Expert: Steve Holleran - 5/21/2007

Question
hi i am trying to work these problems (1) State the value of lim ('small' delta x)-->0 [sin 'small' delta x]/['small' delta x]  NB from now 'small' delta x denoted 'd' (2)Given that sin2(x+'d'x)-sin2x=2cosAsinB. Find A and B in terms of x and/or 'd'x. For qu 1 i definitely need help. I am not sure where nor how to begin it. for (2) this is what i tried: sin2(x+'d'x)-sin2x=2cosAsinB; sin2(x+'d'x)-sin2x=2sin(x+'d'x)cos(x+'d'x)-2sinxcosx= 2[sinxcos'd'x+cosxsin'd'x][cosxcos'd'x-sinxsin'd'x]-2sinxcosx.  I am not sure where to go from here or if i am even going right.plz help me thankz.i appreciate it.

Answer
Hello D,

Okay, on part 1, this limit is a basic trig limit that comes up in calculus.  Since it just asks that you state the value, the answer is 1.

The proof of it is involved with triangles and the unit circle, but I don't think you want to get involved with that here--it can be found in almost any calc text.  The basic idea is

                 lim(A-->0) [sin A / A] = 1

For part 2, I had to go to an obscure trig identity.  
It's called a "sum to product" formula, and it goes like this:

 sin u - sin v = 2 * sin[(u - v)/2] * cos[(u + v)/2]

Here, u = 2(x + dx) and v = 2x, so you have:

              sin[2(x + dx)] - sin[2x]

=  2 * sin[(2(x + dx) - 2x)/2] * cos[(2(x + dx) + 2x )/ 2]

=  2 * sin[2 dx / 2]  * cos [ (4x + 2 dx) / 2]

=  2 * sin(dx) * cos(2x + dx) = 2 cos A sin B

so that A = 2x + dx and B = dx


I hope this is what you were looking for.

Steve Holleran