Find the period of
(a) `(|sin4x|+|cos 4x|)/(|sin 4x-cos 4x|+|sin 4x+cos 4x|)`
(b) `f(x)="sin"(pi x)/(n!)-"cos"(pi x)/((n+1)!)`
(c ) `f(x)=sin x +"tan"(x)/(2)+"sin"(x)/(2^(2))+"tan"(x)/(2^(3))+ ... +"sin"(x)/(2^(n-1))+"tan"(x)/(2^(n))`

Answer

(a) `pi//8` (b) `2(n+1)!` (c ) `2^(n)pi`

Explanation

Period of `|sin 4x|+|cos 4x|" is " (pi)/(8)`
Period of `|sin 4x-cos 4x|+|sin 4x+cos 4x|=(pi)/(8)`
So, the period of given function is `(pi)/(8).`
(b) `f(x)="sin"(pi)/(n!)-"cos"(pi x)/((n+1)!)`
Period of `"sin"(pi)/(n!) " is " (2pi)/((pi)/(n!))=2n!` and period of ` "cos"(pi x)/((n+1)!) " is " (2pi)/((pi )/((n+1)!))=2(n+1)!`
Hence, period of `f(x)=L.C.M" of "{2n!,29n+1)!}=2(n+1)!`
(c ) `f(x)=sin x +"tan"(x)/(2)+"sin"(x)/(2^(2))+"tan"(x)/(2^(3))+ ... +"sin"(x)/(2^(n-1))+"tan"(x)/(2^(n))`
Period of `sin x " is " 2pi.`
Period of ` "tan"(x)/(2) " is " 2pi.`
Period of ` "sin"(x)/(2^(2)) " is " 8pi.`
Period of ` "tan"(x)/(2^(3)) " is " 8pi`.
Period of `"tan"(x)/(2^(n)) " is " 2^(n) pi.`
Hence, period of `f(x)=L.C.M. " of " (2pi,8pi, ..., 2^(n) pi)=2^(n) pi`