Question

Find $\sin t$and $\cos t$ for the values of$t$ whose terminal points are shown on the unit circle in the figure. In Exercise 3, $t$ increases in increments of $\frac{\pi }{4}$.

Short answer

All the terminal point is $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$.

Step-by-step solution

Given information.

The figure is given below as,

Explanation.

Now, find the value of $\sin t,\cos t$.

role="math" localid="1648578384745" $$\begin{gathered} \left(\sin t,\cos t\right)=\left(\sin \frac{\pi }{4},\cos \frac{\pi }{4}\right) \\ =\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right) \end{gathered}$$

As, t increase of increment of $\frac{\pi }{4}$.

Now, find the all value of terminal point’s increase of increment.

role="math" localid="1648578457536" $\begin{array}{c}t=\frac{\pi }{2}\\ \left(\sin t,\cos t\right)=\left(1,0\right)\end{array}$

And,

role="math" localid="1648578493243" $\begin{array}{c}t=\frac{3\pi }{4}\\ \left(\sin \left(\frac{3\pi }{4}\right),\cos \left(\frac{3\pi }{4}\right)\right)=\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)\end{array}$

And,

role="math" localid="1648578590479" $\begin{array}{c}t=\pi \\ \left(\sin \left(\pi \right),\cos \left(\pi \right)\right)=\left(0,-1\right)\end{array}$

And,

role="math" localid="1648578614093" $\begin{array}{c}t=\frac{5\pi }{4}\\ \left(\sin \left(\frac{5\pi }{4}\right),\cos \left(\frac{5\pi }{4}\right)\right)=\left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)\end{array}$

And,

role="math" localid="1648578646741" $\begin{array}{c}t=\frac{3\pi }{2}\\ \left(\sin \left(\frac{3\pi }{2}\right),\cos \left(\frac{3\pi }{2}\right)\right)=\left(-1,0\right)\end{array}$

And,

role="math" localid="1648578672814" $\begin{array}{c}t=\frac{7\pi }{4}\\ \left(\sin \left(\frac{7\pi }{4}\right),\cos \left(\frac{7\pi }{4}\right)\right)=\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\end{array}$

And,

role="math" localid="1648578705691" $\begin{array}{c}t=2\pi \\ \left(\sin \left(2\pi \right),\cos \left(2\pi \right)\right)=\left(0,1\right)\end{array}$