# Thread: computing sin of an angle (difficult problem)

1. ## computing sin of an angle (difficult problem)

what is sine of 2004 degrees without using a calculator?

this was an extra credit problem given in an honors precalc class and i am curious on how to solve it. since it is precal, you can't use taylor polynomials. 2004 degrees is the same as 204 degrees but that is not one of the "special" angles like 30, 60, or 90.

2. It is the same as -sin(24).

3. Originally Posted by oblixps
what is sine of 2004 degrees without using a calculator?

this was an extra credit problem given in an honors precalc class and i am curious on how to solve it. since it is precal, you can't use taylor polynomials. 2004 degrees is the same as 204 degrees but that is not one of the "special" angles like 30, 60, or 90.
You can find it, but it really is a pain. Why would he give this on a test?

4. Ok, so I have been thinking about this and it really seems to just boil down to solving a few LDEs.

Problem: Compute $\sin\left(2000^{\circ}\right)$

Solution:

$\sin\left(2000^{\circ}\right)=\sin\left(\frac{10\p i}{9}\right)$. Now, obviously $\sin\left(\frac{10\pi}{9}\right)=\sin\left(\pi+\fr ac{\pi}{9}\right)=\sin\left(\pi\right)\cos\left(\f rac{\pi}{9}\right)+\cos(\pi)\sin\left(\frac{\pi}{9 }\right)=-\sin\left(\frac{\pi}{9}\right)$. So, now we must merely solve the LDE $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{9}\Le ftrightarrow 9xy+9zx+9yz=xyz$. Clearly we must have that $xyz\equiv 0\text{ mod }9$. So, let's think about this. We only want values of $x,y,z$ that we can work with ( $1,2,3,4,6,9\cdots$). So, we automatically can throw out a few triplets considering, for example, $1\cdot 4\cdot6\not\equiv0\text{ mod }9$. A little experimentation shows that $x=-9,y=-9,z=3$ works.

So we have that $\sin\left(\frac{\pi}{9}\right)=\sin\left(\frac{\pi }{3}+\frac{-2\pi}{9}\right)=\sin\left(\frac{\pi}{3}\right)\cos \left(\frac{-2\pi}{9}\right)+\cos\left(\frac{\pi}{3}\right)\sin \left(\frac{-2\pi}{9}\right)$. A little simplification shows this equal to $\frac{\sqrt{3}}{2}\left[1-2\sin^2\left(\frac{\pi}{9}\right)\right]-\sin\left(\frac{\pi}{9}\right)\cos\left(\frac{\pi} {9}\right)\quad{\color{red}\star}$

But, $\cos\left(\frac{\pi}{9}\right)=\cos\left(\frac{\pi }{3}+\frac{-2\pi}{9}\right)=\cos\left(\frac{\pi}{3}\right)\cos \left(\frac{-2\pi}{9}\right)-2\sin\left(\frac{\pi}{3}\right)\sin\left(\frac{-2\pi}{9}\right)$. Once again, a little simplification shows that this is $\cos\left(\frac{\pi}{9}\right)=\frac{1}{2}\left[1-2\sin^2\left(\frac{\pi}{9}\right)\right]-\frac{\sqrt{3}}{2}\cos\left(\frac{\pi}{9}\right)\s in\left(\frac{\pi}{9}\right)$. Solving for $\cos\left(\frac{\pi}{9}\right)$ gives $\cos\left(\frac{\pi}{9}\right)=\frac{1-2\sin^2\left(\frac{\pi}{9}\right)}{2-\sqrt{3}\sin\left(\frac{\pi}{9}\right)}$

Subbing this back into $\color{red}\star$ and letting $z=\sin\left(\frac{\pi}{9}\right)$ gives us $z=\frac{\sqrt{3}}{2}\left[1-2z^2\right]-\frac{1-2z^2}{2-\sqrt{3}z}$. Which, with some effort (not going to bore you with the algebra) is equal to $z^3-\frac{3}{4}z=\frac{-\sqrt{3}}{8}$

Let $z=x+y$ such that $3xy+\frac{-3}{4}=0$. Substitution and subsequent simplification leads us to (look up depressed cubic) $z=\sqrt[3]{\frac{-\sqrt{3}}{16}+\sqrt{\frac{3}{256}-\frac{1}{64}}}+\sqrt[3]{\frac{-\sqrt{3}}{16}+\sqrt{\frac{-3}{256}-\frac{-1}{64}}}$

So that we see that $\sin\left(2000^{\circ}\right)=-\sqrt[3]{\frac{-\sqrt{3}}{16}+\sqrt{\frac{3}{256}-\frac{1}{64}}}+\sqrt[3]{\frac{-\sqrt{3}}{16}+\sqrt{\frac{-3}{256}-\frac{-1}{64}}}$

Remark: After writing all of that I just realized that $\sin\left(3\varphi\right)=3\sin\left(\vartheta\rig ht)-4\sin^3\left(\vartheta\right)$. Letting $\vartheta=\frac{\pi}{9}$ gives us the equivalent depressed cubic.

Pre-calc test, huh?

5. $\sin{(2004)} = \sin{(\frac{2004 \pi}{180})} = \sin{(\frac{167 \pi}{15})}$

Since the sine function is periodic over $2 \pi$, this is equivalent to $\sin{(\frac{17 \pi}{15})}$

You can use the Taylor approximation series for the sine function :

$\sin{(x)} = x - \frac{x^3}{3!} + \frac{x^5}{5!}$ - ...

Apply this to your problem :

$\sin{(\frac{17 \pi}{15})} = \frac{17 \pi}{15} - \frac{{(\frac{17 \pi}{15})}^3}{3!} + \frac{{(\frac{17 \pi}{15})}^5}{5!}$ - ...

Let's do this like ten times. We get that :

$\sin{(\frac{17 \pi}{15})} \approx -0,406736$

EDIT : damn I forgot you weren't allowed to use Taylor polynomials

6. Oh my god. I didn't see that four. It makes it easier though.

$\sin\left(2000^{\circ}\right)=\sin\left(\frac{17\p i}{15}\right)$. We make two subsequent observations

1. $\frac{17}{15}=1+\frac{2}{15}$

2. Solving $\frac{1}{x}+\frac{1}{y}$ for "nice" answers gives $x=-5,y=3$.

Combining these two we see that

$\sin\left(\frac{17\pi}{15}\right)=\sin\left(\pi+\f rac{2\pi}{15}\right)=\sin\left(\pi\right)\cos\left (\frac{2\pi}{15}\right)+\cos\left(\pi\right)\sin\l eft(\frac{2\pi}{15}\right)=-\sin\left(\frac{2\pi}{15}\right)$.

But,

$\sin\left(\frac{2\pi}{15}\right)=\sin\left(\frac{\ pi}{3}+\frac{-\pi}{5}\right)=\sin\left(\frac{\pi}{3}\right)\cos\ left(\frac{-\pi}{5}\right)+\sin\left(\frac{-\pi}{5}\right)\cos\left(\frac{\pi}{3}\right)$.

It is fairly easy to verify that $\sin\left(5\vartheta\right)=5\sin\left(\vartheta\r ight)-20\sin^3\left(\vartheta\right)+16\sin^5\left(\vart heta\right)$

Setting $\vartheta=\frac{\pi}{5}$ gives us

$\sin\left(\pi\right)=0=5\sin\left(\vartheta\right)-20\sin^3\left(\frac{\pi}{5}\right)+16\sin^5\left(\ frac{\pi}{5}\right)$.
]

We may clear out the extraneous $\sin\left(\frac{\pi}{5}\right)$ since it is clearly not equal to zero. We are left then with an equation of the form

$16\zeta^4-20\zeta^2+5=0$
.

Thus, by the quadratic formula,

$\zeta^2=\frac{5\pm\sqrt{5}}{8}$
.

But, since $\sin(x)$ is increasing we must have that $\sin\left(\frac{\pi}{5}\right)\leqslant\sin\left(\ frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}$ where it follows that

$\zeta^2=\frac{5-\sqrt{5}}{8}\implies \zeta=\sqrt{\frac{5-\sqrt{5}}{8}}$.

Also, we know that

$\cos\left(\frac{\pi}{5}\right)=\sqrt{1-\zeta^2}=\sqrt{\frac{3+\sqrt{5}}{8}}$.

We may now then conclude that

$-\sin\left(2004^{\circ}\right)=\frac{\sqrt{3}}{2}\s qrt{\frac{3+\sqrt{5}}{8}}-\frac{1}{2}\sqrt{\frac{5-\sqrt{5}}{8}}$

Ta-da!

7. Awesome ! But you still need a calculator to compute that xD

8. Originally Posted by Bacterius
Awesome ! But you still need a calculator to compute that xD