Math, asked by mysagoud82, 7 months ago

the range of y= 1-sin x is​

Answers

Answered by senboni123456
0

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]

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