Question

the range of y= 1-sin x is​

Answer 1

Step-by-step explanation:

We have,

$y = 1 - \sin(x)$

We know,

$- 1 \leqslant \sin(x) \leqslant 1$

$\implies1 \geqslant - \sin(x) \geqslant - 1$

$\implies2 \geqslant 1 - \sin(x ) \geqslant 0$

Hence,

$range(f) \in [0 , 2]$