x<y ----------(1)
=> sinx <siny -------(2)
now in (0,pi/2)
both x-sinx and y-siny are increaisng functions
=> x-sinx<y-siny
x<y ----------(1)
=> sinx <siny -------(2)
now in (0,pi/2)
both x-sinx and y-siny are increaisng functions
=> x-sinx<y-siny
As x<y,
sinx<siny, sinx is increasing in (0,∩/2)
From the graph,
x>sinx
or,x-sinx>0
=>∂(x-sinx)/∂x=1-cosx >0 for x in (0,∩/2)
Therefore x-sinx is increasing funtion
So, as x<y
x-sinx<y-siny
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Or simply consider defined by
f(t)=t−sint
For t∈(0,π/2),
f′(t)=1−cost>0
Hence, f is strictly increasing. As such for
0<x<y<π/2
Just for the heck of it one more solution, this one without calculus.
Since x,y∈[0,2π], we have 2y−x,2y+x∈[0,2π]
Now siny−sinx=2sin2y−xcos2y+x<2 (2y−x)cos2y+x<y−x
Hence x - sin x < y - sin y