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![[Post New]](/templates/default/images/icon_minipost_new.gif) 13 Jun 2007 21:28:36 IST
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who answers correctly first shall be accepted as correct ok here is the problem cos  /7 . cos 2/7 . cos 4/7............ answer with steps pls.... I tried it but i'm goin wrong somewhere so i'm not gettin the right answer....... oops guys ...... typed the q wrong !!!! its only cos /7 . cos 2/7 . cos 4/7
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In the process of learnin..............blunders do happen !!! |
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13 Jun 2007 22:37:08 IST
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not sure but is it tending to zero?? on simplifying u get sin (2n+1pi/7) ----------- first generalising for n terms. . 2n sin(pi/7) now as n tends to infi. this expression tends to zero. as sin ( ) is going to be from [-1,1] but 2n will tend to infinitely large value so the whole expression wud tend to zero.
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14 Jun 2007 08:39:01 IST
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y=cos(pi/7).cos(2pi/7).cos(4pi/7)............
for n terms
2^n * sin(pi/7).y=2^n * sin(pi/7).cos(pi/7).cos(2pi/7).cos(4pi/7)....cos(2^npi/7)
2^n * sin(pi/7).y=2^n-1 * sin(2pi/7).cos(2pi/7).cos(4pi/7)....cos(2^npi/7)
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2^n *sin(pi/7).y = 2sin2^n(pi/7).cos2^n(pi/7)
2^n *sin(pi/7).y = sin2^n+1(pi/7)
y=[sin2^n+1(pi+7)]/2^n*sin(pi/7)
now as n -> inf.
lim y= (sin2^n+1(pi/7))
n->inf ---------------- ---->0
(2^n) .sin(pi/7)
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Your attitude determines your altitude |
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14 Jun 2007 10:07:22 IST
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Let y = cos /7 cos2 /7 cos4/7 .....
y(2sin/7) = (2sin/7cos/7) cos2/7 cos4/7 .....upto n terms
y(2sin/7) = sin2/7 cos2/7 cos4/7 .....upto n terms
y(22sin/7) = (2sin2/7 cos2/7) cos4/7 .....upto n terms
y(22sin/7) = sin4/7 cos4/7 .....upto n terms
Similarly
y(2nsin/7) = sin2n+1/7
y= sin2n+1/7/2nsin/7
The required sum is
S = n  y = n sin2n+1/7/2nsin/7 = 0
As the Nr is a bounded quantity lying between -1 and 1. So it has a finite value. The Dr increases without bound. Hence the quotient approaches zero.
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14 Jun 2007 12:15:33 IST
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good work everyone .
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14 Jun 2007 13:29:02 IST
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oops guys ...... typed the q wrong !!!! its only cos /7 . cos 2/7 . cos 4/7
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In the process of learnin..............blunders do happen !!! |
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14 Jun 2007 15:52:06 IST
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Cos  /7 . cos 2 /7 . cos 4 / 7 Remember there is a formula such as : CosA.Cos2A.Cos22A.Cos23A..................Cos2n-1A =Sin2n A / 2n Sin A so then we will get where "A"= /7 =Sin 2 3 /7 / 8 sin /7 =Sin 8 /7 / 8 Sin /7 =1/8 Therefore the answer is 1/8 plZ rate me for my efforts Cheers !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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Always available for help !
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14 Jun 2007 16:35:04 IST
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can you do it without the formula ? cos this prob is given before the formula is given n the ans is -1/8
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In the process of learnin..............blunders do happen !!! |
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14 Jun 2007 16:50:54 IST
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i am not able to understand the answers related to this question.can anyone explain it clearly
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all the people are invited to answer my queries!!!!!!! |
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14 Jun 2007 17:12:57 IST
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impossible the answer has to be 1/8 and nothing else
I advise u to check the problem again
or
the answer given is wrong.
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Always available for help !
But Remember Don't hesitate to ask a good Question but
Be damn serious for Questioning a weak one.
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@waterdemon...it has to b -1/8
u 4got dat (sin 8pi/7)/(sin pi/7) = -1 not 1!!!
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NO SIGNATURE |
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15 Jun 2007 01:28:06 IST
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look it's easy. multiply & divide by 2sin /7. we get [sin2 /7 cos2 /7cos4 /7]/2sin /7 multiply ÷ by2 again by 2 {sin8 /7}/8sin /7 = -1/8
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16 Jun 2007 18:12:36 IST
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even according to formula its 1/8!!
sin 8pi/7 = -sin pi/7
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3 Sep 2007 17:00:59 IST
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@ him ............ thanx for makin it way toooooo easy !!
good ans !!
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