Mathematics - Functions Question with Solution | TestHub

For $0<x<\frac{\pi }{2}$:

$$\therefore [x]=\{\begin{array}{ll}0& if0<x<1\\ 1& if1\le x<\frac{\pi }{2}\end{array}$$

$$\Rightarrow \sin [x]=\{\begin{array}{ll}\sin 0=0& if0<x<1\\ \sin 1& if1\le x<\frac{\pi }{2}\end{array}$$

We have $0<\sin x<1$ when $0<x<\frac{\pi }{2}$.

$$\therefore [\sin x]=0for0<x<\frac{\pi }{2}$$

$$\therefore \sin [x]+[\sin x]=\{\begin{array}{ll}0& if0<x<1\\ \sin 1& if1\le x<\frac{\pi }{2}\end{array}$$