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Question Number 7031 by Tawakalitu. last updated on 07/Aug/16

Commented by FilupSmith last updated on 07/Aug/16

$$myattempt$$$$0<x,y<\frac{\pi }{2}$$$$x<y$$$$x-\sin \left(x\right)<y-\sin \left(y\right)$$$$$$$$y=x+n$$$$0<x<x+n<\frac{\pi }{2}$$$$\sin \left(t\right)\mathrm{is}\mathrm{positive}\mathrm{for}2\pi n<\mathrm{t}<2\pi n+\pi ,n\in \mathbb{Z}$$$$\therefore \mathrm{\forall }x,0<x<\frac{\pi }{2},\sin \left(x\right)>0$$$$\therefore x+n=y>x$$$$\therefore \sin \left(y\right)>\sin \left(x\right)$$$$$$$$x-\sin \left(x\right)<x+n-\sin (x+n)$$$$\therefore x-\sin \left(x\right)<y-\sin \left(y\right)$$

Commented by Tawakalitu. last updated on 07/Aug/16

$$Iappreciate.Thankyou.$$

Commented by prakash jain last updated on 07/Aug/16

$$\sin \mathrm{Y}-\sin \mathrm{X}=2\sin \frac{\mathrm{Y}-\mathrm{X}}{2}\cos \frac{\mathrm{Y}+\mathrm{X}}{2}$$$$\sin \frac{\mathrm{Y}-\mathrm{X}}{2}<\frac{\mathrm{Y}-\mathrm{X}}{2}$$$$\cos \frac{\mathrm{Y}+\mathrm{X}}{2}⩽1$$$$2\sin \frac{\mathrm{Y}-\mathrm{X}}{2}\cos \frac{\mathrm{Y}+\mathrm{X}}{2}<\frac{\mathrm{Y}-\mathrm{X}}{2}\times 2=\mathrm{Y}-\mathrm{X}$$$$\sin \mathrm{Y}-\sin \mathrm{X}<\mathrm{Y}-\mathrm{X}$$$$\Rightarrow \mathrm{X}-\sin \mathrm{X}<\mathrm{Y}-\sin \mathrm{Y}$$

Commented by prakash jain last updated on 07/Aug/16

$$\mathrm{To}\mathrm{prove}\mathrm{the}\mathrm{relation}\mathrm{geometrically}\mathrm{you}$$$$\mathrm{can}\mathrm{use}\mathrm{the}\mathrm{same}\mathrm{contruction}\mathrm{as}\mathrm{used}\mathrm{to}$$$$\mathrm{prove}\sin \mathrm{C}-\sin \mathrm{D}\mathrm{formula}.$$

Commented by Tawakalitu. last updated on 08/Aug/16

$$Thankssomuchforyourhelp$$

Answered by Yozzii last updated on 08/Aug/16

$$Checkforananswerincomments.$$

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