Solution:

Part a) Convert $135°$ to radians

Multiply by $\frac{\pi}{180}$:

$135° \times \frac{\pi}{180} = \frac{135\pi}{180} = \frac{3\pi}{4} \text{ radians}$

Part b) Convert $\frac{5\pi}{6}$ radians to degrees

Multiply by $\frac{180}{\pi}$:

$\frac{5\pi}{6} \times \frac{180}{\pi} = \frac{5 \times 180}{6} = \frac{900}{6} = 150°$

Answers:

• a) $\frac{3\pi}{4}$ radians
• b) $150°$

Solution:

Part (a): To convert degrees to radians, multiply by $\frac{\pi}{180}$:

$225° \cdot \frac{\pi}{180} = \frac{225\pi}{180} = \frac{5\pi}{4}$ radians

Part (b): To convert radians to degrees, multiply by $\frac{180}{\pi}$:

$\frac{5\pi}{6} \cdot \frac{180}{\pi} = \frac{5 \cdot 180}{6} = \frac{900}{6} = 150°$

Part (c): $\frac{5\pi}{6}$ is in Quadrant II (between $\frac{\pi}{2}$ and $\pi$).

Reference angle: $\pi - \frac{5\pi}{6} = \frac{6\pi - 5\pi}{6} = \frac{\pi}{6}$

In Quadrant II: sine is positive, cosine and tangent are negative.

Using reference angle $\frac{\pi}{6}$:

$\sin\frac{5\pi}{6} = +\sin\frac{\pi}{6} = \frac{1}{2}$

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

$\tan\frac{5\pi}{6} = \frac{\sin\frac{5\pi}{6}}{\cos\frac{5\pi}{6}} = \frac{1/2}{-\sqrt{3}/2} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}$

Solution:

Part a) $\sin\left(\frac{7\pi}{6}\right)$

Step 1: Determine the quadrant. $\frac{7\pi}{6}$ is between $\pi$ and $\frac{3\pi}{2}$, so it's in Quadrant III.

Step 2: Find the reference angle. $\frac{7\pi}{6} - \pi = \frac{7\pi}{6} - \frac{6\pi}{6} = \frac{\pi}{6}$

Step 3: Determine the sign. In Quadrant III, sine is negative.

Step 4: Evaluate. $\sin\left(\frac{7\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}$

Part b) $\cos\left(\frac{5\pi}{4}\right)$

Quadrant III, reference angle: $\frac{5\pi}{4} - \pi = \frac{\pi}{4}$

Cosine is negative in Quadrant III.

$\cos\left(\frac{5\pi}{4}\right) = -\cos\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}$

Part c) $\tan\left(\frac{5\pi}{3}\right)$

Quadrant IV, reference angle: $2\pi - \frac{5\pi}{3} = \frac{\pi}{3}$

Tangent is negative in Quadrant IV.

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

Answers:

• a) $-\frac{1}{2}$
• b) $-\frac{\sqrt{2}}{2}$
• c) $-\sqrt{3}$

Solution:

Part (a): The $x$-coordinate is negative and the $y$-coordinate is positive.

This means $P$ is in Quadrant II.

Part (b): On the unit circle, the coordinates of a point are $(\cos\theta, \sin\theta)$.

Therefore:

• $\cos\theta = -\frac{3}{5}$
• $\sin\theta = \frac{4}{5}$
• $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{4/5}{-3/5} = -\frac{4}{3}$

Part (c): Reciprocal functions:

$\sec\theta = \frac{1}{\cos\theta} = \frac{1}{-3/5} = -\frac{5}{3}$

$\csc\theta = \frac{1}{\sin\theta} = \frac{1}{4/5} = \frac{5}{4}$

Solution:

Given:

• Radius: $r = 8$ cm
• Central angle: $\theta = \frac{2\pi}{3}$ radians

Arc Length:

Using $s = r\theta$:

$s = 8 \cdot \frac{2\pi}{3} = \frac{16\pi}{3} \text{ cm}$

$s \approx 16.76 \text{ cm}$

Sector Area:

Using $A = \frac{1}{2}r^2\theta$:

$A = \frac{1}{2}(8)^2 \cdot \frac{2\pi}{3}$

$A = \frac{1}{2} \cdot 64 \cdot \frac{2\pi}{3}$

$A = 32 \cdot \frac{2\pi}{3} = \frac{64\pi}{3} \text{ cm}^2$

$A \approx 67.02 \text{ cm}^2$

Answers:

• Arc length: $\frac{16\pi}{3} \approx 16.76$ cm
• Sector area: $\frac{64\pi}{3} \approx 67.02$ cm²