Question

1. find the exact value of each.

a) \\( \cos \left( - \frac { 8 \pi } { 3 } \
ight) \\)
5a)_________________
b) \\( \tan \left( \frac { 15 \pi } { 4 } \
ight) \\)
5b)_________________

1. find the exact value of the given expression. simplify your answer.

\\( \frac { \cot \left( \frac { \pi } { 3 } \
ight) } { \sec \left( \frac { \pi } { 4 } \
ight) + \csc \left( \frac { \pi } { 6 } \
ight) } \\)
6)_________________

1. find the exact coordinates of the points \\( p _ { 1 } \\) and \\( p _ { 2 } \\), given that these points are on

the circumference of a unit circle (a circle of radius one centred at the origin) and:
a) \\( p _ { 1 } \\) is on the terminal ray of the angle \\( \frac { 7 \pi } { 6 } \\)
\\( p _ { 1 } \\)_________________
b) \\( p _ { 2 } \\) is on the terminal ray of the angle \\( \frac { 7 \pi } { 4 } \\)
\\( p _ { 2 } \\)_________________

Explanation:

Step1: Calculate \(\cos(-\frac{8\pi}{3})\)

Use the property \(\cos(-\alpha)=\cos\alpha\), so \(\cos(-\frac{8\pi}{3})=\cos\frac{8\pi}{3}\).
Then \(\cos\frac{8\pi}{3}=\cos(2\pi + \frac{2\pi}{3})\).
Since \(\cos(2k\pi+\alpha)=\cos\alpha\) (\(k\in\mathbb{Z}\)), \(\cos(2\pi+\frac{2\pi}{3})=\cos\frac{2\pi}{3}\).
And \(\cos\frac{2\pi}{3}=-\frac{1}{2}\).

Step2: Calculate \(\tan(\frac{15\pi}{4})\)

\(\tan\frac{15\pi}{4}=\tan(4\pi-\frac{\pi}{4})\).
Use the property \(\tan(2k\pi - \alpha)=-\tan\alpha\) (\(k\in\mathbb{Z}\)), so \(\tan(4\pi-\frac{\pi}{4})=-\tan\frac{\pi}{4}\).
Since \(\tan\frac{\pi}{4} = 1\), then \(\tan\frac{15\pi}{4}=-1\).

Step3: Calculate \(\frac{\cot(\frac{\pi}{3})}{\sec(\frac{\pi}{4})+\csc(\frac{\pi}{6})}\)

We know that \(\cot\frac{\pi}{3}=\frac{\sqrt{3}}{3}\), \(\sec\frac{\pi}{4}=\sqrt{2}\), \(\csc\frac{\pi}{6} = 2\).
Substitute these values into the expression: \(\frac{\frac{\sqrt{3}}{3}}{\sqrt{2}+2}=\frac{\sqrt{3}}{3(\sqrt{2} + 2)}=\frac{\sqrt{3}(2-\sqrt{2})}{3(2 + \sqrt{2})(2-\sqrt{2})}\).
Simplify the denominator \((2+\sqrt{2})(2 - \sqrt{2})=4-2 = 2\).
So the expression becomes \(\frac{\sqrt{3}(2-\sqrt{2})}{6}=\frac{2\sqrt{3}-\sqrt{6}}{6}\).

Step4: Find coordinates of \(P_1\)

For a unit - circle \(x=\cos\alpha\), \(y = \sin\alpha\). When \(\alpha=\frac{7\pi}{6}\), \(\cos\frac{7\pi}{6}=-\frac{\sqrt{3}}{2}\), \(\sin\frac{7\pi}{6}=-\frac{1}{2}\). So \(P_1=(-\frac{\sqrt{3}}{2},-\frac{1}{2})\).

Step5: Find coordinates of \(P_2\)

When \(\alpha=\frac{7\pi}{4}\), \(\cos\frac{7\pi}{4}=\frac{\sqrt{2}}{2}\), \(\sin\frac{7\pi}{4}=-\frac{\sqrt{2}}{2}\). So \(P_2=(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})\).

Answer:

5a) \(-\frac{1}{2}\)
5b) \(-1\)

1. \(\frac{2\sqrt{3}-\sqrt{6}}{6}\)

7a) \(P_1=(-\frac{\sqrt{3}}{2},-\frac{1}{2})\)
7b) \(P_2=(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})\)