mathdad
- 1,280
- 0
Evaluate sin 6 without using a calculator? How is this done?
Unit circle?
Unit circle?
The discussion revolves around evaluating the sine of 6 degrees without using a calculator. Participants explore various methods, including approximations and references to the unit circle, while considering both exact and approximate values.
Participants do not reach a consensus on the best method for evaluating $\sin(6^\circ)$, with multiple competing views and approaches presented throughout the discussion.
Some participants note the limitations of using the unit circle for exact evaluations and the dependence on approximations for small angles. There are also mentions of the differences between symbolic and numerical computation methods.
This discussion may be of interest to those studying trigonometry, exploring mathematical approximations, or seeking insights into the use of computational tools like WolframAlpha and Mathematica.
topsquark said:I would imagine that 6 is in degrees. So instead of using 2 pi you would use 360.
-Dan
RTCNTC said:Yes, I meant evaluate sin(6°).
Can the unit circle be used?
convert(sin(Pi/30), radical)
MarkFL said:According to W|A:
$$\sin\left(6^{\circ}\right)=\frac{\sqrt{6(5-\sqrt{5})}-(1+\sqrt{5})}{8}$$
RTCNTC said:How did you arrive at this very impressive value for sin(6°)?
RTCNTC said:How did you arrive at this very impressive value for sin(6°)?
MarkFL said:Using W|A (wolframalpha.com)
RTCNTC said:I heard about wolfram. Who is the creator of wolfram?
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." :)
MarkFL said:Using W|A (wolframalpha.com)
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?
RTCNTC said:I heard about wolfram. Who is the creator of wolfram?
That is not entirely true.Joppy said:Stephen Wolfram is the creator of WolframAlpha. His arrogance is unparalleled...
MarkFL said:According to W|A:
$$\sin\left(6^{\circ}\right)=\frac{\sqrt{6(5-\sqrt{5})}-(1+\sqrt{5})}{8}$$
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
Krylov said:
RTCNTC said:Everyone has been so kind to me in terms of this question. Thank you.
RTCNTC said:- - - Updated - - -
Hope I did not start a war here. I have been curious about the wolfram site for years.