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November 19th, 2007, 03:11 PM   #1
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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 A-B)
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November 19th, 2007, 03:25 PM   #2
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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!
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November 19th, 2007, 03:54 PM   #3
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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.
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November 20th, 2007, 08:34 AM   #4
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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
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December 7th, 2007, 06:46 AM   #5
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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)=cosAcosB-sinAsinB=-(sqrt 2+sqrt 11)/6

b) sin(A+B)=sinAcosB+cosAsinB=(sqrt(22)-1)/6

c) cos2A=1-2(sinA)^2=1/3

d) cos(A-B)=cosAcosB+sinAsinB=(sqrt 11-sqrt2)/6
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