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testing sin, cos, tan of random(0,90,120,135,150,180,210,225,240,270,300,315,330) degrees but in radians

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Often we prefer to work with exact values rather than approximations from a calculator. In this question we require you input your answer without decimals and without entering the words sin, cos or tan. For example, to input the exact value of $\\sin\\left(\\frac{\\pi}{3}\\right)$, which is $\\dfrac{\\sqrt{3}}{2}$, you would input sqrt(3)/2

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The exact value of $\\sin\\left(\\var{theta[1]}\\right)$ is [[0]].

The exact value of $\\cos\\left(\\var{theta[1]}\\right)$ is [[1]].

Is $\\tan\\left(\\var{theta[1]}\\right)$ defined? [[2]]

The exact value of $\\tan\\left(\\var{theta[1]}\\right)$ is [[3]].

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Yes

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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 $\\dfrac{\\pi}{3}$.

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

	$\\dfrac{\\pi}{6}$	$\\dfrac{\\pi}{4}$	$\\dfrac{\\pi}{3}$

$\\sin$	$\\dfrac{1}{2}$	$\\dfrac{1}{\\sqrt{2}}$	$\\dfrac{\\sqrt{3}}{2}$

$\\cos$	$\\dfrac{\\sqrt{3}}{2}$	$\\dfrac{1}{\\sqrt{2}}$	$\\dfrac{1}{2}$

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

That combined with the unit circle definitions:

$\\cos\\theta$ is the $x$ coordinate of the point on the unit circle $\\theta$ radians anti-clockwise from the positive $x$-axis
$\\sin\\theta$ is the $y$ coordinate of the point on the unit circle $\\theta$ radians anti-clockwise from the positive $x$-axis
$\\tan\\theta$ is the slope of the line through the origin and the point on the unit circle $\\theta$ radians anti-clockwise from the positive $x$-axis, or equivalently, $\\tan\\theta=\\dfrac{\\sin\\theta}{\\cos\\theta}$

and some understanding of congruent triangles:

allows us to work out $\\sin$, $\\cos$ and $\\tan$ for certain angles regardless of what quadrant the point is in. Because whatever angle we are asked about, we can always use the triangle in the first quadrant to determine the side lengths and then consider the signs of the coordinates separately.

For example, to determine $\\sin\\left(\\frac{7\\pi}{6}\\right)$, $\\cos\\left(\\frac{7\\pi}{6}\\right)$ and $\\tan\\left(\\frac{7\\pi}{6}\\right)$ we first draw the following:

From this diagram, we can see that $\\cos\\left(\\frac{7\\pi}{6}\\right)=-\\cos\\left(\\frac{\\pi}{6}\\right)$, and $\\sin\\left(\\frac{7\\pi}{6}\\right)=-\\sin\\left(\\frac{\\pi}{6}\\right)$ since the triangles are congruent and we are in the 3rd quadrant where both the $x$ and $y$ values (and hence the $\\cos$ and $\\sin$ values) are negative.

But given we know these exact values, we can conclude \\[\\cos\\left(\\frac{7\\pi}{6}\\right)=-\\cos\\left(\\frac{\\pi}{6}\\right)=-\\dfrac{\\sqrt{3}}{2},\\] \\[\\sin\\left(\\frac{7\\pi}{6}\\right)=-\\sin\\left(\\frac{\\pi}{6}\\right)=-\\dfrac{1}{2},\\] and finally \\[\\tan\\left(\\frac{7\\pi}{6}\\right)=\\dfrac{\\sin\\left(\\frac{7\\pi}{6}\\right)}{\\cos\\left(\\frac{7\\pi}{6}\\right)}=\\dfrac{-\\frac{1}{2}}{-\\frac{\\sqrt{3}}{2}}=\\dfrac{1}{\\sqrt{3}}.\\]

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