# How do you find the domain and range of sqrt(1-sinx) ?

Mar 28, 2016

Whatever value $x$ takes $\sin x$ will be between $- 1 \mathmr{and} + 1$
So the argument ($1 - \sin x$) will between $0 \mathmr{and} + 2$
Conclusion: the domain is unlimited $\left(- \infty , + \infty\right)$
The range falls between $\sqrt{0} \mathmr{and} \sqrt{2}$ inclusive (see above)
Or $\left[0 , \sqrt{2}\right]$