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 November 19th, 2007, 03:11 PM #1 Newbie   Joined: Oct 2007 Posts: 19 Thanks: 0 challenge problems If sin A=1/squareroot 3 with 0
 November 19th, 2007, 03:25 PM #2 Senior Member   Joined: May 2007 Posts: 402 Thanks: 0 Could you explain the restrictions of numbers A and B? It's like saying that two is two but is should be between 1 and 3!
 November 19th, 2007, 03:54 PM #3 Newbie   Joined: Nov 2007 Posts: 6 Thanks: 0 If sin A= 1/√3 with 0
 November 20th, 2007, 08:34 AM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms I'll give you numeric solutions so you can check yourself. cos(a+b) -0.78847305878808248298610357681339746 sin(a+b) 0.61506929330390492576093835225741105 cos(a+a) 0.33333333333333333333333333333333333 cos(a-b) 0.31706853799705080005220733541016477
 December 7th, 2007, 06:46 AM #5 Newbie   Joined: Dec 2007 From: PRC Posts: 1 Thanks: 0 answer a) because 0

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### solution of sin(a-b)=sina.cosb-cosa.sinb

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