Chapter 4.3, Problem 74E

Solution

To identify: The why cos $\cos \left(t+\frac{\pi }{2}\right)=-\sin t$ .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 $\cos \left(t+\frac{\pi }{2}\right)=-\sin t$

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

  $\cos \left(t+\frac{\pi }{2}\right)=-\sin t$

First evaluate $\cos t$

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

  $\begin{array}{l}\sin t=\frac{y}{r}\\ \sin t\frac{b}{1}\mathrm{...}\left(1\right)\\ \sin t=b\end{array}$

Now evaluate $\sin \left(t+\frac{\pi }{2}\right)$

For this determine reference triangle

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

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

So reference triangle is in 2 quadrant with sides length 1 a and b

  $x=-b,y=a,r=1$

See that

  $\begin{array}{l}\sin \left(t+\frac{\pi }{2}\right)=\frac{x}{r}\\ \sin \left(t+\frac{\pi }{2}\right)=\frac{-b}{1}\\ \sin \left(t+\frac{\pi }{2}\right)=-b\end{array}$

Hence equation (1)

  $\cos \left(t+\frac{\pi }{2}\right)=-\sin t$