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if period of function cos(nx).sin(5x/n) is 3 pi, then number of integral values of n are ?
hi subhadeep
the function can be written in the form sin (A+B) + sin (A-B) using trigonometric identity
then this is of the form of sum of two functions then find the L.C.M of the two arguments
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jitender ''
askiitians expert
your approach is correct but how to solve further......taking the LCM and finding values of n ???
that is my question........even i already knew the approach of sin (a+b)+sin(a-b)........if possible tell me the method after that .............i am stuck in taking the LCM.........and what is the aswer u r getting???
the answer is no integral solution
actually we dont have to use the product to sum identity in trigonometry
the product of any two periodic function is also periodic
we just have to find L.C.M of their fundamental period
fundamnetal period of cos(nx) is T1 = 2 pie / n -------1
and other function the fundamental period is T2 = 2npie/5------------2
their ratio (T1/T2) = 5 / n^2
n^2 T1 = 5 T2 = T = 3 pie ------ given (T is the composite fundamental period )
T1 = 3 pie / n^2 T2 = 3 pie / 5
putting the values of T1 and T2 we get n = 3/2
so no integral solution
why did u take the ratio of T1 and T2 and then cross multiplying equated to T = 3 pi ..........??
the logic ???
if c is L.C.M of a , b
a, b can be written as
c = ma , c = nb
where m,n are integes
so c = ma = nb
that is why i took the ratio of T1 , T2 and wrote in terms of T
and also it was necessary as we had to find the value of n
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jitender
but the answer given is 8 integral solutions ............
pls provide those integral values if they are given in the answer
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