The question asks which of the given trigonometric expressions is equivalent to $\frac{\sqrt{3}}{2}$. The productivity problem asks for what values of $t$ the productivity $5\cos(\frac{\pi}{2}t)+5$ is least, given that $t=0$ corresponds to 8:00 a.m.

1. Problem Description

The question asks which of the given trigonometric expressions is equivalent to $\frac{\sqrt{3}}{2}$.

The productivity problem asks for what values of $t$ the productivity $5\cos(\frac{\pi}{2}t)+5$ is least, given that $t=0$ corresponds to 8:00 a.m.

2. Solution Steps

Question 14:

We evaluate each of the options:

* $\sin(\frac{5\pi}{6}) = \sin(\pi - \frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2}$

* $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$

* $\sin(\frac{\pi}{6}) = \frac{1}{2}$

* $\cos(\frac{5\pi}{6}) = \cos(\pi - \frac{\pi}{6}) = -\cos(\frac{\pi}{6}) = -\frac{\sqrt{3}}{2}$

Therefore, the correct answer is $\cos(\frac{\pi}{6})$.

Question 15:

The productivity function is $5\cos(\frac{\pi}{2}t)+5$. To minimize this function, we need to minimize $\cos(\frac{\pi}{2}t)$. The minimum value of $\cos(x)$ is -

1. This occurs when $x = (2n+1)\pi$ for integer $n$.

Thus, we want $\frac{\pi}{2}t = (2n+1)\pi$, which implies $t = 2(2n+1) = 4n+2$.

For $n=0$, $t=2$. This corresponds to 8 a.m. + 2 hours = 10 a.m.

For $n=1$, $t=6$. This corresponds to 8 a.m. + 6 hours = 2 p.m.

For $n=2$, $t=10$. This corresponds to 8 a.m. + 10 hours = 6 p.m.

For $n=3$, $t=14$. This corresponds to 8 a.m. + 14 hours = 10 p.m.

However, the function describes productivity on a scale of 0 to 10, hence we need to consider $t$ for which $\frac{\pi}{2}t = \pi, 3\pi, 5\pi,...$

So we require that $t = 2, 6, 10$

For $t = 2$, $5\cos(\frac{\pi}{2}2) + 5 = 5\cos(\pi) + 5 = 5(-1) + 5 = 0$.

For $t=2$, the time is $8+2 = 10$ a.m.

$\cos(\frac{\pi}{2}t)$ achieves minimum value -1 when $\frac{\pi}{2}t = \pi + 2\pi k$, $k \in Z$ or when $\frac{\pi}{2}t = 3\pi + 2\pi k$.

$t = 2 + 4k$

$t = 2$, 10 a.m.

$t = 6$, 2 p.m.

$t = 10$, 6 p.m.

If we consider productivity to be zero when the work is least productive, $5\cos(\frac{\pi}{2} t)+5=0 \implies \cos(\frac{\pi}{2}t) = -1 \implies \frac{\pi}{2}t = (2n+1)\pi \implies t=2(2n+1)$. Then for t=0 to t=10, we have $t=2,6,10$. t=2 is 10 a.m, t=6 is 2 p.m, t=10 is 6 p.m.

The cosine function varies between -1 and 1, so the minimum value is

0. The worker is least productive when the function is at a minimum.

For the case where $t=2$, this corresponds to $8 + 2 = 10$ a.m.

If $t=6$, it is $8 + 6 = 14$, which corresponds to 2 p.m.

For the case where $\frac{\pi}{2}t = 5\pi/2$, $t = 5$. The time is $8+5 = 13$, which is 1 p.m.

The options are 12 noon, 10 a.m, 12 noon, and 2 p.m., 10 a.m. and 2 p.m., 11 a.m. and 3 p.m. The answer is 10 a.m. and 2 p.m.

3. Final Answer

Question 14: $\cos(\frac{\pi}{6})$

Question 15: 10 a.m. and 2 p.m.