8A F11 Q17

Question: Compute the following trig ratios: a) $\sec {\frac {3\pi }{4}}$       b) $\tan {\frac {11\pi }{6}}$       c) $\sin(-120)$

Foundations
1) How is secant related to either sine or cosine?
2) What quadrant is each angle in? What is the reference angle for each?	Answer:
1) $sec(x)={\frac {1}{cos(x)}}$
2) a) Quadrant 2, b) Quadrant 4, c) Quadrant 3. The reference angles are: ${\frac {\pi }{4}},{\frac {\pi }{6}}$, and 60 degrees or ${\frac {\pi }{3}}$

Solution:

Final Answer A:
Since $\sec(x)={\frac {1}{\cos(x)}}$, and the angle is in quadrant 2, $\sec({\frac {3\pi }{4}})={\frac {1}{\cos({\frac {3\pi }{4}})}}={\frac {1}{\frac {-1}{\sqrt {2}}}}=-{\sqrt {2}}$

Final Answer B:
The reference angle is ${\frac {\pi }{6}}$ and is in the fourth quadrant. So tangent will be negative. Since the angle is 30 degees, using the 30-60-90 right triangle, we can conclude that $\tan({\frac {11\pi }{6}})=-{\frac {\sqrt {3}}{3}}$

Final Answer C:
Sin(-120) = - sin(120). So you can either compute sin(120) or sin(-120) = sin(240). Since the reference angle is 60 degrees, or ${\frac {\pi }{3}}$, So $sin(-120)={\frac {\sqrt {3}}{2}}$

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