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Reconciling definitions of sine
By George Walker on Wednesday, December 11, 2002 - 11:14 am:

I may be being a little stupid, but in our analysis course in the last lecture we defined the sine function (its definition is the power series; sinx=x-x3/3!+... )

What I was wondering is how possibly the ratio of the opposite and hypotenuse in a right angled triangle is in any way derived from this...since when we are 12 we learn that sine is "opposite over hypotenuse" !

By Demetres Christofides on Wednesday, December 11, 2002 - 12:18 pm:

This is not a stupid but a very natural question to ask. The whole point is that this definition is actually equivalent to "opposite over hypotenuse" that we learn at school. I'll try to give some hints on why this is so.
Firstly we need to explain how we measure the angle. The degrees are not helpful any more so we will use radians (I hope it will become clearer later why we choose to do so).
Draw a circle of unit radius and look at O=(0,0), A=(1,0) and B=(x,y) where B is on the circumference of the circle on the first quadrant. Convince yourself that the angle AOB is the length of the arc of the circle from A to B (that's where radians come in).
We aim to formalize this:
Define g:[-1,1]:®Â2 by f(t)=(Ö(1-t2),t) [We just take the positive square root]
Let O=(0,0), A=g(0), B=g(t) where t>=0 and define the angle AOB to be the length of g from 0 to t. Write q(t) for this. [Convince yourself that is what we want the angle to be]
Show that q(t)=ò0 t1/Ö(1-x2)dx.
This can also be extended for negative t.
Define p to be equal to 2q(1) [Check that it agrees with what what we understand by p].
Now some more analysis:
Show that q:[-1,1]®[-p/2,p/2] is strictly increasing. So q-1 exists. Call it sin ! [Note that for 0<y<1 if g(y)=(x,y), then sin(q(y))=y, which is what we want it to be! (Draw a picture)]
Extend its domain to the whole of  so that it is periodic.
Show that sin is continuous twice differentiable with sin''(x)+sin(x)=0 for all real x.
Solve this differential equation (using power series) to recover your lecturer's definition of sin.


Demetres

By George Walker on Thursday, December 12, 2002 - 10:04 am:

Thank you, I had done the last bit (power series solution of the differential equation), it was just formalising the first part. Good answer