Evaluate sin 6 without using a calculator?

  • MHB
  • Thread starter mathdad
  • Start date
  • Tags
    Sin
In summary: Updated - - -That is not entirely true.Sorry, I could not resist. (Smirk)Hope I did not start a war here. I have been curious about the wolfram site for years.Wolfram has been around for a long while with the Mathematica program, which I will easily say is almost an unbelievable program. A number of "patents" have awarded for the methods used (how do you patent a method of integration??) What I have to wonder, though, is how much of this from the staff or Wolfram himself?My studies in QFT have shown me that even Mathematica has limits on what it can do...
  • #1
mathdad
1,283
1
Evaluate sin 6 without using a calculator? How is this done?
Unit circle?
 
Mathematics news on Phys.org
  • #2
I don't know how to evaluate it exactly.

You could notice that $\sin{6} = \sin{(6 - 2\pi)}$ by periodicity. Now, $6 - 2\pi = -0.283185308\ldots$ and we have $\sin{6} \approx 6 - 2\pi$ in first approximation.

For comparison, if you type $\sin{6}$ in your calculator, you obtain a value of $-0.2794154982\ldots$. Behind the scenes your calculator also gives a (better) approximation, perhaps by including higher order terms in the series expansion I linked to above. In any case, you see that the first order approximation was not too bad: It gave the right sign and was accurate to two decimal places. This is because $6 - 2\pi$ is relatively small.
 
  • #3
I would imagine that 6 is in degrees. So instead of using 2 pi you would use 360.

-Dan
 
  • #4
topsquark said:
I would imagine that 6 is in degrees. So instead of using 2 pi you would use 360.

-Dan

Ah, yes, that could be, thank you. In that case I still don't know how to evaluate it, but the OP could use the same crude approximation method. (The approximation will be a bit better, too.)
 
  • #5
Yes, I meant evaluate sin(6°).

Can the unit circle be used?
 
  • #6
RTCNTC said:
Yes, I meant evaluate sin(6°).

Can the unit circle be used?

I do not think you can use the unit circle to get an exact evaluation.

However, you can draw the unit circle and indicate a 6 degree angle with the horizontal in the first quadrant to get intuition for the fact that $\sin{x}$ is well-approximated by $x$ when $x$ is small: The hypotenuse and the base of the resulting triangle will be approximately of equal length for small values of $x$.
 
  • #7
Ok. I will try to play with this question as you suggested.
 
  • #8
According to W|A:

\(\displaystyle \sin\left(6^{\circ}\right)=\frac{\sqrt{6(5-\sqrt{5})}-(1+\sqrt{5})}{8}\)
 
  • #9
Nice!

That equals $0.1045284634\ldots$ while $\frac{\pi}{30} = 0.1047197551\ldots$ where $\frac{\pi}{30}$ rads is $6$ degrees.
So here you have a perfect evaluation and a reasonable first order approximation.

P.S. Does anyone see an easy manipulation to arrive at MarkFL's expression?

P.P.S. In Maple, the command
Code:
convert(sin(Pi/30), radical)
gives the same result.
 
Last edited:
  • #10
MarkFL said:
According to W|A:

\(\displaystyle \sin\left(6^{\circ}\right)=\frac{\sqrt{6(5-\sqrt{5})}-(1+\sqrt{5})}{8}\)

How did you arrive at this very impressive value for sin(6°)?
 
  • #11
RTCNTC said:
How did you arrive at this very impressive value for sin(6°)?

Well, it's been a while on this material, but $60 = 2^{2}\cdot F_{0}\cdot F_{1} = 4\cdot 3\cdot 5$. Thus, we should be able to construct (with standard Euclidean construction tools) a Regular 60-gon. This suggests that a 6º angle should be expressible as a two-tower - just square roots and square roots of square roots and so on - quadratic surds abound!
 
  • #12
I'm not there yet. MarkFL, good job.
 
  • #13
RTCNTC said:
How did you arrive at this very impressive value for sin(6°)?

Using W|A (wolframalpha.com)
 
  • #14
MarkFL said:
Using W|A (wolframalpha.com)

I heard about wolfram. Who is the creator of wolfram?
 
  • #15
RTCNTC said:
I heard about wolfram. Who is the creator of wolfram?

I don't know...the guy who introduced it to me shortly after I got involved in the online math communities told me it was "google for nerds." :)
 
  • #16
MarkFL said:
I don't know...the guy who introduced it to me shortly after I got involved in the online math communities told me it was "google for nerds." :)

I have 5 personal questions for you. Check your PM in 10 minutes.
 
  • #17
MarkFL said:
Using W|A (wolframalpha.com)

I am in the trigonometry sections of the David Cohen book. Should I post my questions in the MHB trigonometry forum or continue posting here?
 
  • #18
RTCNTC said:
I am in the trigonometry sections of the David Cohen book. Should I post my questions in the MHB trigonometry forum or continue posting here?

Personally, I would post trig. questions in the trig. forum. :)
 
  • #19
Look for my trig questions in the trig group. All trig questions come from my precalculus textbook.
 
  • #20
RTCNTC said:
I heard about wolfram. Who is the creator of wolfram?

Stephen Wolfram is the creator of WolframAlpha. His arrogance is unparalleled...
 
  • #21
Joppy said:
Stephen Wolfram is the creator of WolframAlpha. His arrogance is unparalleled...
That is not entirely true.

Sorry, I could not resist. (Smirk)
 
  • #22
MarkFL said:
According to W|A:

\(\displaystyle \sin\left(6^{\circ}\right)=\frac{\sqrt{6(5-\sqrt{5})}-(1+\sqrt{5})}{8}\)

$6^\circ= \frac{\pi}{5}-\frac{\pi}{6}$ and sin and cos for these values are computable and hence well known and can be used to compute
 
  • #23
kaliprasad said:
$6^\circ= \frac{\pi}{5}-\frac{\pi}{6}$ and sin and cos for these values are computable and hence well known and can be used to compute
I would not have known $\sin{\frac{\pi}{5}}$ and $\cos{\frac{\pi}{5}}$ just like that, but maybe that is merely an indication that I need to return to pre-calculus class. (Giggle)
 
  • #24
Everyone has been so kind to me in terms of this question. Thank you.

- - - Updated - - -

Krylov said:
That is not entirely true.

Sorry, I could not resist. (Smirk)

Hope I did not start a war here. I have been curious about the wolfram site for years.
 
  • #25
Wolfram has been around for a long while with the Mathematica program, which I will easily say is almost an unbelievable program. A number of "patents" have awarded for the methods used (how do you patent a method of integration??) What I have to wonder, though, is how much of this from the staff or Wolfram himself?

My studies in QFT have shown me that even Mathematica has limits on what it can do...The program simply isn't of much value to a theorist. However at some point Wolfram did come up with a Notebook for QFT and I am unable to buy the upgrade to find out if it's any good. Has anyone used it?

-Dan
 
  • #26
I know that paying wolfram leads to a step by step solution for questions but, like Dan stated, wolfram is not perfect. Sometimes, I type simple linear equations like 2x + 5 = 40, and it gives me an answer far from the true value of x.
 
  • #27
RTCNTC said:
Everyone has been so kind to me in terms of this question. Thank you.

You asked a question that is very simple but had a solution that was - at least to me - unexpected and led to a nice discussion, so thank you for that.

RTCNTC said:
- - - Updated - - -

Hope I did not start a war here. I have been curious about the wolfram site for years.

No, of course you didn't.

Wolfram Alpha is a web front-end to part of Mathematica, a software environment for symbolic and numerical computation. An example of a symbolic result is in post #8, while an example of a (very simple) numerical result is in post #2. The mathematics underlying symbolic algorithms is quite different from numerical mathematics, but you don't need to know too much about either in order to start using the software productively.

There are other packages and libraries for symbolic computation, some of which also have numerical functionality. A well-established competitor of Mathematica is Maple, originally developed by the University of Waterloo in Canada. Maple plays nicely together with MATLAB, which in turn is an environment for numerics. Neither of the two is free, but students can usually get a license for a significantly reduced price, sometimes through their college or university.

There exist free alternatives, too. Of those, I am only familiar with SymPy, a Python library that works quite well, but cannot (yet?) match the comprehensiveness of the commercial packages. Another well-known offering is Maxima, but I have never used it myself.

Ultimately, the choice is probably a matter of "upbringing" and preference. I think physicists tend to use Mathematica, while I was "raised" with Maple. One of the free packages could be a good starting point if you do not want to invest too much initially. Otherwise, it is probably smartest to adopt the software that already dominates in your neighborhood (school, work, etc.) because it makes it easier to do assignments and get help.

Finally, no matter how sophisticated it may be, all this software still relies on sensible input and critical interpretation of output. It cannot "magically" solve your problems, although sometimes it can seem that way. (Wink)
 
Last edited:
  • #28
Interesting information here. Thanks.
 

FAQ: Evaluate sin 6 without using a calculator?

How can I evaluate sin 6 without using a calculator?

To evaluate sin 6 without using a calculator, you can use the Taylor series expansion of sin(x) = x - x^3/3! + x^5/5! - ..., where x is in radians. By plugging in 6 for x and performing the calculations, you can find the approximate value of sin 6.

Why is it important to evaluate sin 6 without a calculator?

Evaluating sin 6 without a calculator can help build a better understanding of trigonometric functions and their properties. It also allows for a more precise calculation, as calculators often round off decimal values.

Can I use a calculator to evaluate sin 6 in degrees?

No, since the Taylor series expansion of sin(x) is based on radians, you cannot use a calculator to directly evaluate sin 6 in degrees. You would first need to convert 6 degrees to radians by multiplying it by π/180.

What is the most accurate method to evaluate sin 6 without a calculator?

The most accurate method is to use the Maclaurin series expansion of sin(x) = x - x^3/3! + x^5/5! - ..., where x is in radians, but this may require more terms to be calculated. Another accurate method is to use a trigonometric identity, such as sin(a+b) = sin(a)cos(b) + cos(a)sin(b), to reduce the angle to a value that is easier to work with, such as sin 6 = sin 5(cos 1) + cos 5(sin 1).

Are there any other methods to evaluate sin 6 without a calculator?

Yes, there are other methods such as using a unit circle, geometric constructions, or interpolation techniques. However, these methods may require more time and effort compared to using the Taylor series or trigonometric identities.

Similar threads

Replies
5
Views
3K
Replies
1
Views
821
Replies
3
Views
1K
Replies
5
Views
1K
Replies
6
Views
2K
Replies
2
Views
980
Replies
1
Views
995
Back
Top