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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<A<pi/2, and cos B= 1/2squareroot3, with pi/2<B<pi, calculate a)cos(A+B) b) sin(A+B) c) cos2A d) cos AB) 
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<A<pi/2, and cos B= 1/2√3, with pi/2<B<pi, calculate... I hope that this is clearer to read. Just in case, it says sin A= 1 over squareroot 3 with 0 is less than A and less than pi over 2, and cos B= negative one over 2 squareroot 3 with restrictions of pi over 2 is less than B and less than pi. 
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(ab) 0.31706853799705080005220733541016477 
December 7th, 2007, 06:46 AM  #5 
Newbie Joined: Dec 2007 From: PRC Posts: 1 Thanks: 0  answer
a) because 0<A<pi/2,sin A=1/squareroot 3 cosA=sqrt(1(sinA)^2)=sqrt(2/3) pi/2<B<pi, cosB=1/2sqrt 3 sinB=sqrt(1(cosB)^2)=11/(2*sqrt 3) cos(A+B)=cosAcosBsinAsinB=(sqrt 2+sqrt 11)/6 b) sin(A+B)=sinAcosB+cosAsinB=(sqrt(22)1)/6 c) cos2A=12(sinA)^2=1/3 d) cos(AB)=cosAcosB+sinAsinB=(sqrt 11sqrt2)/6 

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