Question

Just the answer please.
1: http://imgur.com/zZrwi0n
2: http://imgur.com/vJ0OUTg

Answer 1

Answer on Question #58895 – Math – Trigonometry

Question

1. Just the answer please.

\cos \frac {3 \pi}{4} = \underline {{} $$- \frac {\sqrt {2}}{2}$$ $$\frac {1}{2}$$ $$\frac {\sqrt {2}}{2}$$ $$\frac {\sqrt {3}}{2}$$

Solution

$$\cos \left(\frac {3 \pi}{4}\right) = \cos \left(\frac {4 \pi - \pi}{4}\right) = \cos \left(\pi - \frac {\pi}{4}\right) = - \cos \left(\frac {\pi}{4}\right) = - \frac {\sqrt {2}}{2}.$$

Answer: $\cos \left(\frac {3 \pi}{4}\right) = - \frac {\sqrt {2}}{2}$ .

Question

2. Just the answer please

Check all that apply. $\frac{\pi}{6}$ is the reference angle for:

$$\begin{array}{l} \frac{3\pi}{6} \\ \frac{8\pi}{6} \\ \frac{5\pi}{6} \\ \frac{13\pi}{6} \\ \end{array}$$

Solution

It is necessary to subtract $360{}^{\circ}$ ($2\pi$ radians) from the angle greater than $360{}^{\circ}$ ($2\pi$ radians) until it lies between 0 and $360{}^{\circ}$ ($2\pi$ radians).

It is necessary to add $360{}^{\circ}$ ($2\pi$ radians) to the negative angle until it lies between 0 and $360{}^{\circ}$ ($2\pi$ radians). Next step is to define which quadrant the angle is in.

Depending on the quadrant, the reference angle is given in the following table.

Angles $\frac{8\pi}{6}, \frac{5\pi}{6}$ lie in the third and second quadrants respectively. Angle $\frac{13\pi}{6}$ is greater than $2\pi$.

a) $\frac{3\pi}{6} = \frac{\pi}{2}$;

b) $\frac{8\pi}{6} = \frac{4\pi}{3} = \frac{3\pi + \pi}{3} = \pi + \frac{\pi}{3}$;

c) $\frac{5\pi}{6} = \frac{6\pi - \pi}{6} = \pi - \frac{\pi}{6}$;

d) $\frac{13\pi}{6} = \frac{12\pi + \pi}{6} = 2\pi + \frac{\pi}{6}$.

Accordingly, $\frac{\pi}{6}$ is the reference angle for $5\frac{\pi}{6}$ and $13\frac{\pi}{6}$.

**Answer:**

$\frac{\pi}{6}$ is the reference angle for $5\frac{\pi}{6}$ and $13\frac{\pi}{6}$.

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