Chapter 5.2, Problem 3E

To determine

  $\sin t,\cos t$ for values t.

Answer to Problem 3E

For $t=0$ $\boxed{\begin{array}{l}\sin t=0\\ \cos t=1\end{array}}$

For $t=0+\frac{\pi }{4}=\frac{\pi }{4}$ $\boxed{\begin{array}{l}\sin t=\frac{1}{\sqrt{2}}\\ \cos t=\frac{1}{\sqrt{2}}\end{array}}$

For $t=\frac{\pi }{4}+\frac{\pi }{4}=\frac{\pi }{2}$ $\boxed{\begin{array}{l}\sin t=1\\ \cos t=0\end{array}}$

For $t=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}$ $\boxed{\begin{array}{l}\sin \frac{3\pi }{4}=\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}\\ \cos \frac{3\pi }{4}=\cos \frac{\pi }{4}=-\frac{1}{\sqrt{2}}\end{array}}$

For $t=\frac{3\pi }{4}+\frac{\pi }{4}=\pi$ $\boxed{\begin{array}{l}\sin \pi =\sin 0=0\\ \cos \pi =\cos 0=-1\end{array}}$

For $t=\pi +\frac{\pi }{4}=\frac{5\pi }{4}$ $\boxed{\begin{array}{l}\sin \frac{5\pi }{4}=\sin \frac{\pi }{4}=-\frac{1}{\sqrt{2}}\\ \cos \frac{5\pi }{4}=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}\end{array}}$

For $t=\frac{5\pi }{4}+\frac{\pi }{4}=\frac{3\pi }{2}$ $\boxed{\begin{array}{l}\sin \frac{3\pi }{2}=\sin \frac{\pi }{2}=-1\\ \cos \frac{3\pi }{2}=\cos \frac{\pi }{2}=0\end{array}}$

For $t=\frac{3\pi }{2}+\frac{\pi }{4}=\frac{7\pi }{4}$ $\boxed{\begin{array}{l}\sin \frac{7\pi }{4}=\sin \frac{7\pi }{4}=-\frac{1}{\sqrt{2}}\\ \cos \frac{3\pi }{2}=\cos \frac{\pi }{2}=-\frac{1}{\sqrt{2}}\end{array}}$

Explanation of Solution

Given information:

  

Formula used:

Reference angles:

1^st quadrant: ${\theta }_{ref}=\theta$

2^nd quadrant: ${\theta }_{ref}=\pi -\theta$

3^rd quadrant: ${\theta }_{ref}=\theta -\pi$

4^th quadrant: ${\theta }_{ref}=2\pi -\theta$

For $t=0$ $\boxed{\begin{array}{l}\sin t=0\\ \cos t=1\end{array}}$

For $t=0+\frac{\pi }{4}=\frac{\pi }{4}$ $\boxed{\begin{array}{l}\sin t=\frac{1}{\sqrt{2}}\\ \cos t=\frac{1}{\sqrt{2}}\end{array}}$

For $t=\frac{\pi }{4}+\frac{\pi }{4}=\frac{\pi }{2}$

  $\begin{array}{l}\sin t=1\\ \cos t=0\end{array}$

For $t=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}$

As angle lies in 2^nd quadrant

  $\begin{array}{l}{\theta }_{ref}=\pi -\theta \\ {\theta }_{ref}=\pi -\frac{3\pi }{4}\\ {\theta }_{ref}=\frac{4\pi }{4}-\frac{3\pi }{4}\\ {\theta }_{ref}=\frac{\pi }{4}\end{array}$

As $\sin \theta$ is positive and $\cos \theta$ is negative in 2^nd quadrant

  $\boxed{\begin{array}{l}\sin \frac{3\pi }{4}=\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}\\ \cos \frac{3\pi }{4}=\cos \frac{\pi }{4}=-\frac{1}{\sqrt{2}}\end{array}}$

For $t=\frac{3\pi }{4}+\frac{\pi }{4}=\pi$

As angle lies in 2^nd quadrant

  $\begin{array}{l}{\theta }_{ref}=\pi -\theta \\ {\theta }_{ref}=\pi -\pi \\ {\theta }_{ref}=0\end{array}$

As $\sin \theta$ is positive and $\cos \theta$ is negative in 2^nd quadrant

  $\boxed{\begin{array}{l}\sin \pi =\sin 0=0\\ \cos \pi =\cos 0=-1\end{array}}$

For $t=\pi +\frac{\pi }{4}=\frac{5\pi }{4}$

As angle lies in 3^rd quadrant

  $\begin{array}{l}{\theta }_{ref}=\theta -\pi \\ {\theta }_{ref}=\frac{5\pi }{4}-\pi \\ {\theta }_{ref}=\frac{5\pi }{4}-\frac{4\pi }{4}\\ {\theta }_{ref}=\frac{\pi }{4}\end{array}$

As $\sin \theta$ is negative and $\cos \theta$ is positive in 3^rd quadrant

  $\boxed{\begin{array}{l}\sin \frac{5\pi }{4}=\sin \frac{\pi }{4}=-\frac{1}{\sqrt{2}}\\ \cos \frac{5\pi }{4}=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}\end{array}}$

For $t=\frac{5\pi }{4}+\frac{\pi }{4}=\frac{3\pi }{2}$

As angle lies in 3^rd quadrant

  $\begin{array}{l}{\theta }_{ref}=\theta -\pi \\ {\theta }_{ref}=\frac{3\pi }{2}-\pi \\ {\theta }_{ref}=\frac{3\pi }{2}-\frac{2\pi }{2}\\ {\theta }_{ref}=\frac{\pi }{2}\end{array}$

As $\sin \theta$ is negative and $\cos \theta$ is positive in 3^rd quadrant

  $\boxed{\begin{array}{l}\sin \frac{3\pi }{2}=\sin \frac{\pi }{2}=-1\\ \cos \frac{3\pi }{2}=\cos \frac{\pi }{2}=0\end{array}}$

For $t=\frac{3\pi }{2}+\frac{\pi }{4}=\frac{7\pi }{4}$

As angle lies in 4^th quadrant

  $\begin{array}{l}{\theta }_{ref}=2\pi -\theta \\ {\theta }_{ref}=2\pi -\frac{7\pi }{4}\\ {\theta }_{ref}=\frac{8\pi }{4}-\frac{7\pi }{4}\\ {\theta }_{ref}=\frac{\pi }{4}\end{array}$

As $\sin \theta$ is negative and $\cos \theta$ is negative in 4^th quadrant

  $\boxed{\begin{array}{l}\sin \frac{7\pi }{4}=\sin \frac{7\pi }{4}=-\frac{1}{\sqrt{2}}\\ \cos \frac{3\pi }{2}=\cos \frac{\pi }{2}=-\frac{1}{\sqrt{2}}\end{array}}$