Trouble developing trig expression

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Homework Help Overview

The discussion revolves around proving a trigonometric identity involving the cosine of 36 degrees. The original poster is attempting to manipulate the expression \((\cos(36º) + 1)²/\cos(36º)\) to show that it equals \(5\cos(36º)\) without directly using the right-hand side of the equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss various methods to develop the expression and question how to simplify certain terms. Some suggest plotting the function to investigate its behavior across different values of x, while others reference the geometric significance of the angle involved.

Discussion Status

The conversation is ongoing, with participants offering suggestions for exploration and referencing external resources that may aid in understanding the problem. There is no explicit consensus on the approach, but several lines of reasoning are being examined.

Contextual Notes

Some participants note the importance of considering the angle in both degrees and radians, as well as the potential implications of the identity being true only for specific angles rather than universally.

frkCarl
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Homework Statement


It's not a problem per se, I'm just trying something, so there's no statement. What I'm trying to do it's to prove the forllowing equation but without using the member of the right.

(cos(36º) + 1)²/(cos(36º)) = 5cos(36º)

There's the trivial answer, using both members, that it's multiply both sides by cos(36º), call it x and solve the 2nd degree equation and get the positive result. I'm not trying to do that. I'm trying to develop (cos(36º) + 1)²/(cos(36º)) and turn it into 5cos(36º).

Homework Equations


I'm not sure.

The Attempt at a Solution


For simplicity sake, let 36º = u

(cos(u) + 1)²/(cos(u))
Developing and using sin^2(u) + cos^2(u) = 1 -> [cos^2(u) + 2cos(u) + 1]/cos(u)⇔ 2cos^2(u)+2cos(u)+sin^2(x)/cos(u)

Using the double angle for cosine, cos(2x) = cos^2(u) - sin^2(u):
3cos^2(u)-cos(2u)+2cos(u) / cos(u) ⇔ (3cos^2(u)/cos(u)) + (-cos(2u)+2cos(u))/cos(u)

Here's where I got stuck. I got everything in terms of cosine but the (-cos(2u)+2cos(u))/cos(u)) is giving me some trouble. How do I handle it?
 
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Perhaps try plotting y = (cos(x) + 1)²/(cos(x)) - 5cos(x) and see if it crosses the x axis. If so, is it true for all x or just some values of x ?
Many software packages (including Excel) will treat the argument of cosine() as in radians. So keep that in mind.

If it is true for all x, then yes you can use Trig manipulation to make one expression into the other. If it is only true for some angles, then no you cannot.
 
scottdave said:
Perhaps try plotting y = (cos(x) + 1)²/(cos(x)) - 5cos(x) and see if it crosses the x axis. If so, is it true for all x or just some values of x ?
Many software packages (including Excel) will treat the argument of cosine() as in radians. So keep that in mind.

If it is true for all x, then yes you can use Trig manipulation to make one expression into the other. If it is only true for some angles, then no you cannot.
The equation you wrote above isn't an identity, but it is true for an angle of 36° (or ##\frac \pi 5##).
 
Written a regular decadon into a circle of radius r, it is easy to prove that the side of the decagon is ##a = \frac{1}{2}\left(-1+\sqrt{5}\right)##
Consider the isoceles triangle ABO. The angle at the vertex is 36° and the angles at the base are of 72°. Halve one of the 72°angle, draw the ray, it crosses the side AO at point C. ABO triangle is similar to ABC triangle, so ##\frac {r-a}{a}=\frac{a}{r}##, which results in ##a = \frac{1}{2}\left(-1+\sqrt{5}\right)##
##\cos 36°= \frac{r}{2a} =\frac {1+\sqrt 5 }{4}##

upload_2018-7-14_16-21-53.png
 

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