If f(x) = {{:(-x-pi/2" , " x le - pi/2),(-cos x" , " -pi/2 lt x le 0",""then"),(x-1", "0lt x le 1),(ln x", "x gt 1):}
![sin(pi/2-x) + sin(pi-x)]^2+[(cos((3pi)/2-x)+cos(2pi-x)]^2=](https://d10lpgp6xz60nq.cloudfront.net/ss/web/63533.jpg "sin(pi/2-x) + sin(pi-x)]^2+[(cos((3pi)/2-x)+cos(2pi-x)]^2=")
sin(pi/2-x) + sin(pi-x)]^2+[(cos((3pi)/2-x)+cos(2pi-x)]^2=
&line2=\mathrm{Solution:\:}\mathrm{Phase}:\frac{1}{2},\:\mathrm{Vertical}:0 "shift 20cos(pi x-(pi)/2) - Symbolab")
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![The range of f(x) = sin\ pi/2 [x] is, (Here [x] denotes greatest integer lt= x)](https://d10lpgp6xz60nq.cloudfront.net/ss/web/208221.jpg "The range of f(x) = sin\ pi/2 [x] is, (Here [x] denotes greatest integer lt= x)")
The range of f(x) = sin\ pi/2 [x] is, (Here [x] denotes greatest integer lt= x)
![lim_[x- gtpi/2]cosx/[x-pi/2]` - YouTube](https://i.ytimg.com/vi/_GCXQ4Jqzp8/maxresdefault.jpg "lim_[x- gtpi/2]cosx/[x-pi/2]` - YouTube")
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