At first appearance we should apply identities property $\displaystyle \sin^{-1}(\sin \theta)=\theta$ for $\displaystyle -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$\\ Because $\displaystyle \frac{5\pi}{8}$ is not in the interval $\displaystyle \Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]$ we can not apply identities property right away. First we find an angle $\displaystyle \theta$ in $\displaystyle \Big[-\frac{\pi}{2},\frac{\pi}{2}\Big]$ for which $\displaystyle \sin \theta =\sin \frac{5\pi}{8}$. Using illustration below we have that $\displaystyle \sin \frac{5\pi}{8}=\sin \frac{3 \pi}{8}$

Now applying identity property we get $$\displaystyle \sin^{-1}\Big(\sin \frac{5\pi}{8}\Big)=\sin^{-1}\Big(\sin \frac{3\pi}{8}\Big)=\frac{3\pi}{8}$$

a. Let $\displaystyle \theta =\sin ^{-1}\Big(\frac{2}{3}\Big)$ which is equivalent to $\displaystyle \sin \theta =\frac{2}{3}$ when $-\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}$ by definition of inverse of sine. The range $\displaystyle -\frac{\pi}{2}\leq \theta \leq \frac{\pi}{2}$ corresponds to quadrants I and IV, but since $\sin \theta =\frac{2}{3}$ is positive we know that $\theta$ is an angle in quadrant I.
b. Draw angle $\theta$ in quadrant I and label the sides known from the sine value $\displaystyle \sin \theta =\frac{2}{3}=\frac{\text{opposite}}{\text{hypotenuse}}$.

c. Find the unknown side length a by Pythagorean theorem therefore $a=\sqrt{5}$

d. Find $\displaystyle \cos\Big[\sin^{-1}\Big(\frac{2}{3}\Big)\Big]=\cos (\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\sqrt{5}}{3}$

Let $\theta =\tan^{-1}u$ therefore, $\tan\theta=u=\frac{u}{1}$. Notice that $u$ cab be positive or negative. Since the range of the inverse tangent function is $\Big(-\frac{\pi}{2}, \frac{\pi}{2}\Big)$, sketch the angle $\theta$ in both quadrants I and IV and draw the corresponding two right triangles. Recalling that the tangent ratio is opposite over adjacent, we label those corresponding sides with $u$ and 1, respectively. Then solving for the hypotenuse using the Pythagorean theorem gives $\sqrt{u^2+1}$

Substitute $\theta =\tan^{-1}u$ into $\cos(\tan^{-1}u)$ we get $$\displaystyle \cos(\tan^{-1}u)=\cos \theta=\frac{1}{\sqrt{u^2+1}}$$ Rationalize the denominator $$\displaystyle \cos(\tan^{-1}u)=\cos \theta=\frac{1}{\sqrt{u^2+1}}\cdot \frac{\sqrt{u^2+1}}{\sqrt{u^2+1}}=\frac{\sqrt{u^2+1}}{u^2+1}$$

The height of the point begins at the lowest value, 0, increases to the highest value of 28 inches, and continues to oscillate above and below a center height of 14 inches.

First, we write height above ground as a function of the angle of rotation, $\theta$ $$h(\theta)=-14\cos (\theta)+14$$

In this situation, $x$ is representing a linear distance the wheel has traveled, corresponding to an arclength along the circle. Since arclength and angle can be related by $s=r \theta$, in this case we can write $x=14 \theta$, which allows us to express the angle in terms of $x$: $\theta(x)=\frac{x}{14}$
Plugging $\theta(x)=\frac{x}{14}$ into equation above yields, $$h(x)=h(\theta(x))=-14\cos \Big(\frac{x}{14}\Big)+14=-14\cos \Big(\frac{1}{14}x\Big)+14$$ To check we can calculate the period of this function which would be $P=\frac{2\pi}{B}=\frac{2\pi}{\frac{1}{14}}=2\pi \cdot 14=28\pi$ the circumference of the circle.