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March 5, 2015

March 5, 2015

Posted by **Sam** on Wednesday, November 13, 2013 at 11:23am.

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 11:58amUsing 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 11:59amTwo of the standard conversion formulas are:

sinA - sinB = 2sin((A-B)/2) cos( (A+B)/2)

and

cosA - cosB = - 2sin( (A-B)/2) sin( (A+B)/2)

see:

near bottom of page under:

Sum-to-Product Formulas

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

RS

= -(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)

= cot((X+Y)/2)

= LS

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