Calculate: $\sin9°$ I found this question in the olympiad book. But I could not find the solution.
The question is to calculate the following real number: $$\sin{9°}$$

Let $x = 18$ then $5 x = 90$ so $2x = 90 - 3x$.
Now
\begin{align}
\sin(2x) &= \sin(90 - 3x)\\
2 \sin x \cos x &= \cos 3x\\
2\sin x \cos x  &= 4\cos^3x - 3\cos x\\
2\sin x &= 4\cos^2x - 3\\
2\sin x&= 4 - 4\sin^2x-3\\
4\sin^2x + 2\sin x - 1 &=0
\end{align}
Solving this quadratic equation, we get 
$$\sin x = \frac{-1+\sqrt{5}}{4}$$
Also 
$$\cos x = \sqrt{1 - \sin^2x} = 0.951$$
Now 
$$\cos x = 2\cos^2\frac{x}{2}-1$$
So that $$\cos \frac{x}{2} = \sqrt{\frac{0.951+1}{2}} = 0.987$$
Finally, since $\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$ so that 
$$\sin 9 = \sin \frac{x}{2} = \frac{\sin x}{2\cos \frac{x}{2}} = 0.156$$
Note: Since we are dealing with $x=18$ so we are in the first quadrant, therefore all the time we take the positive values.
Also
\begin{align}
\cos 3x = \cos(2x+x) &= \cos 2x \cos x - \sin 2x \sin x\\
&= (\cos^2x-\sin^2x)\cos x - 2\sin x\cos x\sin x \\
&= \cos^3x-\sin^2x\cos x - 2\sin^2x \cos x\\
&=\cos^3x - 3\sin^2x\cos x\\
&=\cos^3x-3(1-\cos^2x)\cos x\\
&=4\cos^3 x - 3\cos x
\end{align}
