Chapter 4.3, Problem 75E

Solution

To identify: The why $\sin \left(t+\frac{\pi }{2}\right)=\cos t$ .refer t is an angle with radian measure $0<t<\frac{\pi }{2}$ Draw a similar figure for an angle with radian measure $\frac{\pi }{2}<t<\pi$

  

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

Given information

The given is t is an angle such that $\frac{\pi }{2}<t<\pi$

And P is a point on the terminal side of angle $t+\frac{\pi }{2}$

  

Explanation:

Consider the given information

It is needed to show

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

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

  $\begin{array}{l}\cos t=\frac{x}{r}\\ \cos t=\frac{-a}{1}\mathrm{...}\left(1\right)\\ \cos t=-a\end{array}$

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

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

Hence equation (1)

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