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Trignometry

Onkar's Avatar
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4 Mar 2009 12:13:00 IST
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find the value of sin 7.5
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find the value of sin 7.5


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manas mahapatra's Avatar

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Joined: 21 Feb 2009
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6 Mar 2009 02:02:15 IST
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2sin^2{@/2}= 1-cos 2@, take @  =  7.5, cos 15= [root3+1]divded by 2rt2. get the ans. if undstod pls rate


Blazing goIITian

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20 Mar 2009 21:48:05 IST
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THE VALUE OF SIN(7.5)

IS     

revenge's Avatar

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Joined: 22 Feb 2009
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20 Mar 2009 21:50:21 IST
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its  equal to  sin  367.5


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20 Mar 2009 21:59:05 IST
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REVENGE,

I THINK UR ALSO OUT OF UR MIN DS .  HE ASKED THE VALUE OF SIN (7.5).

NOT AN EQUIVALENNT  TRIGO. EQUATION.

THESE SMILEYS SYMBOLISES UR AMBIGUOS ANSWER.

TEJASWI V SHETTY.'s Avatar

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21 Mar 2009 20:33:50 IST
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I AGREE WITH MANAS.

USING D SAME METHOD WHERE @ OF MANAS IS 15, THE ANS U GET IS  .

U CAN ALSO USE  DOUBLE ANGLE FORMULA OF SIN BUT THAT WILL B MORE LENGTHY.

HOPE U UNDERSTOOD.


Blazing goIITian

Joined: 7 May 2007
Posts: 1724
21 Mar 2009 20:40:24 IST
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THE VALUE OF SIN(7.5)

IS     

 

is this answer not correct.

if not, tell where it is wrong.

TEJASWI V SHETTY.'s Avatar

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22 Mar 2009 11:27:41 IST
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i think u hav seen d ans i hav posted earlier for this question which has included d method i hav used.

if u want to know ur fault then plz put the method u hav used in solving this prob...

Sagar Saxena's Avatar

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Joined: 8 Oct 2008
Posts: 7221
23 Mar 2009 08:32:47 IST
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Sin 150  is known.u can get it by applying the formula sin (450   - 300)

 

u can obtain Sin 7.50 by applying half angle formula in Tan and than convert it into Sin


Blazing goIITian

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24 Mar 2009 17:56:09 IST
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TEJASWI SHETTY,

U WANTED ME TO POST MY SOLUTION.

SIN( @ / 2 ) = [(1 - COS @)/2]^(1/2)

FROM THIS FORMULA,

SIN 15 = SIN ( 30 / 2 )

THEN

SIN (7.5) = SIN ( 15 / 2 )

NOW NOTIFY.?

TEJASWI V SHETTY.'s Avatar

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25 Mar 2009 19:26:06 IST
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kapil cvs,

u r absolutely right in ur approach.  Even i used  d same method.

i think there is a prob. in ur calculations. check that. i think u would be able to come accross ur mistake.

have a nice day!


New kid on the Block

Joined: 31 Oct 2011
Posts: 1
9 Feb 2012 19:06:11 IST
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thanx



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