testing sin, cos, tan of angles that are negative or greater than 2pi radians that result in nice exact values.

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

In particular, the angle of $\\displaystyle\\var[fractionnumbers]{omega_pi}\\pi$ is a complete revolution ($2\\pi$) and then an additional $\\displaystyle\\var{thetadisplay}$ anticlockwiseis two complete revolutions ($2\\times 2\\pi=4\\pi$) and then an additional $\\displaystyle\\var{thetadisplay}$ anticlockwiseis $\\displaystyle\\var[fractionnumbers]{abs(omega_pi)}\\pi$ clockwise is a complete revolution ($2\\pi$) clockwise and then an additional $\\displaystyle\\var[fractionnumbers]{abs(omega_pi+2)}\\pi$ clockwise is two complete revolutions ($2\\times 2\\pi=4\\pi$) clockwise and then an additional $\\displaystyle\\var[fractionnumbers]{abs(omega_pi+4)}\\pi$ clockwiseis three complete revolutions ($3\\times 2\\pi$) clockwise which puts the point {textquadrant}.

Therefore, for trigonometry, working with $\\displaystyle\\var[fractionnumbers]{omega_pi}\\pi$ would be the same as working with $\\displaystyle\\var{thetadisplay}$.

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

$\\displaystyle\\cos\\left(\\var{thetadisplay}\\right)=\\var{cos({thetarad})}$, and

$\\displaystyle\\tan\\left(\\var{thetadisplay}\\right)=\\dfrac{\\displaystyle\\sin\\left(\\var{thetadisplay}\\right)}{\\displaystyle\\cos\\left(\\var{thetadisplay}\\right)}=\\dfrac{\\var{sin({thetarad})}}{\\var{cos({thetarad})}}\\var{if(theta=90 or theta=270, \" which is undefined\",\" = \" + precround(tan(theta*pi/180),0))}.$

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

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

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

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

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

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

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