Science:Math Exam Resources/Courses/MATH110/December 2013/Question 05 (a)/Solution 1

We evaluate these values by following three steps:

1. Reduce the parameter to be in the interval $[0,2\pi )$.
2. Apply the relevant special triangle, in this case
   
3. Check the sign of the result by remembering the $\cos(t)$ gives the $x$-coordinate and $\sin(t)$ gives the $y$-coordinate.

• $\sin \left({\frac {5\pi }{3}}\right)$.

  1. The angle ${\frac {5\pi }{3}}$ is in the desired interval.

  2. The angle ${\frac {5\pi }{3}}$ is ${\frac {\pi }{3}}$ below the $x$-axis. Applying the special triangle, we get $\sin \left({\frac {5\pi }{3}}\right)=\pm {\frac {\sqrt {3}}{2}}$.

  3. We are in the fourth quadrant so the $y$-coordinate is negative.

  We have: $\sin \left({\frac {5\pi }{3}}\right)=-{\frac {\sqrt {3}}{2}}$.

• $\cos \left(-{\frac {\pi }{3}}\right)$

  1. The angle $-{\frac {\pi }{3}}$ is not in the desired interval. So we apply periodicity:

     ${\begin{aligned}\cos \left(-{\frac {\pi }{3}}\right)&=\cos \left(-{\frac {\pi }{3}}+2\pi \right)\\&=\cos \left(-{\frac {\pi }{3}}+{\frac {6\pi }{3}}\right)\\&=\cos \left({\frac {5\pi }{3}}\right)\end{aligned}}$

  2. The angle ${\frac {5\pi }{3}}$ is ${\frac {\pi }{3}}$ below the $x$-axis. Applying the special triangle, we get $\cos \left({\frac {5\pi }{3}}\right)=\pm {\frac {1}{2}}$.

  3. We are in the fourth quadrant so the $x$-coordinate is positive.

  We have: $\cos \left(-{\frac {\pi }{3}}\right)={\frac {1}{2}}.$

• $\tan \left({\frac {11\pi }{3}}\right)$.

  1. The angle ${\frac {11\pi }{3}}$ is not in the desired interval. So we apply periodicity:

     ${\begin{aligned}\tan \left({\frac {11\pi }{3}}\right)&=\tan \left({\frac {11\pi }{3}}-2\pi \right)\\&=\tan \left({\frac {11\pi }{3}}-{\frac {6\pi }{3}}\right)\\&=\tan \left({\frac {5\pi }{3}}\right)\end{aligned}}$

  2. The angle ${\frac {5\pi }{3}}$ is ${\frac {\pi }{3}}$ below the $x$-axis. Applying the special triangle, we get $\tan \left({\frac {5\pi }{3}}\right)=\pm {\sqrt {3}}$.

  3. We are in the fourth quadrant and since $\tan(t)={\frac {\sin(t)}{\cos(t)}}$, we get that $\tan(t)$ is negative.

  Alternative, we can make use of the previous parts to get:

  $\tan \left({\frac {11\pi }{3}}\right)=\tan \left({\frac {5\pi }{3}}\right)={\frac {\sin \left({\frac {11\pi }{3}}\right)}{\cos \left({\frac {11\pi }{3}}\right)}}={\frac {\frac {-{\sqrt {3}}}{2}}{\frac {1}{2}}}$

  Either way, we have: $\tan \left({\frac {11\pi }{3}}\right)=-{\sqrt {3}}$