Re: really easy trig question

.. wrote:
> <matt271829-news@xxxxxxxxxxx> wrote in message
> news:1125486982.505779.162180@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> >
> > . wrote:
> >> Hi All,
> >>
> >> I used to know this, but my brain has got rusty with age.
> >>
> >> Why is it that sin (Pi/3) = (root 3)/2 ?
> >>
> >> Thanks for your help
> >>
> >> Michael
> >
> > Draw an equilateral triangle, and drop a perpendicular from the apex to
> > the base. Then a bit of Pythagoras and you should be done.
> >
>
> Sorry, I must have worded my question badly.
>
> If I do as you have said for a unit equalateral triangle I get ~0.866 ,
> which ~(root 3)/2
> But if I didn't know 0.866 ~(root 3)/2, how would I come up with the idea of
> trying (root 3)/2 ?

I don't entirely understand how you got 0.866... without knowing how
you got it. Unless you did this with physical measurement?? I didn't
mean to do that.

Take an equilateral triangle ABC of side length 1. Drop a perpendicular
from A to BC, to intersect BC at point D. ADC is a right-angled
triangle. Angle ACD is pi/3.

We have that AC = 1, and, by symmetry, DC = 1/2. By Pythagoras'
theorem, AD^2 + DC^2 = AC^2, so AD = sqrt(AC^2 - DC^2) = sqrt(1 - 1/4)
= sqrt(3)/2.

Then from the definition of the sine, sin(pi/3) = AD/AC = sqrt(3)/2.

(This shows that sin(pi/3) IS equal to sqrt(3)/2, but whether it shows
WHY sin(pi/3) is equal to sqrt(3)/2 is another matter!)

.

Relevant Pages

  • Re: really easy trig question
    ... and drop a perpendicular from the apex to ... > Take an equilateral triangle ABC of side length 1. ... Sin =sqrt/2 because of the definition of the sine function. ...
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  • Re: really easy trig question
    ... >>> Draw an equilateral triangle, and drop a perpendicular from the apex to ... > Take an equilateral triangle ABC of side length 1. ...
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