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

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 $\\sin\\left(\\dfrac{\\pi}{3}\\right)$, which is $\\dfrac{\\sqrt{3}}{2}$, you would input sqrt(3)/2

In particular, the angle of $\\var{thetadisplay}$ puts the point {textquadrant}.

Since this point falls on an axis and the point is on the unit circle, it is clear that its coordinates are $(\\var{cos({thetarad})}, \\var{sin({thetarad})})$. From these, we can conclude that

$\\tan\\left(\\var{thetadisplay}\\right)=\\dfrac{\\sin\\left(\\var{thetadisplay}\\right)}{\\cos\\left(\\var{thetadisplay}\\right)}=\\dfrac{\\var{sin({thetarad})}}{\\var{cos({thetarad})}}\\var{if(theta=90 or theta=270, \" which is undefined\",\" = \" + precround(tan(thetarad),0))}.$

The triangle {textquadrant} has the same side lengths as the related triangle in the first quadrant (they are congruent). Therefore, we can recall or use right-angled triangle trigonometry to determine the lengths of the triangle in the first quadrant and then change the signs as needed. Recall:

By drawing the following triangles, we can determine the exact values of $\\sin$, $\\cos$ and $\\tan$ 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

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

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

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

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

From the above, the triangle in the first quadrant tells us that:

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

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

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

Since $\\theta=\\var{thetadisplay}$ puts us {textquadrant}, the $x$-coordinate (the cosine value) is positive,negative, and the $y$-coordinate (the sine value) is positivenegative. That is:

$\\sin\\left(\\var{thetadisplay}\\right)=\\phantom{-}\\sin\\left(\\var{phiraddisplay}\\right)=\\;\\;\\phantom{-}\\dfrac{1}{2}$$\\sin\\left(\\var{thetadisplay}\\right)=\\phantom{-}\\sin\\left(\\var{phiraddisplay}\\right)=\\phantom{-}\\dfrac{1}{\\sqrt{2}}$$\\sin\\left(\\var{thetadisplay}\\right)=\\phantom{-}\\sin\\left(\\var{phiraddisplay}\\right)=\\phantom{-}\\dfrac{\\sqrt{3}}{2}$

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

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

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

$\\tan\\left(\\var{thetadisplay}\\right)=\\phantom{-}\\tan\\left(\\var{phiraddisplay}\\right)=\\phantom{-}\\dfrac{1}{\\sqrt{3}}$$\\tan\\left(\\var{thetadisplay}\\right)=\\phantom{-}\\tan\\left(\\var{phiraddisplay}\\right)=\\phantom{-}\\;\\;1$$\\tan\\left(\\var{thetadisplay}\\right)=\\phantom{-}\\tan\\left(\\var{phiraddisplay}\\right)=\\phantom{-}\\sqrt{3}$

$\\cos\\left(\\var{thetadisplay}\\right)=\\phantom{-}\\cos\\left(\\var{phiraddisplay}\\right)=\\phantom{-}\\dfrac{\\sqrt{3}}{2}$$\\cos\\left(\\var{thetadisplay}\\right)=\\phantom{-}\\cos\\left(\\var{phiraddisplay}\\right)=\\phantom{-}\\dfrac{1}{\\sqrt{2}}$$\\cos\\left(\\var{thetadisplay}\\right)=\\phantom{-}\\cos\\left(\\var{phiraddisplay}\\right)=\\;\\;\\phantom{-}\\dfrac{1}{2}$

An alternative approach is to use the mnemonic \"All Stations To Central\" or \"ASTC\" to recall which trig functions are positive in each quadrant (and hence which are negative):