Given the angle \( \theta \) that belongs to Quadrant I. Then the signs of \( \sec (-\theta) \) and \( \csc (\theta) \) are:

1) \( 8 \sin \left(\frac{\pi}{2}-x\right)-1=4 \sin ^{2} x \) 2) \( 1-3 \cos ^{2} x-\sin x \cos x \geq 0 \) 3) \( (\sqrt{3} \sin x-\cos x-1) \cdot(2 \cos (2 x)+1) \geq 0 \) 4) \( \frac{\sin x}{\cos \left(\frac{x}{2}\right)-\frac{1}{2}} \geq 0 \)

Si \( f(x)=\frac{3}{2}+\cos x \) para \( x \in(0,2 \pi) \) entonces la función \( f(x) \) es creciente en: a. \( (0, \pi / 6) \cup(5 \pi / 6,2 \pi) \) b. \( (5 \pi / 6,2 \pi) \) c. \( (0, \pi / 6) \) d. \( (\pi / 6,5 \pi / 6) \) Retroalimentación:

4. Verify the identity: \( \sec \theta-\cos \theta=\sin \theta \tan \theta \) \( \sec =\tan +1 \)

Given \( \cos (\theta)=-\frac{2}{3} \) and \( \theta \) belongs to Quadrant II. Then the values of \( \sec (\theta) \) and \( \sin (\theta) \) are:

Use the power-reducing formulas to rewrite the expression in terms of first powers of the cosines of multiple angles. \[ 5 \cos ^{4}(x) \] \( \frac{15}{8}+\frac{5}{2} \cos (2 x)+\frac{5}{6} \cos (4 x) \quad \) Nice work.