Inequality

If 0<x<y<Ï€/2, show that : x - sin x < y - sin y

5 Answers

24
eureka123 ·

x<y ----------(1)
=> sinx <siny -------(2)

now in (0,pi/2)
both x-sinx and y-siny are increaisng functions
=> x-sinx<y-siny

24
eureka123 ·

i hope i have not committed any blunder [3]

1
Kalyan Pilla ·

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

[339]

66
kaymant ·

Or simply consider f:[0,\pi/2]\longrightarrow \mathbb{R} defined by
f(t)=tsint
For t(0,π/2),
f(t)=1cost>0
Hence, f is strictly increasing. As such for
0<x<y<π/2
f(x)>f(y)\quad\Rightarrow \ x-\sin x>y-\sin y

341
Hari Shankar ·

Just for the heck of it one more solution, this one without calculus.

Since x,y[0,2π], we have 2yx,2y+x[0,2π]

Now sinysinx=2sin2yxcos2y+x<2 (2yx)cos2y+x<yx

Hence x - sin x < y - sin y

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