Coming out with clean values for arccos, arcsine and arctan

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The discussion focuses on the concept of "clean" values for inverse trigonometric functions like arccos, arcsine, and arctan. It highlights that angles with tidy sine and cosine values typically correspond to multiples of 30° and 45°, derived from specific right triangles. For example, arccos(1/√2) can be expressed as π/4 due to its relation to these angles. However, not all angles yield clean expressions; arccos(3/5) is noted as a case where a simple fractional representation does not exist, leading to the need for decimal approximations. Ultimately, the conversation emphasizes the connection between trigonometric ratios and the simplicity of angle measures.
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Coming out with "clean" values for arccos, arcsine and arctan

Homework Statement


Not so much a homework problem just something that bugs me and it's taken until first year Uni math before I asked it.

In the text or in lectures when taking the arccos of say, 1/root2, the lecturer won't write 0.7853..., but rather a very nice looking pi/4.

How does s/he know this? It seems like they know a "nice" way of writing every arc function value.

At the moment I need arccos(3/5), what about this one, is there a nice way of writing this? Instead of 0.927...

Cheers.
 
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Really it's all about playing around with the unit circle (http://en.wikipedia.org/wiki/Unit_circle). The circle contains some simple angles (30,45,60, and 90 degrees, along with 90 degree shifts of all of these angles all of the way around the circle) represented in radians and the value of sine and cosine at those angles. You can derive the values of sine and cosine by considering the 45-45-90 and 30-60-90 right triangle, and looking at the trigonometric properties that follow.

For other angles, the answer may not always be "clean", though you can often write the solution as a sum of "clean" values by using some basic trigonometric identities (search around, they're all over the web).

-------------------------
theUndergrad

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To elaborate a bit on theUndergrad's reply, the angles that have the tidiest sines, cosines, etc. are those which are multiples of 30º and 45º. The right triangle with two 45º angles is a right isoceles triangle, which can be assigned sides of 1, 1, and sqrt(2), so the trig functions for 45º are built out of the possible ratios of those numbers. The 30º-60º-90º triangle is a bisected equilateral triangle; if we take its sides to be 2 , then the sides of our triangle will be 1, sqrt(3), and 2 , so all the trig functions for 30º and 60º are built out of all the possible ratios of this set of numbers.

The inverse trig functions then give the simple divisions of pi radians (pi/6, pi/4, and their multiples) for the ratios involving 1, sqrt(2), sqrt(3), and 2. Other even subdivisions of pi will be related to somewhat more complicated ratios through the application of various trigonometric identities.

So it is unfortunately the case (the ancients -- particularly the Pythagoreans -- would have been horrified!) that the simplest angles are connected with ratios involving irrational numbers and, by contrast, that simple ratios of integers are connected through the trig functions to not-at-all-simple angles. Your particular inverse trig function relates to the 3-4-5 triangle, which does not have angles which are simple fractions of 180º or pi radians, so values like arccos(3/5) don't have "clean" expressions, and instead we have to live with decimal approximations.
 


dynamicsolo said:
values like arccos(3/5) don't have "clean" expressions,
IMHO, arccos(3/5) is already a "clean" expression. :smile:
 
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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