MHB Evaluate Trig Expressions....Part 2

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The discussion focuses on evaluating trigonometric expressions, specifically cos 4π/3 and sec 4π/3, using a method outlined in the textbook. For cos 4π/3, the reference angle is calculated as π/6, leading to a value of -1/2, while the evaluation of sec 4π/3 also results in a reference angle of π/6, yielding -2√3/3. The participants note that both angles are in Quadrant 3, where cosine values are negative. There is a preference expressed for using algebraic methods over graphing for finding reference angles. Overall, the discussion emphasizes the importance of understanding reference angles in trigonometric evaluations.
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Evaluate the trig expressions using the method shown in the textbook. Steps A through C show the method given in the textbook.

1. cos 4π/3

A. We are told to graph cos 4π/3. We are in Quadrant 3.

B. Find the reference number r.

r = 3π/2 - 4π/3

r = π/6

C. Evaluate r.

cos π/6 = -sqrt{3}/2

Book's answer for r is -1/2.

2. sec 4π/3

A. We are told to graph sec 4π/3.
We are in Quadrant 3.

B. Find the reference number r.

r = 3π/2 - 4π/3

r = π/6

C. Evaluate r.

sec π/6 = -2sqrt{3}/3.

Book's answer for r is -2.
 
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$\dfrac{4\pi}{3}$ is in quad III $\implies$ cosine is negative

reference angle is $\dfrac{4\pi}{3} - \pi = \dfrac{\pi}{3} \implies \cos\left(\dfrac{4\pi}{3}\right) = -\cos\left(\dfrac{\pi}{3}\right) = - \dfrac{1}{2}$
 
I will practice more on finding reference angles using the algebraic method you provided in Part 1, I believe. It's a very easy concept but using the graph can be a bit tricky. I prefer using algebra over graphing any time.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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