Question

1. consider the functions ( y=cos x ) and ( y=cos ^{-1} x ).

(a) fill in the table below for the sine function on the restricted interval (0, pi).

(b) using your table in part (a) construct a table for ( y=cos ^{-1} x ).

(c) plot and label the points from both tables on the axes below.

Explanation:

Step1: Calculate $\cos(\frac{\pi}{2})$

We know that the cosine of $\frac{\pi}{2}$ radians is $0$.
$$\cos(\frac{\pi}{2}) = 0$$

Step2: Construct the table for $y = \cos^{-1}(x)$

For the inverse - cosine function $y=\cos^{-1}(x)$, when $x = 0$, since $\cos(\frac{\pi}{2})=0$, then $\cos^{-1}(0)=\frac{\pi}{2}$.

Answer:

(a) The completed table for $y = \cos(x)$:

$x$	$\cos(x)$
$\frac{\pi}{2}$	$0$
$\pi$	$- 1$

(b) The completed table for $y=\cos^{-1}(x)$:

$x$	$\cos^{-1}(x)$
$0$	$\frac{\pi}{2}$
$-1$	$\pi$

(c) For the function $y = \cos(x)$:

• When $x = 0,y = 1$, so the point is $(0,1)$.
• When $x=\frac{\pi}{2},y = 0$, so the point is $(\frac{\pi}{2},0)$.
• When $x=\pi,y=-1$, so the point is $(\pi,-1)$.

For the function $y=\cos^{-1}(x)$:

• When $x = 1,y = 0$, so the point is $(1,0)$.
• When $x = 0,y=\frac{\pi}{2}$, so the point is $(0,\frac{\pi}{2})$.
• When $x=-1,y=\pi$, so the point is $(-1,\pi)$.

To plot these points:

• Mark the point $(0,1)$ for $y = \cos(x)$ on the $y$ - axis.
• Mark the point $(\frac{\pi}{2},0)$ (approximate $\frac{\pi}{2}\approx1.57$ on the $x$ - axis and $y = 0$).
• Mark the point $(\pi,-1)$ (approximate $\pi\approx3.14$ on the $x$ - axis and $y=-1$).
• Mark the point $(1,0)$ on the $x$ - axis for $y=\cos^{-1}(x)$.
• Mark the point $(0,\frac{\pi}{2})$ (on the $y$ - axis, $y\approx1.57$) for $y=\cos^{-1}(x)$.
• Mark the point $(-1,\pi)$ (on the $x=-1$ and $y\approx3.14$) for $y=\cos^{-1}(x)$.