Trigonometry

1. a) Multiply by $\dfrac{\pi}{180°}$:

$150° \times \frac{\pi}{180°} = \frac{150\pi}{180} = \frac{5\pi}{6} \text{ radians}$

b) Multiply by $\dfrac{180°}{\pi}$:

$\frac{5\pi}{4} \times \frac{180°}{\pi} = \frac{5 \times 180°}{4} = \frac{900°}{4} = 225°$

c) The arc length formula is $s = r\theta$, where $\theta$ is in radians:

$6 = 4\theta \implies \theta = \frac{6}{4} = \frac{3}{2} \text{ radians}$

Convert to degrees:

$\frac{3}{2} \times \frac{180°}{\pi} = \frac{270°}{\pi} \approx 85.94°$

2. a) $\dfrac{2\pi}{3}$ is in Quadrant II ($120°$). The reference angle is $\pi - \frac{2\pi}{3} = \frac{\pi}{3}$.

$\sin\left(\frac{2\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

(Sine is positive in Quadrant II.)

b) $\dfrac{7\pi}{4}$ is in Quadrant IV ($315°$). The reference angle is $2\pi - \frac{7\pi}{4} = \frac{\pi}{4}$.

$\cos\left(\frac{7\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$

(Cosine is positive in Quadrant IV.)

c) $\dfrac{5\pi}{6}$ is in Quadrant II ($150°$). The reference angle is $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$.

$\tan\left(\frac{5\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$

(Tangent is negative in Quadrant II.)

d) $\dfrac{3\pi}{2}$ corresponds to the point $(0, -1)$ on the unit circle.

$\sin\left(\frac{3\pi}{2}\right) = -1$

3. a) The rider starts at the lowest point, which is $40 - 30 = 10$ feet above the ground. The general form is:

$h(t) = -A\cos\left(\frac{2\pi}{T} t\right) + D$

where $A = 30$ (radius), $T = 60$ (period in seconds), and $D = 40$ (center height).

The negative cosine is used because the rider starts at the bottom ($\cos(0) = 1$, so $-\cos(0) = -1$, giving the minimum).

$h(t) = -30\cos\left(\frac{2\pi}{60} t\right) + 40 = -30\cos\left(\frac{\pi}{30} t\right) + 40$

Verification: $h(0) = -30(1) + 40 = 10$ feet (lowest point) $\checkmark$; $h(30) = -30(-1) + 40 = 70$ feet (highest point) $\checkmark$.

b) At $t = 15$:

$h(15) = -30\cos\left(\frac{\pi}{30} \cdot 15\right) + 40 = -30\cos\left(\frac{\pi}{2}\right) + 40 = -30(0) + 40 = 40 \text{ feet}$

The rider is at the center height — level with the axle of the wheel.

c) Set $h(t) = 55$:

$-30\cos\left(\frac{\pi}{30} t\right) + 40 = 55$

$-30\cos\left(\frac{\pi}{30} t\right) = 15$

$\cos\left(\frac{\pi}{30} t\right) = -\frac{1}{2}$

The cosine equals $-\frac{1}{2}$ at $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ (in $[0, 2\pi]$):

$\frac{\pi}{30} t = \frac{2\pi}{3} \implies t = 20 \text{ seconds}$

$\frac{\pi}{30} t = \frac{4\pi}{3} \implies t = 40 \text{ seconds}$

The rider is at $55$ feet at $t = 20$ seconds and $t = 40$ seconds during the first revolution.

4. a) July corresponds to $m = 7$:

$T(7) = 25\sin\left(\frac{\pi}{6}(7 - 4)\right) + 55 = 25\sin\left(\frac{3\pi}{6}\right) + 55 = 25\sin\left(\frac{\pi}{2}\right) + 55 = 25(1) + 55 = 80°\text{F}$

b) The function $T(m) = 25\sin\left(\frac{\pi}{6}(m-4)\right) + 55$ has amplitude $25$ and midline $55$.

• Maximum: $55 + 25 = 80°$F, when $\sin\left(\frac{\pi}{6}(m-4)\right) = 1$, i.e., $\frac{\pi}{6}(m-4) = \frac{\pi}{2}$, so $m - 4 = 3$, meaning $m = 7$ (July).

• Minimum: $55 - 25 = 30°$F, when $\sin\left(\frac{\pi}{6}(m-4)\right) = -1$, i.e., $\frac{\pi}{6}(m-4) = \frac{3\pi}{2}$, so $m - 4 = 9$, meaning $m = 13$, which wraps to $m = 1$ (January).

c) The period is $\frac{2\pi}{\pi/6} = 12$ months. This means the temperature pattern repeats every 12 months — i.e., annually — which makes physical sense for seasonal temperature variation.

5. Write the function in the standard form $f(x) = A\cos(B(x - C)) + D$:

$f(x) = -3\cos\left(2\left(x - \frac{\pi}{6}\right)\right) + 1$

Reading off the parameters:

• Amplitude: $|A| = |-3| = 3$
• Period: $\dfrac{2\pi}{|B|} = \dfrac{2\pi}{2} = \pi$
• Phase shift: $C = \dfrac{\pi}{6}$ to the right
• Vertical shift: $D = 1$ (upward)

Transformations from $y = \cos(x)$:

1. Horizontal compression by a factor of $\frac{1}{2}$ (period changes from $2\pi$ to $\pi$)
2. Reflection across the $x$-axis (the negative sign on $A$)
3. Vertical stretch by a factor of $3$ (amplitude changes from $1$ to $3$)
4. Horizontal shift $\frac{\pi}{6}$ units to the right
5. Vertical shift $1$ unit upward