Tag Archives: sin(18)

October 9, 2025 · 6:37 am

Trigonometric Exact Values

Find exactly $sin(18^\circ)$

We must be able to find an arithmetic combination of the exact values we knew to find $18$ .

$\begin{equation*}90=5\times 18\end{equation}$

$\begin{equation*}90-3(18)=2(18)\end{equation}$

I re-arranged as above, so I could take advantage of $cos(90)=0$ and $sin(90)=1$

sin(2x)=2sin(x)cos(x)

cos(2x)=cos^2(x)-sin^2(x)=1-2sin^2(x)=2cos^2(x)-1

$\begin{equation*}sin(90-3(18))=sin(2(18))\end{equation}$

$\begin{equation*}sin(90)cos(3(18))-sin(3(18))cos(90)=2sin(18)cos(18)\end{equation}$

$\begin{equation*}cos(3(18))=2sin(18)cos(18)\end{equation}$

$\begin{equation*}4cos^3(18)-3cos(18)=2sin(18)cos(18)\end{equation}$

$\begin{equation*}4cos^3(18)-3cos(18)-2sin(18)cos(18)=0\end{equation}$

$\begin{equation*}cos(18)(4cos^2(18)-3-2sin(18))=0\end{equation}$

Hence,

$\begin{equation*}4cos^2(18)-2sin(18)-3=0\end{equation}$

$\begin{equation*}4-4sin^2(18)-2sin(18)-3=0\end{equation}$

$\begin{equation*}-4sin^2(18)-2sin(18)+1=0\end{equation}$

Use the quadratic equation formula

$\begin{equation*}sin(18)=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{equation}$

$\begin{equation*}sin(18)=\frac{2 \pm \sqrt{4-4(-4)(1)}}{-8}\end{equation}$

$\begin{equation*}sin(18)=\frac{2 \pm \sqrt{20}}{-8}\end{equation}$

$\begin{equation*}sin(18)=\frac{-2 \mp 2\sqrt{5}}{8}\end{equation}$

$\begin{equation*}sin(18)=\frac{-1 \mp \sqrt{5}}{4}\end{equation}$

As $sin(18)>0$ , $sin(18)=\frac{-1+\sqrt{5}}{4}$

sin(18)=\frac{-1+\sqrt{5}}{4}

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Filed under Addition and Subtraction Identities, Algebra, Identities, Quadratic, Quadratics, Solving, Solving Equations, Solving Trig Equations, Trigonometry, Year 11 Specialist Mathematics

Tagged as exact values, sin(18), trig exact values, trigonometric identities