Calculate: $\sin9°$

Question

I found this question in the olympiad book. But I could not find the solution.

The question is to calculate the following real number: $$\sin{9°}$$

Answer 1

hint

$$\sin (5x)=\sin (2x+3x) $$ $$\sin (45^\circ)=\frac {\sqrt {2}}{2} $$

Answer 2

Determine $\cos(72^\circ)+i\sin(72^\circ)$; it's one of the roots of $z^4+z^3+z^2+z+1=0$. From this, you get $\cos(18^\circ)$ and $\sin(18^\circ)$. It is now easy to get $\sin(9^\circ)$.

Answer 3

$\sin18^{\circ}=\frac{\sqrt5-1}{4}$, which says that $$\cos18^{\circ}=\sqrt{1-\left(\frac{\sqrt5-1}{4}\right)^2}=\frac{1}{4}\sqrt{10+2\sqrt5}.$$ Id est, $$\sin9^{\circ}=\sqrt{\frac{1-\frac{1}{4}\sqrt{10+2\sqrt5}}{2}}$$

Answer 4

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}

Answer 5

For an approximation of sin 9 degrees: first, expand sin with the Maclaurin series, then evaluate the series for x = pi/20 (9 degrees). Computing the series to n=4 will give a remainder of (pi/20)^9 X 1/9! X cos(y x pi/20) which represents the error, where y is a value between 0 and 1. Since pi/20^9 X 1/9! is greater than the remainder term and can be easily evaluated it can be said that the error is less than this term. The more terms you compute the more accurate the result.