what is the domain and range of (absolute value of sinx)^2

Question

solomonzelman · Answer

$\large\color{black}{   \left|  \sin x \right| ^2    }$.
Well, for all we know, $\large\color{black}{  -1\le \sin x\le1    }$.
(this is also true about the cosine of x)

So, the range of $\large\color{black}{   \left|  \sin x \right| \le1    }$ would be just $\large\color{black}{   [0,1]   }$ .
$\normalsize\color{black}{   \bullet   }$ Because negatives are not allowed, so $\normalsize\color{blue}{     \rm from~zero              }$.
$\normalsize\color{black}{   \bullet   }$ $\normalsize\color{blue}{     \rm Till~1                }$, Because sin(x) can't be equal to a number greater than 1.

solomonzelman · Answer

the range of $\large\color{black}{   \left|  \sin x \right|^2    }$ would not differ (from $\large\color{black}{   \left|  \sin x \right|    }$ ).      (the $\large\color{black}{   \left|  \sin x \right|^2    }$ would have corners but $\large\color{black}{   \left|  \sin x \right|    }$ would not)

solomonzelman · Answer

Now, the domain can be any number, because there is no restriction on what you are plugging in for x.

$\large\color{black}{   \left|  \sin x \right|^2    }$ is defined for all values of  $\large\color{black}{ x  }$.

So the domain is $\large\color{black}{   (-\infty ,+\infty) }$.