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a) The sine and cosine functions accept any real number input. Regardless of the value of $\\simplify{{inp}}$, the function $\\simplify{{out}}$ will output a number. That is, the domain of $\\simplify{{out}}$ is the set of all real numbers. We can write this as

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\$\\text{dom}(\\simplify{{out}})=\\mathbb{R}\$

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or as the open interval

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\$\\text{dom}(\\simplify{{out}})=(-\\infty,\\infty).\$

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b) Recall that $\\tan(x)=\\frac{\\sin(x)}{\\cos(x)}$ by definition and so is not defined when $\\cos(x)=0$ since division by zero is undefined. These $x$ values correspond to asymptotes. There are an infinite number of asymptotes to $y=\\tan(x)$, equally spaced with $\\pi$ radians between each adjacent asymptote. The asymptotes closest to the origin are $x=-\\dfrac{\\pi}{2}$ and $x=\\dfrac{\\pi}{2}$. We can write the set of asymptotes as $\\left\\{\\dfrac{\\pi}{2}+k\\pi:\\, k=\\ldots,-1,0,1,\\ldots\\right\\}$, or $\\left\\{\\dfrac{\\pi}{2}+k\\pi:\\, k\\in\\mathbb{Z}\\right\\}$ (where $\\mathbb{Z}$ is the set of integers). This means the domain of $\\tan(x)$ would be all the real numbers except for the asymptotes, which we can write as $\\mathbb{R}\\setminus \\left\\{\\dfrac{\\pi}{2}+k\\pi:\\, k\\in\\mathbb{Z}\\right\\}$, or $\\mathbb{R}\\setminus \\left\\{\\dfrac{\\pi}{2}+k\\pi:\\, k=\\ldots,-1,0,1,\\ldots\\right\\}$, or $\\left\\{x\\in\\mathbb{R}:\\,x\\ne \\frac{\\pi}{2}+k\\pi \\text{ for } k\\in \\mathbb{Z}\\right\\}$, or $\\left\\{x\\in\\mathbb{R}:\\,x\\ne \\frac{\\pi}{2}+k\\pi \\text{ for } k\\ldots,-1,0,1,\\ldots\\right\\}$.

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c) The tan function doesn't accept inputs of $\\dfrac{\\pi}{2}+k\\pi$ for $k\\in\\mathbb{Z}$. Since in the function {hardtan}, tan acts on $\\simplify[fractionNumbers]{{b2}{inp}+{c2}pi}$, we require that

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\$\\simplify[fractionNumbers]{{b2}{inp}+{c2}pi}\\ne \\dfrac{\\pi}{2}+k\\pi \\text{ for } k\\in\\mathbb{Z}.\$

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To get a better description of the domain we rearrange for $\\simplify{{inp}}$:

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\\begin{align}&&\\simplify[fractionNumbers]{{b2}{inp}+{c2}pi}&\\ne \\dfrac{1}{2}\\pi+k\\pi \\text{ for } k\\in\\mathbb{Z}\\\\ &\\implies&\\simplify[fractionNumbers]{{b2}{inp}}&\\ne \\simplify[simplifyFractions,fractionNumbers]{{1/2-c2}pi}+k\\pi \\text{ for } k\\in\\mathbb{Z}\\\\ &\\implies&\\simplify{{inp}}&\\ne \\simplify[fractionNumbers,simplifyFractions]{{(1/2-c2)/b2}pi+{1/{b2}}pi*k} \\text{ for } k\\in\\mathbb{Z} \\\\&\\implies&\\simplify{{inp}}&\\ne \\left(\\simplify[fractionNumbers,simplifyFractions]{{(1/2-c2)/b2}}+\\simplify[fractionNumbers,simplifyFractions]{{1/b2}}k\\right)\\pi \\text{ for } k\\in\\mathbb{Z} \\end{align}

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and so we can write the domain as $\\left\\{\\simplify{{inp}}\\in\\mathbb{R}:\\,\\simplify{{inp}}\\ne\\left(\\simplify[fractionNumbers,simplifyFractions]{{(1/2-c2)/b2}}+\\simplify[fractionNumbers,simplifyFractions]{{1/b2}}k\\right)\\pi \\text{ for } k\\in\\mathbb{Z} \\right\\}$.

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d) Recall that $\\sec(x)=\\dfrac{1}{\\cos(x)}$ by definition and so $\\sec(x)$ is not defined when $\\cos(x)=0$ since division by zero is undefined. Note this is the same condition that was on $\\tan(x)$ and so $\\sec(x)$ and $\\tan(x)$ have the same domain. In particular, for {seccsc}:

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\\begin{align}\\text{dom}(\\simplify{{out2}})&=\\left\\{\\simplify{{inp2}}\\in\\mathbb{R}:\\,\\simplify{cos({inp2})}\\ne 0\\right\\}\\\\&=\\left\\{\\simplify{{inp2}}\\in\\mathbb{R}:\\,\\simplify{{inp2}}\\ne \\frac{\\pi}{2}+k\\pi \\text{ for } k\\in \\mathbb{Z}\\right\\}\\end{align}

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d) Note, $\\csc$ is also denoted $\\text{cosec}$ by some people. Recall that $\\csc(x)=\\dfrac{1}{\\sin(x)}$ by definition and so $\\csc(x)$ is not defined when $\\sin(x)=0$ since division by zero is undefined. In particular, for {seccsc}:

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\\begin{align}\\text{dom}(\\simplify{{out2}})&=\\left\\{\\simplify{{inp2}}\\in\\mathbb{R}:\\,\\simplify{sin({inp2})}\\ne 0\\right\\}\\\\&=\\left\\{\\simplify{{inp2}}\\in\\mathbb{R}:\\,\\simplify{{inp2}}\\ne k\\pi \\text{ for } k\\in \\mathbb{Z}\\right\\}\\end{align}

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e) Recall that $\\cot(\\theta)=\\dfrac{\\cos(\\theta)}{\\sin(\\theta)}$ by definition and so $\\cot(\\theta)$ is not defined when $\\sin(\\theta)=0$ since division by zero is undefined. In particular, for $f(\\theta)=\\simplify[fractionNumbers]{{b2}cot(theta)+{a2}}$:

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\\begin{align}\\text{dom}(f)&=\\left\\{\\theta\\in\\mathbb{R}:\\,\\sin(\\theta)\\ne 0\\right\\}\\\\&=\\left\\{\\theta\\in\\mathbb{R}:\\,\\theta\\ne k\\pi \\text{ for } k\\in \\mathbb{Z}\\right\\}\\end{align}

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Multiple choice questions. Given randomised trig functions select the possible ways of writing the domain of the function.

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expression(random(['pi/3','pi/2','pi/4','pi/6','-pi/3','-pi/2','-pi/4','-pi/6']))

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Which of the following represents the domain of {sincos}?

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$\\mathbb{R}$

", "

$(-\\infty,\\infty)$

", "

$[\\var{d},\\infty)$

", "

$[\\var{a},\\infty)$

", "

$\\left[\\simplify{{-c}/{d}},\\infty\\right)$

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$\\{\\simplify{{inp}}\\in\\mathbb{R}:\\,\\simplify{{b}<={inp}<={c}}\\}$

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Which of the following represents the domain of {easytan}?

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$\\mathbb{R}$

", "

$\\left\\{\\simplify{{inp2}}\\in\\mathbb{R}:\\,\\simplify{{inp2}}\\ne \\frac{\\pi}{2}+k\\pi \\text{ for } k\\in \\mathbb{Z}\\right\\}$

", "

$\\left\\{\\simplify{{inp2}}\\in\\mathbb{R}:\\,\\simplify{{inp2}}\\ne \\frac{\\pi}{2}+k\\pi \\text{ for } k=\\ldots, -1,0,1,\\ldots\\right\\}$

", "

$\\mathbb{R}\\setminus\\left\\{\\frac{\\pi}{2}+k\\pi:\\, k\\in \\mathbb{Z}\\right\\}$

", "

$\\left\\{\\simplify{{inp2}}\\in\\mathbb{R}:\\,\\simplify{{inp2}}\\ne \\pm\\frac{\\pi}{2}\\right\\}$

", "

$\\{\\simplify{{inp2}}\\in\\mathbb{R}:\\,\\simplify{{inp2}>={d2}}\\}$

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$\\mathbb{R}\\setminus\\left\\{k\\pi:\\, k\\in \\mathbb{Z}\\right\\}$

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Given the real functions below (which are defined in terms of radians), you should be able to determine their domain.

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