5
$\begingroup$

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

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

$\endgroup$

closed as off-topic by Namaste, Thomas Russell, Dietrich Burde, Simply Beautiful Art, kingW3 Jun 7 '17 at 16:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Thomas Russell, Dietrich Burde, Simply Beautiful Art, kingW3
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Approximately or symbolically? $\endgroup$ – Shinaolord Jun 7 '17 at 13:14
  • 2
    $\begingroup$ You need to use sum and difference formulas for sine and cosine, starting with values you know, like $30^\circ$ and $45^\circ$. $\endgroup$ – GEdgar Jun 7 '17 at 13:15
  • $\begingroup$ Please edit question.I know english very bad. only with radicals. $\endgroup$ – user452578 Jun 7 '17 at 13:16
  • $\begingroup$ @GEdgar has suggested to you a likely method. Let us know how it goes? This is similar to a homework question so I don't feel comfortable personally answering. $\endgroup$ – Shinaolord Jun 7 '17 at 13:19
  • 2
    $\begingroup$ Figure out $\cos 36^\circ$. If you are into olympiad math you know how to construct a regular 10-gon, so that's not a problem. Then use the angle-halving formulas. $\endgroup$ – Jyrki Lahtonen Jun 7 '17 at 13:19
6
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ Thank you good answer..+1 $\endgroup$ – user452578 Jun 7 '17 at 13:49
  • $\begingroup$ @Student If you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. $\endgroup$ – Ahmed Jun 7 '17 at 14:41
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Perfect!!..Please ,explain me, Why did my question downvote?İs ıt bad question? $\endgroup$ – user452578 Jun 7 '17 at 13:40
  • $\begingroup$ @Student Because you need to show us your trying. By the way, I think that your question is nice. $\endgroup$ – Michael Rozenberg Jun 7 '17 at 13:41
  • $\begingroup$ Did you mean the experiment I made to solve the question? $\endgroup$ – user452578 Jun 7 '17 at 13:43
  • $\begingroup$ @Student Yes, of course. You need to show your trying. $\endgroup$ – Michael Rozenberg Jun 7 '17 at 13:44
  • $\begingroup$ Hmmm..i understood.Ok.Thank you Rozenberg for help!!.. $\endgroup$ – user452578 Jun 7 '17 at 13:45
1
$\begingroup$

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 $\sin(9^\circ)$.

$\endgroup$
0
$\begingroup$

hint

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Please use MathJax to make your post more readable. $\endgroup$ – M. Winter Jun 7 '17 at 14:43