Why do we restrict the domain of sine function to dh interval[-pi/2,pi/2] and not [-3pi/2,-pi/2]

Because it's easier to state, and we're going from a value of -1 to 1. the other would go from a value of 1 to -1

Sine function is a circular function.
It's ordinates always lie in the interval[-1,1]
It's abscissae lie in the interval
[-m(n+(-pi/2)),-m(n+(p/2))]
where,m = 1 when n =pi(1+2k)
and m = -1 when n = 2piek
where 'k' is a whole number.

Report Abuse

when, m = -1 and n = 0(k=0)
we get the interval[-pi/2,pi/2]
when m = 1 and n = pie(k=0)
we get[-pi/2,-3pi/2]
which is taken as[-3pi/2,-pi/2] and hence the image[-1,1] is a mirror image.

Moreover,sin(x) = -sin(x - pi)
therefore,sin(-pi/2) = -sin(-3pi/2)
and,sin(pi/2) = -sin(-pi/2)

Report Abuse

Hence you have infinite intervals to obtain the image interval[-1,1]
Hence,Sine function is a many-one functioin which prohibits it's inverse.
To validate it's inverse and to make it a one-one function the interval [-pi/2,pi/2] is deliberately choosen from those infiniteis.

Report Abuse

Why only [-pi/2,pi/2]?
The interval [-pi/2,pi/2] is the simplest and passes through the origin making it the easist to study and to refer


Hope this helps..

Report Abuse