The Origins Of Triginometry

jasonbe said:

sin(x) = 2t/(1+t^2) and cos(x) = (1 - t^2)/(1 + t^2) are at the bottom of List of trigonometric identities - Wikipedia, the free encyclopedia. I don’t know if they are accurate, but the Wikipedia article shows how to derive them from the half-angle formula – which is shown on the same page. I suppose one could confirm them by looking up the half-angle formula and making an algebraic substitution using t = tan(x/2). I've seen θ used instead of x for angles.

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trennonix said:

What's "a"? How did we get its values ( a2=-0.166.. )???
and, so, there's absolutely no formula that isn't a form of a series?

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jasonbe said:

I became interested in precision because of what trennonix wrote.

I'm not sure what you meant when you said that there would be something else besides what I was expecting to be sin(a) and cos(a).

Am I correct in thinking that there are different ways that precision is determined for answers after adding, multiplying, and raising a number to a power? I thought that there was, and thought that the ways for cosine of x and sine of x might be the same as those for multiplication - because from what I understand multiplication has a lot in common with division. Sine and cosine are the result of dividing - a side of a triangle by the hypotenuse of the triangle. However, I was guessing that this wouldn't apply when calculating a derivative or an integral of a sine or cosine function because I think that the sides of a triangle that are used to calculate the quotient change at different rates according to the angle in the function. For example, I think that the rate that a sine function changes near zero degrees is at its maximum - measured to be one, and the rate that a sine function changes after a quarter turn is zero.

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MrAl said:

Hi,


Lets call sin(x) 'Sx', and cos(x) 'Cx', then we have again:

Sx^2+Cx^2=y

and for any x, y will always be equal to 1.

Lets look at this another way, this time we'll call Sx^2 simply s, and
Cx^2 simply c:

s+c=1

What this says is that whatever s and c are, they must
*ALWAYS* add up to 1. This is an identity.

Now if they dont add up to 1, then s and c can not be defined
as sin(x) squared and cos(x) squared respectively.

If you change the accuracy of s and c, then at the least we would
end up with an error term for one or both like this:

y=(Sx+Serr)^2+(Cx+Cerr)^2

and so the equation is not the same anymore.

If we force Serr and Cerr to go to zero,
then we end up with this:

1=Sx^2+Cx^2

because s+c can only add up to 1 and nothing else.
Note that that '1' is exactly 1, and not 1.0000000001 or 0.9999999999.

The domain of x is real or complex, and the range of sin or cos
(what i think you are interested in) is {-1 to +1} including
the complex numbers. What this sort of means is that sin and
cos are taken to be perfectly accurate, in that you can calculate
any precision required as long as you have the storage space.

BTW:

d(A*sin(x))/dx=A*cos(x)

and

d(A*cos(x))/dx=-A*sin(x)

so except for the minus sign, the derivatives are the opposite
functions, and another way of looking at this is that the
derivatives are 90 degrees out of phase with the functions.

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MrAl said:

Hi,

For sin(x)^2+cos(x)^2=y, y can only be equal to 1. It's not a general
equation for y like y=2*x+1 or something like that.

If you make the precision of sin(a) different than cos(a) then you
no longer have sin and cos, you have something else.

What is your purpose for making the precision different?

BTW, if we want to prove those identities for t=tan(x/2)
instead of tan(x/2) we would substitute cos(x/2)/sin(x/2) and
reduce the result. The result would be sin(x) or cos(x) depending
on which of those two equations in t were chosen.

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jasonbe said:

If I am using and expanding upon your notation correctly, given 1 - Aerr =(Sx+Serr)^2+(Cx+Cerr)^2, how would Aerr, Serr, and Cerr relate? What about with (d(sin(x + Serr))/dx)^2 + (d(cos(x + Cerr))/dx)^2 = 1 - Aerr. I'm just guessing that it might be piecewise.

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jasonbe said:

What would a proof of those identities look like?

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MrAl said:

Hi,


From your questions i think that you would do well to get some books on Algebra, Trig, and geometry and
start to learn some of this so that you can answer some of these questions yourself. If you have that
many questions like this you would be happy to learn some of this stuff.





Simply:

Aerr=2*Serr*Sx+Serr^2+2*Cerr*Cx+Cerr^2






Here is a nice graphical proof:

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