Rewrite the cosine function \( f(x)=-3 \cos \left(x-\frac{\pi}{8}\right)+2 \) as a translated sine function. \( \begin{array}{l}\text { (1 point) } \\ f(x)=-3 \sin \left(x+\frac{\pi}{4}\right)+2 \\ f(x)=-3 \sin \left(x+\frac{\pi}{24}\right)+2 \\ f(x)=-3 \sin \left(x+\frac{3 \pi}{8}\right)+2 \\ f(x)=-3 \sin \left(x-\frac{5 \pi}{8}\right)+2\end{array} \)

\( \tan \theta=\frac{8}{15} \), given that \( \theta \) is in quadrant III

\( 2 \cos ^ { 2 } x - 1 \geq 0 \)

\( \theta=\frac{2 \pi}{7} \) radians, \( \omega=\frac{2 \pi}{3} \) radian per minute Use formula \( \omega=\frac{\theta}{t} \) to find the value of the missing variable.

\( 1 - \tan ^ { 2 } x \geq 0 \)

As part of her assignment, Devi has to use her clinometer to measure the height of an office tower \( T A \). From a point \( B \), she observes that angle \( T B A=34^{\circ} \). She then walks 25 m away from the office tower to a point \( C \) and observes that angle \( T C A=26^{\circ} \). Find the height of the sfice tover.