domain of some function











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I need domain the following functions.
Can you help me?
$f(x)= tan x .cot x$ and $g(x)= sqrt {1-sin^2 x}$ and $h(x)= sqrt {(1- sin^2 x)^2}$.






calculus







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    I need domain the following functions.
    Can you help me?
    $f(x)= tan x .cot x$ and $g(x)= sqrt {1-sin^2 x}$ and $h(x)= sqrt {(1- sin^2 x)^2}$.






    calculus







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      up vote
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      down vote

      favorite











      I need domain the following functions.
      Can you help me?
      $f(x)= tan x .cot x$ and $g(x)= sqrt {1-sin^2 x}$ and $h(x)= sqrt {(1- sin^2 x)^2}$.






      calculus







      share|cite|improve this question









      New contributor




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      I need domain the following functions.
      Can you help me?
      $f(x)= tan x .cot x$ and $g(x)= sqrt {1-sin^2 x}$ and $h(x)= sqrt {(1- sin^2 x)^2}$.






      calculus




      calculus






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      edited 2 days ago









      Bernard

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          2 Answers
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          HINT





          • $f(x)= tan x cdot cot x$ we need $sin xneq 0 , land ,cos x neq 0$


          • $g(x)= sqrt {1-sin^2 x}$ we need $1-sin^2 xge 0$


          • $h(x)= sqrt {(1-sin^2 x)^2}$ we need $(1-sin^2 x)^2ge 0$






          share|cite|improve this answer




























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            • We have $f(x) = frac{sin(x)}{cos(x)}frac{cos(x)}{sin(x)}$. We have $cos(x) = 0$ if and only if $x in { frac{pi}{2} + kpi mid k in mathbb{Z}}$, and $sin(x) = 0$ if and only if $x in { kpi mid k in mathbb{Z}}$. Hence domain of $f$ is $mathbb{R}setminus { frac{pi}{2} + kpi , kpimid k in mathbb{Z} } = mathbb{R}setminus { frac{kpi}{2} mid k in mathbb{Z} }$.


            • $h(x)$ is defined as long as $(1-sin^2(x))^2 geq 0$, which always holds. Hence domain of $h$ is $mathbb{R}$.


            • $g(x)$ is defined as long as $1-sin^2(x) geq 0$. We have for every $x in mathbb{R}$ that $sin(x) in [-1,1]$, hence $sin^2(x) in [0,1]$, hence $1-sin^2(x) in [0,1]$. We thus have that domain of $g$ is $mathbb{R}$.






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              2 Answers
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              2 Answers
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              up vote
              0
              down vote













              HINT





              • $f(x)= tan x cdot cot x$ we need $sin xneq 0 , land ,cos x neq 0$


              • $g(x)= sqrt {1-sin^2 x}$ we need $1-sin^2 xge 0$


              • $h(x)= sqrt {(1-sin^2 x)^2}$ we need $(1-sin^2 x)^2ge 0$






              share|cite|improve this answer

























                up vote
                0
                down vote













                HINT





                • $f(x)= tan x cdot cot x$ we need $sin xneq 0 , land ,cos x neq 0$


                • $g(x)= sqrt {1-sin^2 x}$ we need $1-sin^2 xge 0$


                • $h(x)= sqrt {(1-sin^2 x)^2}$ we need $(1-sin^2 x)^2ge 0$






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  HINT





                  • $f(x)= tan x cdot cot x$ we need $sin xneq 0 , land ,cos x neq 0$


                  • $g(x)= sqrt {1-sin^2 x}$ we need $1-sin^2 xge 0$


                  • $h(x)= sqrt {(1-sin^2 x)^2}$ we need $(1-sin^2 x)^2ge 0$






                  share|cite|improve this answer












                  HINT





                  • $f(x)= tan x cdot cot x$ we need $sin xneq 0 , land ,cos x neq 0$


                  • $g(x)= sqrt {1-sin^2 x}$ we need $1-sin^2 xge 0$


                  • $h(x)= sqrt {(1-sin^2 x)^2}$ we need $(1-sin^2 x)^2ge 0$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 days ago









                  gimusi

                  86k74294




                  86k74294






















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                      • We have $f(x) = frac{sin(x)}{cos(x)}frac{cos(x)}{sin(x)}$. We have $cos(x) = 0$ if and only if $x in { frac{pi}{2} + kpi mid k in mathbb{Z}}$, and $sin(x) = 0$ if and only if $x in { kpi mid k in mathbb{Z}}$. Hence domain of $f$ is $mathbb{R}setminus { frac{pi}{2} + kpi , kpimid k in mathbb{Z} } = mathbb{R}setminus { frac{kpi}{2} mid k in mathbb{Z} }$.


                      • $h(x)$ is defined as long as $(1-sin^2(x))^2 geq 0$, which always holds. Hence domain of $h$ is $mathbb{R}$.


                      • $g(x)$ is defined as long as $1-sin^2(x) geq 0$. We have for every $x in mathbb{R}$ that $sin(x) in [-1,1]$, hence $sin^2(x) in [0,1]$, hence $1-sin^2(x) in [0,1]$. We thus have that domain of $g$ is $mathbb{R}$.






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                        up vote
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                        down vote














                        • We have $f(x) = frac{sin(x)}{cos(x)}frac{cos(x)}{sin(x)}$. We have $cos(x) = 0$ if and only if $x in { frac{pi}{2} + kpi mid k in mathbb{Z}}$, and $sin(x) = 0$ if and only if $x in { kpi mid k in mathbb{Z}}$. Hence domain of $f$ is $mathbb{R}setminus { frac{pi}{2} + kpi , kpimid k in mathbb{Z} } = mathbb{R}setminus { frac{kpi}{2} mid k in mathbb{Z} }$.


                        • $h(x)$ is defined as long as $(1-sin^2(x))^2 geq 0$, which always holds. Hence domain of $h$ is $mathbb{R}$.


                        • $g(x)$ is defined as long as $1-sin^2(x) geq 0$. We have for every $x in mathbb{R}$ that $sin(x) in [-1,1]$, hence $sin^2(x) in [0,1]$, hence $1-sin^2(x) in [0,1]$. We thus have that domain of $g$ is $mathbb{R}$.






                        share|cite|improve this answer








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                          up vote
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                          up vote
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                          • We have $f(x) = frac{sin(x)}{cos(x)}frac{cos(x)}{sin(x)}$. We have $cos(x) = 0$ if and only if $x in { frac{pi}{2} + kpi mid k in mathbb{Z}}$, and $sin(x) = 0$ if and only if $x in { kpi mid k in mathbb{Z}}$. Hence domain of $f$ is $mathbb{R}setminus { frac{pi}{2} + kpi , kpimid k in mathbb{Z} } = mathbb{R}setminus { frac{kpi}{2} mid k in mathbb{Z} }$.


                          • $h(x)$ is defined as long as $(1-sin^2(x))^2 geq 0$, which always holds. Hence domain of $h$ is $mathbb{R}$.


                          • $g(x)$ is defined as long as $1-sin^2(x) geq 0$. We have for every $x in mathbb{R}$ that $sin(x) in [-1,1]$, hence $sin^2(x) in [0,1]$, hence $1-sin^2(x) in [0,1]$. We thus have that domain of $g$ is $mathbb{R}$.






                          share|cite|improve this answer








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                          • We have $f(x) = frac{sin(x)}{cos(x)}frac{cos(x)}{sin(x)}$. We have $cos(x) = 0$ if and only if $x in { frac{pi}{2} + kpi mid k in mathbb{Z}}$, and $sin(x) = 0$ if and only if $x in { kpi mid k in mathbb{Z}}$. Hence domain of $f$ is $mathbb{R}setminus { frac{pi}{2} + kpi , kpimid k in mathbb{Z} } = mathbb{R}setminus { frac{kpi}{2} mid k in mathbb{Z} }$.


                          • $h(x)$ is defined as long as $(1-sin^2(x))^2 geq 0$, which always holds. Hence domain of $h$ is $mathbb{R}$.


                          • $g(x)$ is defined as long as $1-sin^2(x) geq 0$. We have for every $x in mathbb{R}$ that $sin(x) in [-1,1]$, hence $sin^2(x) in [0,1]$, hence $1-sin^2(x) in [0,1]$. We thus have that domain of $g$ is $mathbb{R}$.







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                          answered 2 days ago









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