I needed to compute $\sin 18^{\circ}$.

Now, these two relations hold for every $x$:
$\cos 5x=16\cos^5x-20\cos^3x+5\cos x$
$\sin5x=16\sin^5x-20\sin^3x+5\sin x$, which can be easily proved using the multiple angle formulae.

Now, one thing to observe is : $\sin5(18^{\circ})=1$ and $\cos5(18^{\circ})=0$
$$\begin{align}16\cos^518^{\circ}-20\cos^318^{\circ}+5\cos18^{\circ}&=0 \\\implies16\cos^418^{\circ}-20\cos^218^{\circ}+5&=0 \tag{1} \end{align}$$ And, $$\begin{align}16\sin^518^{\circ}-20\sin^3+5\sin18^{\circ}&=1\\ \implies16\sin^418^{\circ}-20\sin^218^{\circ}+5&=\dfrac{1}{\sin18^{\circ}} \tag{2} \end{align}$$

So, before moving forward with my computation, I would like to ask: Is there any problem in these two equations?

Now, equation (1) consists of squares of cosines, which clearly means that (1) can be represented consisting of sines like this:
$$\begin{align}16\cos^418^{\circ}-20\cos^218^{\circ}+5 &=0\\ \implies16(1-\sin^218^{\circ})^2 -20(1-\sin^218^{\circ})+5&=0 \\ \implies 16\sin^418^{\circ}-12\sin^218+1&=0 \end{align}$$

Now, using the above equation, we can easily deduce that $\sin^218=\dfrac{12\pm4\sqrt5}{32}=\dfrac{3\pm\sqrt5}{8}$.

Now, if I put this value in equation (2), I get something like this:
$\dfrac{1}{\sin18^{\circ}}=16\dfrac{(3\pm\sqrt5)^2}{64}-20\dfrac{(3\pm\sqrt5)}{8}+5$, which reduces to $1\mp\sqrt5$, which upon solving does not give me the correct value for $\sin18^{\circ}$. So, what mistake am I doing here?

Edit: By solving, I mean that the actual value of $\sin18^{\circ}=\dfrac{\sqrt5-1}{4}$, but in my computation, the negative $1$ and the denominator $4$ is missing.


Your expression got reduced to $1\mp\sqrt5$. But remember, this expression is the value of $\dfrac{1}{\sin18^{\circ}}$. Since $\sin18^{\circ}$ is not negative, it's reciprocal cannot be negative. Thus, we discard the value $1-\sqrt{5}$. Thus, the correct value is -

$\dfrac{1}{\sin18^{\circ}} = 1+\sqrt{5}$

$\sin18^{\circ} = \dfrac{1}{1+\sqrt{5}} = \dfrac{1}{\sqrt{5}+1}*\dfrac{\sqrt{5}-1}{\sqrt{5}-1}$

$\sin18^{\circ} = \dfrac{\sqrt{5}-1}{4}$

  • $\begingroup$ It is still bugging me as to why I didn't proceed with the computation..... thnx a lot @Akash Bajaj $\endgroup$ – codetalker Jul 21 '16 at 17:12

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