Question

Trigonometric Identities and Applications? help

Answer 1

$\bf h=-15cos\left( \frac{2\pi }{5}t \right)\qquad \boxed{h=8}\qquad 8=-15cos\left( \frac{2\pi }{5}t \right) \\\\\\ \cfrac{8}{-15}=cos\left( \frac{2\pi }{5}t \right)\implies cos^{-1}\left( \frac{8}{-15}\right)=cos^{-1}\left[ cos\left( \frac{2\pi }{5}t \right) \right] \\\\\\ cos^{-1}\left( \frac{8}{-15}\right)=\cfrac{2\pi }{5}t\implies \cfrac{5}{2\pi }\cdot cos^{-1}\left( \frac{8}{-15}\right)=t\\\\ -------------------------------$

$\bf cos^{-1}\left( \frac{8}{-15}\right)=\stackrel{radians}{\stackrel{II~quadrant}{\approx 2.13}~~,~~\stackrel{III~quadrant}{\approx 4.15}}\qquad \qquad t= \begin{cases} \frac{5}{2\pi }\cdot 2.13\\\\ \frac{5}{2\pi }\cdot 4.15 \end{cases}$