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

[Radwa Kamal]

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 ?!

 

[Mentallic]

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.

 

[Architeuthis Dux]

Great job so far! You are on the right track. To continue, we can use the double-angle formula for sine, which is: sin 2a = 2sin a.cos a. This will allow us to express sin a in terms of 2a.

sin a = sin(90 - a) = cos 4a
sin a = 2sin 2a.cos 2a = 2(2sin a.cos a)(1 - 2sin^2 a)
sin a = 4sin a.cos a - 8sin^3 a.cos a

Now, let's substitute this into our previous expression for cos 4a:

cos 4a = 1 - 8sin^2 a + 8sin^4 a
cos 4a = 1 - 8sin^2 a + 8sin^3 a - 8sin^4 a

We can now use the Pythagorean identity, sin^2 a + cos^2 a = 1, to rewrite sin^2 a in terms of cos^2 a:

cos 4a = 1 - 8sin^2 a + 8sin^3 a - 8(1 - cos^2 a)^2
cos 4a = 1 - 8sin^2 a + 8sin^3 a - 8(1 - 2cos^2 a + cos^4 a)
cos 4a = 1 - 8sin^2 a + 8sin^3 a - 8 + 16cos^2 a - 8cos^4 a

Now, we can rearrange this equation to solve for cos^4 a:

8cos^4 a = 1 - 8sin^2 a + 8sin^3 a - 16cos^2 a + 8
cos^4 a = (1 - 8sin^2 a + 8sin^3 a - 16cos^2 a + 8)/8

Finally, we can substitute this expression for cos^4 a into our previous expression for sin a:

sin a = 4sin a.cos a - 8sin^3 a.cos a
sin a = 4sin a.cos a - 8sin^3 a(1 - 16cos^2 a + 8)/8
sin a = 4sin