A periodic function passes through the origin and then descends to a minimum of \( -\frac{2}{3} \) when \( x=\frac{\pi}{2} \). It passes through the x-axis again when \( x=\pi \) and rises to a maximum of \( \frac{2}{3} \) when periodic function? (1 point) \( y=-\frac{2}{3} \cos x \) \( y=\frac{2}{3} \sin x \) \( y=\frac{4}{3} \sin x \) \( y=-\frac{2}{3} \sin x \)

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

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

\( \tan \theta=-\frac{3}{4} \), given that \( \theta \) is in quadrant IV

Дан треугольник \( A B C \), в котором \( \angle A+\angle B=90^{\circ} \), а \( \sin B=\frac{2 \sqrt{5}}{10 \sqrt{10}} \) Найди \( \cos ^{2} B \).

Question 36, 3.4.7 Part 1 of 3 Suppose that point \( P \) is on a circle with radius \( r \), and ray \( O P \) is rotating with angular speed \( \omega \). Complete parts (a) through (c). \( \mathrm{r}=8 \mathrm{~cm}, \omega=\frac{\pi}{8} \) radian per sec, \( t=3 \mathrm{sec} \)