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multiple choice testing sin, cos, tan of random(pi/6, pi/4, pi/3) radians
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Often we prefer to work with exact values rather than approximations from a calculator.
", "advice": "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 $\\dfrac{\\pi}{3}$.
\nSince we are asked about $\\var{distheta}$ radians, we use the triangle on the leftright and the mnemonic SOH CAH TOA to determine:
\n$\\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}$
\nAlternatively, one can memorise the following table:
\n| \n | $\\dfrac{\\pi}{6}$ | \n$\\dfrac{\\pi}{4}$ | \n$\\dfrac{\\pi}{3}$ | \n
| \n | \n | \n | \n |
| $\\sin$ | \n$\\dfrac{1}{2}$ | \n$\\dfrac{1}{\\sqrt{2}}$ | \n$\\dfrac{\\sqrt{3}}{2}$ | \n
| \n | \n | \n | \n |
| $\\cos$ | \n$\\dfrac{\\sqrt{3}}{2}$ | \n$\\dfrac{1}{\\sqrt{2}}$ | \n$\\dfrac{1}{2}$ | \n
| \n | \n | \n | \n |
| $\\tan$ | \n$\\dfrac{1}{\\sqrt{3}}$ | \n$1$ | \n$\\sqrt{3}$ | \n
Since we are asked about $\\var{distheta}$ radians, we use the $\\var{distheta}$ column of the table to determine that:
\n$\\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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