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Question Number 85442 by Power last updated on 22/Mar/20

Commented by Power last updated on 22/Mar/20

$$\left[\mathbf{x}\right]-\mathbf{integer}\mathbf{part}=\mathbf{trunc}\left(\mathbf{x}\right)$$$$\left\{\mathbf{x}\right\}-\mathbf{fractional}\mathbf{part}=\mathbf{frac}\left(\mathbf{x}\right)$$

Answered by mr W last updated on 22/Mar/20

$$since{0}^{0}isnotdefined,$$$$\sin x\ne 0\Rightarrow x\ne k\pi \Rightarrow \cos x\ne \pm 1$$$$\cos x\ne 0\Rightarrow x\ne (k+\frac{1}{2})\pi \Rightarrow \sin x\ne \pm 1$$$$since\sin x\ne \pm 1and\cos x\ne \pm 1,$$$$\Rightarrow \left[\sin x\right]=0,\left[\cos x\right]=0$$$$\Rightarrow \left\{\sin x\right\}=\sin x,\left\{\cos x\right\}=\cos x$$$${\left\{\sin x\right\}}^{\left[\sin x\right]}={\left(\sin x\right)}^0=1$$$${\left\{\cos x\right\}}^{\left[\cos x\right]}={\left(\cos x\right)}^0=1$$$$i.e.{\left\{\sin x\right\}}^{\left[\sin x\right]}={\left\{\cos x\right\}}^{\left[\cos x\right]}istrue.$$$$$$$$\Rightarrow solutionis$$$$x\in R\wedge x\ne (k+\frac{1}{2})\pi \wedge x\ne k\pi$$

Commented by Power last updated on 22/Mar/20

Commented by Power last updated on 22/Mar/20

$$\mathrm{sir}\mathrm{mr}\mathrm{W}?$$

Commented by Power last updated on 22/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Commented by mr W last updated on 22/Mar/20

$${what}^{\prime }syourproblem?$$

Commented by Power last updated on 22/Mar/20

$$\mathrm{why}\left[\mathrm{sinx}\right]=0\left[\mathrm{cosx}\right]=0?\mathrm{please}\mathrm{explain}\mathrm{that}$$

Commented by Power last updated on 22/Mar/20

$$\mathrm{sinx}=-\frac{1}{2}$$

Commented by Power last updated on 22/Mar/20

$$\mathrm{cosx}=\frac{\sqrt{3}}{2}$$

Commented by mr W last updated on 22/Mar/20

$$-1<\sin x<1\Rightarrow \left[\sin x\right]=ipart\left(\sin x\right)=trunc\left(\sin x\right)=0$$$$-1<\cos x<1\Rightarrow \left[\cos x\right]=ipart\left(\cos x\right)=trunc\left(\cos x\right)=0$$

Commented by Power last updated on 22/Mar/20

$$\mathrm{sir}\mathrm{ipart}?$$

Commented by mr W last updated on 22/Mar/20

$$ipart\left(x\right)=integerpartofx$$

Commented by Power last updated on 22/Mar/20

Commented by mr W last updated on 22/Mar/20

$$[-\frac{1}{2}]=ipart(-\frac{1}{2})=0$$$$\{-\frac{1}{2}\}=fpart(-\frac{1}{2})=-\frac{1}{2}$$$$thisiswhatiapplied,seebelow.$$$$youmayacceptitorrefuseit.$$$$i{won}^{\prime }tdiscusswith$$$$you,becausetherearedifferent$$$$definitionsfor[-\frac{1}{2}].ifyoutakean$$$$otherdefinitionthani,youshould$$$$presentyourownsolution.$$

Commented by mr W last updated on 22/Mar/20

Commented by Power last updated on 22/Mar/20

$$\mathrm{okey}\mathrm{thank}\mathrm{you}\mathrm{sir}$$

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