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multiple choice testing sin, cos, tan of  random(pi/6, pi/4, pi/3) radians

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Often we prefer to work with exact values rather than approximations from a calculator.

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By drawing the following triangles we can determine the exact values of $\\sin$, $\\cos$ and $\\tan$ (and their reciprocals $\\csc$, $\\sec$, $\\cot$) for the angles $\\dfrac{\\pi}{6}$, $\\dfrac{\\pi}{4}$ and $\\dfrac{\\pi}{3}$.

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Since we are asked about $\\var{distheta}$ radians, we use the triangle on the leftright and the mnemonic SOH CAH TOA to determine:

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$\\sin\\left(\\var{distheta}\\right)=\\dfrac{\\text{Opposite}}{\\text{Hypotenuse}}=\\;\\;\\dfrac{1}{2}$$\\sin\\left(\\var{distheta}\\right)=\\dfrac{\\text{Opposite}}{\\text{Hypotenuse}}=\\dfrac{1}{\\sqrt{2}}$$\\sin\\left(\\var{distheta}\\right)=\\dfrac{\\text{Opposite}}{\\text{Hypotenuse}}=\\dfrac{\\sqrt{3}}{2}$

\n

$\\cos\\left(\\var{distheta}\\right)=\\dfrac{\\text{Adjacent}}{\\text{Hypotenuse}}=\\dfrac{\\sqrt{3}}{2}$$\\cos\\left(\\var{distheta}\\right)=\\dfrac{\\text{Adjacent}}{\\text{Hypotenuse}}=\\dfrac{1}{\\sqrt{2}}$$\\cos\\left(\\var{distheta}\\right)=\\dfrac{\\text{Adjacent}}{\\text{Hypotenuse}}=\\;\\;\\dfrac{1}{2}$

\n

$\\tan\\left(\\var{distheta}\\right)=\\;\\;\\dfrac{\\text{Opposite}}{\\text{Adjacent}}\\;\\;=\\dfrac{1}{\\sqrt{3}}$$\\tan\\left(\\var{distheta}\\right)=\\;\\;\\dfrac{\\text{Opposite}}{\\text{Adjacent}}\\;\\;=\\;\\;1$$\\tan\\left(\\var{distheta}\\right)=\\;\\;\\dfrac{\\text{Opposite}}{\\text{Adjacent}}\\;\\;=\\sqrt{3}$

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Alternatively, one can memorise the following table: 

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\dfrac{\\pi}{6}$$\\dfrac{\\pi}{4}$$\\dfrac{\\pi}{3}$
 
$\\sin$$\\dfrac{1}{2}$$\\dfrac{1}{\\sqrt{2}}$$\\dfrac{\\sqrt{3}}{2}$
 
$\\cos$$\\dfrac{\\sqrt{3}}{2}$$\\dfrac{1}{\\sqrt{2}}$$\\dfrac{1}{2}$
 
$\\tan$$\\dfrac{1}{\\sqrt{3}}$$1$$\\sqrt{3}$
\n

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Since we are asked about $\\var{distheta}$ radians, we use the $\\var{distheta}$ column of the table to determine that:

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$\\sin\\left(\\var{distheta}\\right)=\\;\\;\\dfrac{1}{2}$$\\sin\\left(\\var{distheta}\\right)=\\dfrac{1}{\\sqrt{2}}$$\\sin\\left(\\var{distheta}\\right)=\\dfrac{\\sqrt{3}}{2}$

\n

$\\cos\\left(\\var{distheta}\\right)=\\dfrac{\\sqrt{3}}{2}$$\\cos\\left(\\var{distheta}\\right)=\\dfrac{1}{\\sqrt{2}}$$\\cos\\left(\\var{distheta}\\right)=\\;\\;\\dfrac{1}{2}$

\n

$\\tan\\left(\\var{distheta}\\right)=\\dfrac{1}{\\sqrt{3}}$$\\tan\\left(\\var{distheta}\\right)=\\;\\;1$$\\tan\\left(\\var{distheta}\\right)=\\sqrt{3}$

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The exact value of $\\sin\\Large($$\\var{distheta}\\Large)$ is

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$\\dfrac{1}{\\sqrt{2}}$

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$\\dfrac{\\sqrt{3}}{2}$

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$\\dfrac{1}{\\sqrt{3}}$

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$\\sqrt{3}$

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$1$

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The exact value of $\\cos\\Large($$\\var{distheta}\\Large)$ is

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$\\dfrac{1}{2}$

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$\\dfrac{1}{\\sqrt{2}}$

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$\\dfrac{\\sqrt{3}}{2}$

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$\\dfrac{1}{\\sqrt{3}}$

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$\\sqrt{3}$

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$1$

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The exact value of $\\tan\\Large($$\\var{distheta}\\Large)$ is

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$\\dfrac{1}{\\sqrt{2}}$

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$\\dfrac{\\sqrt{3}}{2}$

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$\\dfrac{1}{\\sqrt{3}}$

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$\\sqrt{3}$

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$1$

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