle CDE$ is a right triangle. Since the acute angles of a right triangle are complementary and $m\angle CDE = 30^\circ$, $m\angle ECD = 60^\circ$. Since the ratio of the side lengths in a $30^\circ, 60^\circ, 90^\circ$ right triangle is $1: \sqrt{3}: 2$ and $CE = 1$, $DE = \sqrt{3}$ and $CD = 2$.

By the Segment Addition Postulate, $AD = AE + ED = 1 + \sqrt{3}$.

Since $\angle ABD$ is a right angle, $\triangle ABD$ is a right triangle. Since the acute angles of a right triangle are complementary and $m\angle BDA = 30^\circ$, $m\angle BAD = 60^\circ$. Since the ratio of the side lengths in a $30^\circ, 60^\circ, 90^\circ$ right triangle is $1: \sqrt{3}: 2$ and $AD = 1 + \sqrt{3}$, $AB = \dfrac{1 + \sqrt{3}}{2}$ and $BD = \dfrac{3 + \sqrt{3}}{4}$.

By the Angle Addition Postulate, $m\angle BAD = m\angle BAC + m\angle CAE$. Hence, $m\angle BAC = m\angle BAD - m\angle CAE = 60^\circ - 45^\circ = 15^\circ$.

Since $\angle ABC$ is a right angle, $\triangle ABC$ is a right triangle. Since the acute angles of a right triangle are complementary and $m\angle BAC = 15^\circ$, $m\angle ACB = 75^\circ$.

By the Segment Addition Postulate, $BC + CD = BD$. Hence, $BC = BD - CD = \dfrac{3 + \sqrt{3}}{4} - 2 = \dfrac{-1 + \sqrt{3}}{2}$.

We summarize these results in the diagram below.

\begin{align*} \sin(15^\circ) & = \frac{BC}{AC}\\ & = \frac{\frac{-1 + \sqrt{3}}{2}}{\sqrt{2}}\\ & = \frac{-1 + \sqrt{3}}{2\sqrt{2}}\\ & = \frac{-1 + \sqrt{3}}{2\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}\\ & = \frac{-\sqrt{2} + \sqrt{6}}{4}\\ & = \frac{\sqrt{6} - \sqrt{2}}{4} \end{align*} which is equivalent to your answer since \begin{align*} \frac{\sqrt{2 - \sqrt{3}}}{2} & = \frac{\sqrt{2 - \sqrt{3}}}{2} \cdot \frac{2}{2}\\ & = \frac{\sqrt{8 - 4\sqrt{3}}}{4}\\ & = \frac{\sqrt{6 - 4\sqrt{3} + 2}}{4}\\ & = \frac{\sqrt{(\sqrt{6} - \sqrt{2})^2}}{4}\\ & = \frac{\sqrt{6} - \sqrt{2}}{4} \end{align*}

Answer 2

There is another method that starts with a fun and challenging geometry problem. Not sure if you've seen it before, but it is fairly famous (I don't know if it has a specific name though):

Now, once you've solved that purely with elementary geometry, no trig, you can then get some nice trig results from that. $M$ and $N$ are the midpoints of $AB$ and $CD$ respectively:

Let the square have side $2$. You've already shown $\triangle AOB$ is equilateral, so $\angle OAB = 60^{\circ}$. Which gives $OM = AM\tan 60^{\circ} = \sqrt 3$. Then $ON = 2 - \sqrt 3$. You can get $OD$ by Pythagoras': $OD = \sqrt{(2 - \sqrt 3)^2 + 1^2} = 2\sqrt{2 - \sqrt 3}$. So $\sin 15^{\circ} = \frac{ON}{OD} = \frac 12\sqrt{2-\sqrt 3}$.