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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Multiple n submultilple angles!
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[Post New]7 Apr 2008 15:58:09 IST
    Subject: Multiple n submultilple angles!
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m.viddya (101)

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Q.cos3A+cos3(120+A)cos3(120-A)=kcos3A then k=?
 
1.1/4
 
2.1/2
 
3.3/4
 
4.none
    
[Post New]7 Apr 2008 16:17:56 IST
    Subject: Re:Multiple n submultilple angles!
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uday_zingtudor (931)

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Cos@ + cos(120+@) + cos(120-@) = 0

U can get the above result simply by expanding it.

Then let cos@ =a cos(120+@) = b and cos(120-@) = c

a+b+c =0

that implies a3+b3+c3= 3abc

Using MASMA properties (multiplication addition identities), u get

a3+b3+c3 = 3(1/4)cos3@

k= 3/4

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[Post New]7 Apr 2008 18:49:46 IST
    Subject: Re:Multiple n submultilple angles!
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ankitagg (174)

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try this fundamental method;
cos3A=4cos^3A-3cosA
cos^3A=1/4(cos3A+3cosA)
 putting in eq.
kcos3A= 1/4(cos3A+3cosA)+1/4[cos(360+3A)+3cos(120+A)]+1/4[cos(360-3A)+3cos(120-A)]
=1/4[cos3A+3cosA]+1/4[cos3A+3cos(120+A)]+1/4[cos3A+cos(120-A)]
=3/4cos3A+3/4[cosA+cos(120+A)+cos(120-A)]
=3/4cos3A+3/4[cosA+2cos120cosA]
=3/4cos3A+3/4[cosA-cosA]
=3/4cos3A
therefore=3/4
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