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![[Post New]](/templates/default/images/icon_minipost_new.gif) 17 Jul 2008 18:01:06 IST
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sin(45+A)cos(45-B)+cos(45+A)sin(45-B)=cos(A-B)
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17 Jul 2008 18:06:17 IST
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its sin(45+A+45-b)
therefore its sin(90+A-B)
HENCE we get sin[90+(A-B)]=COS(A-B)
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17 Jul 2008 18:10:55 IST
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It is of the form sinAcosB+cosAsinB
We know that sin(A+B)=sinAcosB+cosAsinB
So,on applying this formula we get sin(90+(A-B))
which is equal to cos(A-B)
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17 Jul 2008 19:21:55 IST
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Can you please explain the entire solution in detail
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17 Jul 2008 19:35:51 IST
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sin(A+B)=sinAcosB+cosAsinB
Here A= 45 + A B= 45 - B
sin(45+A)cos(45-B)+sin(45-B)cos(45+B)
=sin(45+A+45-B)
=sin(90+(A-B)
=cos(A-B)
I hope it is clear now
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17 Jul 2008 21:38:30 IST
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Thank you very much
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17 Jul 2008 23:40:39 IST
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Right answer Nitin. Well done.
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Krishna Gopal Singh
B.Tech Chemical Engg
IIT Delhi 2002
Currently doing PhD from IIT Delhi |
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