Proving sin18 = √5-1/4 without Calculator: Double-Angle Formulae Method

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The discussion focuses on proving that sin(18°) equals (√5 - 1)/4 using trigonometric identities without a calculator. Participants explore the application of double-angle formulas and transformations, particularly expressing sin(18°) in terms of cos(4a). The attempt involves manipulating the equation into a quartic form, leading to a challenge in further simplification. A hint suggests utilizing the rational factor theorem to find rational factors for solving the quartic equation. The conversation emphasizes the complexity of the proof and the need for strategic algebraic manipulation.
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HELP! trigonometry

Homework Statement


without using calculator:
prove that: sin18 = √5-1/4

Homework Equations


Double-angle formulae
sin 2a = 2sin a.cos a
cos 2a = cos^2a - sin^2a = 1 - 2sin^2a = 2cos^2a - 1
sin^2 a + cos^2 a=1

The Attempt at a Solution


let 18=a
sin a = sin(90 - a) = cos 4a
cos 4a = 1 - 2sin^2 2a
= 1 - 2(2sin a.cos a)^2
= 1 - 2(4sin^2 a.cos^2 a)
= 1 - 8sin^2 a.cos^2 a
= 1 - 8sin^2a(1-sin^2a)
= 1 - 8sin^2a + 8sin^4a
i can't go further than that ?!
 
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So now what you have is:

sina=1-8sin^2a+8sin^4a

Hint: To solve this quartic, search for rational factors with the rational factor theorem.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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