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[Post New]3 Apr 2007 18:55:57 IST
    Subject: jst wanted steps of solution
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king_009 (0)

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cos 12+cos84+cos156+cos132=?
please give me method how to solve it!
the answer is 1/2
just give me steps of solving
    
[Post New]3 Apr 2007 19:34:30 IST
    Subject: Re:jst wanted steps of solution
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rsharma.ranchi_000 (27)

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THE GIVEN EXPRESSION IS COS12+COS84+COS156+COS132
OR COS (12) + COS (7*12) + COS (13*12) + COS (11*12)
IF WE SUPPOSE 12 = X , WE GET COS X + COS 7X + COS 11X + COS 13X
NOW USE THE FORMULA OF COS A + COS B
 
THEN YOU WILL GET YOUR ANSWER
 
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[Post New]4 Apr 2007 10:13:59 IST
    Subject: Re:jst wanted steps of solution
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KS9290 (13)

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use cosA+cosB FORMULA . HERE IT GOES
(cos12 +cos132)+(cos84+cos156)
=[2cos72cos60)+(2cos120cos36)
=cos72-cos36
=[root(5)-1]/4 +[root(5)+1]/4
=-1/2
 
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[Post New]4 Apr 2007 14:37:28 IST
    Subject: Re:jst wanted steps of solution
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iitkgp_bipin (6144)

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The difference : 1320 - 120 = 1200 

The sum : 840 + 1560 = 2400

Hence  cos120 + cos1320 + cos840 + cos1560

          = 2cos720cos600 + 2cos1200cos360

          = cos720 - cos360

The rest is easy.
 
Bipin Kumar Dubey
Chemical Dept.
IIT Kharagpur

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