Review question

# Can we find the value of $\sin 18^\circ$ as a surd? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6018

## Solution

1. Show that $\theta=18^\circ$ satisfies the equation $\cos 3\theta=\sin 2\theta$.

We know that $\cos(3\times 18^\circ)=\cos 54^\circ=\sin(90^\circ-54^\circ)=\sin 36^\circ=\sin(2 \times 18^\circ).$ So $\theta=18^\circ$ satisfies $\cos 3\theta=\sin2\theta.$

By expressing $\cos 3\theta$ and $\sin 2\theta$ in terms of trigonometrical functions of $\theta$, show that $\sin 18^\circ$ is a root of the equation $4x^2+2x-1=0,$

We can express both $\cos3\theta$ and $\sin2\theta$ in terms of just $s = \sin\theta$ and $c = \cos\theta$.

We’ll use the identities $\sin2\theta=2sc,$ $\cos2\theta=1-2s^2$ and $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta.$

We have $\cos3\theta=\cos(2\theta+\theta)=\cos(2\theta) c-\sin(2\theta) s = (1-2s^2)c - 2scs = c(1-4s^2).$

So if $\sin2\theta-\cos3\theta=0$, we have $2sc-c(1-4s^2)=0$, or $c(4s^2+2s-1)=0.$

Now we know that $\sin(18^\circ)$ is a solution to this equation, and so $\sin(18^\circ)$ is a root of $4x^2+2x-1=0$, since $\cos18^\circ\ne0$.

Hence find its value as a surd.

We complete the square in order to find the roots of $4x^2+2x-1$ in surd form. We have $4x^2+2x-1=\left(2x+\frac{1}{2}\right)^2-\frac{5}{4}=0,$ so that $2x+\frac{1}{2}=\pm\frac{\sqrt{5}}{2},$ and we have $x=\frac{-1\pm\sqrt{5}}{4}.$ Finally, we must choose whether we need a plus or a minus sign to find the root corresponding to $\sin18^\circ$. We know that $\sin18^\circ>0$, so we choose the plus sign. So we have that $\sin18^\circ=\frac{-1+\sqrt{5}}{4}.$

Find all solutions between $0^\circ$ and $360^\circ$ of the equation $4\tan 2x=\cot x.$

We play a similar game here to that of the first part of the question. Firstly, we can write $\tan x=\frac{s}{c}, \quad \cot x=\frac{1}{\tan x}=\frac{c}{s},$ so the equation we need to solve becomes $4\frac{\sin(2x)}{\cos(2x)}=\frac{c}{s} \implies 4\sin(2x)s-\cos(2x)c=0 \implies 8s^2c-(1-2s^2)c=0 \implies (10s^2-1)c=0.$ So we have that $\cos x=0$ solves this, which gives us the values $x=90^\circ, 270^\circ$ between $0^\circ$ and $360^\circ$. We also have that $\sin x=\pm\frac{1}{\sqrt{10}}$ is a solution, so that $x=\sin^{-1}(\pm1/\sqrt{10})$ are also solutions. To work out how many solutions between $0^\circ$ and $360^\circ$, consider

We see that $\sin x=\pm1/\sqrt{10}$ four times between $0^\circ$ and $360^\circ$.

Using a calculator, one finds that $\sin^{-1}(1/\sqrt{10})=18.4^\circ$ to one decimal place. From the symmetries of sine, we see that the other angles are then $180^\circ-18.4^\circ=161.6^\circ$, $180^\circ+18.4^\circ=198.4^\circ$ and $360^\circ-18.4^\circ=341.6^\circ$, giving us six solutions in total.