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

   

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

• $\cos (-\frac{\pi }{3})$

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

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

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

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

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

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

  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: $\tan \left(\frac{11\pi }{3}\right)=-\sqrt{3}$