Mathematics - Functions Question with Solution | TestHub

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

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

This implies:

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

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

Therefore, $[\sin x]=0$ for $0<x<\frac{\pi }{2}$.

Thus, $\sin [x]+[\sin x]$ is:

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