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 4,$t$ increases in increments of $\frac{\pi }{6}$.

Short answer

All the terminal point is $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$.

Step-by-step solution

Given information.

The figure is given below as,

Explanation.

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

$\begin{array}{c}\left(\sin t,\cos t\right)=\left(\sin \frac{\pi }{6},\cos \frac{\pi }{6}\right)\\ =\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)\end{array}$

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

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

role="math" localid="1648579691260" $\begin{array}{c}t=\frac{2\pi }{6}\\ \left(\sin \left(\frac{2\pi }{6}\right),\cos \left(\frac{2\pi }{6}\right)\right)=\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)\end{array}$

And,

role="math" localid="1648579728138" $\begin{array}{c}t=\frac{3\pi }{6}\\ \left(\sin \left(\frac{3\pi }{6}\right),\cos \left(\frac{3\pi }{6}\right)\right)=\left(1,0\right)\end{array}$

And,

role="math" localid="1648579800412" $\begin{array}{c}t=\frac{2\pi }{3}\\ \left(\sin \left(\frac{2\pi }{3}\right),\cos \left(\frac{2\pi }{3}\right)\right)=\left(\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\end{array}$

And,

$\begin{array}{c}t=\frac{5\pi }{6}\\ \left(\sin \left(\frac{5\pi }{6}\right),\cos \left(\frac{5\pi }{6}\right)\right)=\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)\end{array}$

And,

role="math" localid="1648579878011" $\begin{array}{c}t=\pi \\ \left(\sin \left(\pi \right),\cos \left(\pi \right)\right)=\left(0,-1\right)\end{array}$

And,

role="math" localid="1648579909385" $\begin{array}{c}t=\frac{7\pi }{6}\\ \left(\sin \left(\frac{7\pi }{6}\right),\cos \left(\frac{7\pi }{6}\right)\right)=\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)\end{array}$

And,

$\begin{array}{c}t=\frac{4\pi }{3}\\ \left(\sin \left(\frac{4\pi }{3}\right),\cos \left(\frac{4\pi }{3}\right)\right)=\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)\end{array}$

And,

role="math" localid="1648579953488" $\begin{array}{c}t=\frac{3\pi }{2}\\ \left(\sin \left(\frac{3\pi }{2}\right),\cos \left(\frac{3\pi }{2}\right)\right)=\left(-1,0\right)\end{array}$

And,

$\begin{array}{c}t=\frac{5\pi }{3}\\ \left(\sin \left(\frac{5\pi }{3}\right),\cos \left(\frac{5\pi }{3}\right)\right)=\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right)\end{array}$

And,

$\begin{array}{c}t=\frac{11\pi }{6}\\ \left(\sin \left(\frac{11\pi }{6}\right),\cos \left(\frac{11\pi }{6}\right)\right)=\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)\end{array}$

And,

role="math" localid="1648580040995" $\begin{array}{c}t=2\pi \\ \left(\sin \left(2\pi \right),\cos \left(2\pi \right)\right)=\left(0,1\right)\end{array}$