Posted by Sam on Wednesday, November 13, 2013 at 11:23am.
Show that cot((x+y)/2) = - (sin x - sin y)/(cos x - cos y) for all values of x and y for which both sides are defined.
I tried manipulating both sides in terms of trig identities but I don't really have a solution....help would be appreciated, thanks.
Trigonometry - Identities and proof - Steve, Wednesday, November 13, 2013 at 11:58am
Using the sum-to-product formulas,
sinx-siny = 2cos((x+y)/2)sin((x-y)/2)
cosx-cosy = -2sin((x+y)/2)sin((x-y)/2)
now just divide to get
cos((x+y)/2) / sin((x+y)/2) = cot((x+y)/2)
Trigonometry - Identities and proof - Reiny, Wednesday, November 13, 2013 at 11:59am
Two of the standard conversion formulas are:
sinA - sinB = 2sin((A-B)/2) cos( (A+B)/2)
cosA - cosB = - 2sin( (A-B)/2) sin( (A+B)/2)
near bottom of page under:
= -(sinx - siny)/(cosx - cosy)
= - 2sin((X-Y)/2) cos( (X+Y)/2)/- 2sin( (X-Y)/2) sin( (X+Y)/2)
= cos((X+Y)/2) / sin(X+Y)/2)
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