1ok i got the first one....that was quiet silly of me.......but din get the second one though[2]
try these 2 questions :)
1). if z +1z =2cos3°
then find the value of z2009 + 1z2009
2) if x= 4((51/2 +1)(51/4+1)(51/8+1)(51/16 +1))
then find the last digit of (x+1)48.......
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8 Answers
62the first one seems very simple to me...
take z = cos a + i sin a
1/z = cos a - i sin a
z+1/z = 2 cos a
so a=3 degree
now can you solve it?
it will be 2 cos (2009 x 3)
The second one is a not so tough one.. try to write 4 = (5-1) = (51/2-1)(51/2+1)
162see 51/2+1 gets cancelled.. then if you write 51/2-1 = ( 51/4+1 )( 51/4+1 ) then again ( 51/4+1 ) will get cancelled and so on..
so the value of x will be 51/16-1
so x+1 = 51/16
so (x+1)48 = 125
1oh yeah that was again so silly of me................thanx a lot.... :)
11A very similar way for the 2nd sum is to multiply numerator & Denominator by \sqrt[16]{5}-1.......
62@ kaustab.. may be :D
you are also from calcutta.. that is why this confusion :P