34.) a group of students decided to study the sinusoidal nature of tides. values for the depths of the water…

34.) a group of students decided to study the sinusoidal nature of tides. values for the depths of the water level were recorded at various times. at t = 2 hours low tide was recorded at a depth of 1.8 m. at t = 8 hours, high tide was recorded at a depth of 3.6 m. the function can be modeled by the equation d(t)=0.9 cos(π/6(t - 8))+2.7. what is the depth of the water at t = 18 hours? 35.) find the exact value of each of the following. (a) sec(210°) (b) csc(150°) (c) cot(3π/4)

34.) a group of students decided to study the sinusoidal nature of tides. values for the depths of the water level were recorded at various times. at t = 2 hours low tide was recorded at a depth of 1.8 m. at t = 8 hours, high tide was recorded at a depth of 3.6 m. the function can be modeled by the equation d(t)=0.9 cos(π/6(t - 8))+2.7. what is the depth of the water at t = 18 hours? 35.) find the exact value of each of the following. (a) sec(210°) (b) csc(150°) (c) cot(3π/4)

Answer

Explanation:

Step1: Substitute t value

Substitute $t = 18$ into $D(t)=0.9\cos(\frac{\pi}{6}(t - 8))+2.7$. $D(18)=0.9\cos(\frac{\pi}{6}(18 - 8))+2.7$

Step2: Simplify the argument of cosine

Simplify $\frac{\pi}{6}(18 - 8)$. $\frac{\pi}{6}(18 - 8)=\frac{\pi}{6}\times10=\frac{5\pi}{3}$ So $D(18)=0.9\cos(\frac{5\pi}{3})+2.7$

Step3: Find cosine value

We know that $\cos(\frac{5\pi}{3})=\cos(2\pi-\frac{\pi}{3})=\cos(\frac{\pi}{3})=\frac{1}{2}$ So $D(18)=0.9\times\frac{1}{2}+2.7$

Step4: Calculate final result

$0.9\times\frac{1}{2}+2.7 = 0.45+2.7=3.15$

Answer:

$3.15$ m