Chapter 4.3, Problem 72E

Solution

To identify: The If the coordinates of point P are $\left(a,b\right)$ explain why the coordinates of point Q are $\left(-b,a\right)$ .refer to the unit circle in this figure. Point P is on the terminal side of an angle t, and point Q is on the terminal side of an angle $t+\pi /2$ .

  

The coordinates of Q are $\left(-b,a\right)$

Given information

The given refer to the unit circle in this figure. Point P is on the terminal side of an angle t, and point Q is on the terminal side of an angle $t+\pi /2$ .

  

Explanation:

Consider the given information

It is needed to show coordinate of point Q is $\left(-b,a\right)$

Let coordinate of Q is $\left(X,Y\right)$

See the figure above point on the terminal side of angle t is $P\left(a,b\right)$

Since it is a unit circle so distance from P to origin is 1

So, reference triangle is in 1^ST quadrant with sides length 1 a and b

So $x=a,=b,r=1$

See that

  $\begin{array}{l}\cos t=\frac{x}{r}\\ \cos t\frac{a}{1}\\ \cos t=a\\ \sin t=b\end{array}$

See in the triangle QOS

  $\begin{array}{c}\angle QOS=\pi -\left(\frac{\pi }{2}+t\right)\\ =\frac{\pi }{2}-t\end{array}$

Now in the triangle QOS

  $\begin{array}{l}\sin \left(\frac{\pi }{2}-t\right)=\frac{y}{r}\\ \cos t=\frac{Y}{1}\\ a=Y\text{from}\text{equation}\left(\text{1}\right)\\ and\\ \cos \left(\frac{\pi }{2}-t\right)=\frac{x}{r}\\ -\sin t=\frac{X}{1}\\ -b=X\text{from}\text{equation}\left(\text{2}\right)\end{array}$

Hence the coordinates of Q is $\left(-b,a\right)$ .