Exact Value of sec(5pi/6) | Simple Trig Problem Solution

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The exact value of sec(5π/6) can be determined by recognizing that 5π/6 radians corresponds to 150 degrees. This angle is in the second quadrant, where cosine values are negative, specifically cos(150°) = -√3/2. Therefore, sec(5π/6) is the reciprocal of this cosine value, yielding sec(5π/6) = -2/√3. To find exact trigonometric values, familiarity with standard angles and their sine and cosine values is essential. Understanding the unit circle and reference angles can also aid in solving similar problems efficiently.
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Well, I haven't figured out how to do the scripts yet so Ill just type in this problem as best I can.

Give the exact value: sec(5(pie)/6)

I realize that 5(pie)/6 radians is 150*. So 1/cos(150) in my calculator will give me the approx answer. I just don't know how to get the exact.

*edit* ok let me give this thing a go.

\sec\frac{5\pi}{6}
 
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For sin(a), we have: sin(a) = sin(pi-a) so sin(5pi/6) = sin(pi-5pi/6) = sin(pi/6).
Do you understand why? Symmetry wrt y-axis. Now, pi/6 (or 30°) is a standard angle.
 
I don't understand any of what you said. =\
 
Your calculator will be more or less useless for the exact answer. What would really help is to memorize a table of sin, cos, and tan values for the angles 30, 45, and 60 degrees (as well as 0, 90) etc.

But, since you are working in radian measure, it will be helpful to become familiar with many of the "common" radian measures found in textbooks - multiples of pi/6 and multiples of pi/4 all the way around the unit circle to 2Pi.

The question you're facing is one similar to "find the exact value of the tangent of 60 degrees", except you are in a different quadrant.

Another method, albeit slower, is to draw right triangles, one with two 45 degree angles, and label the legs both with lengths of 1. Pythagorean's theorem gives the length of the hypotenuse, and simple trig ratios (opposite over hypotenuse, etc.) will give you the trig values of 45 degrees. For 30 degrees or 60 degrees, draw an equilateral triangle, with sides = 2 units. Draw the altitude (perpendicular bisector) - this gives you two 30-60-90 degree right triangles, with the hypotenuse = 2 units and the shorter leg = 1 unit. Again, use the pythagorean theorem to find the 3rd side and simple trig ratios to find the trig values.

The 45-45-90 or 30-60-90 triangles can be drawn on a set of coordinate axis with the hypotenuse co-terminal with your angle. Allowing for negative values for the sides of your triangles (in the negative x-direction or negative y-direction), you can simply pick off the trig value needed.

But, memorizing those common angles - 0, 30, 45, 60, and 90 degrees is far quicker and easier.

edit: hopefully, you're aware that the value of sin(190 degree) is negative the value of the sin of 10 degrees. i.e. the reference angle for 190 degrees is 10 degrees, and it's located in the 3rd quadrant where the value of sin is negative.
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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