SOLUTION: Find sec^-1 ((-2√3)/(3)) exactly in both radians and degrees

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Question 1161418: Find sec^-1 ((-2√3)/(3)) exactly in both radians and degrees
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

sec%5E%28-1%29%28%28-2sqrt%283%29%29%2F%283%29%29%29

If an angle has a secant of

%28-2sqrt%283%29%29%2F3 

then it has a cosine which is its reciprocal

-3%2F%282sqrt%283%29%29 

We rationalize the denominator:

-3%2F%282sqrt%283%29%29%22%22%2A%22%22sqrt%283%29%2Fsqrt%283%29

-3sqrt%283%29%2F%282%2A3%29

-cross%283%29sqrt%283%29%2F%282%2Across%283%29%29

-sqrt%283%29%2F2

Angles that have this for their cosine are special angles that
we can find on the unit circle, in QII and QIII:

In degrees they are 150° and 210°

In radians they are 5π/6 and 7π/6

Edwin