# Trigonometric identity

• Dec 14th 2009, 02:09 AM
ldx2
Trigonometric identity
Can somebody solve this and give detailed description ?
I get stuck on this part 1/2 sin18..
http://img207.imageshack.us/img207/1634/firstz.jpg
• Dec 14th 2009, 07:22 AM
sin 18 degrees
Hello ldx2

Welcome to Math Help Forum!
Quote:

Originally Posted by ldx2
Can somebody solve this and give detailed description ?
I get stuck on this part 1/2 sin18..
http://img207.imageshack.us/img207/1634/firstz.jpg

I agree: $\displaystyle \sin^224^o-\sin^26^o = \tfrac12\sin 18^o$.

There are various ways of deriving $\displaystyle \sin18^o$. Here's one. And here's my version:
$\displaystyle \sin 18^o = \cos 72^o = 2\cos^236^o-1=2(1-2\sin^218)^2-1$
So if we put $\displaystyle x = \sin18^o, x$ is the root of the equation:
$\displaystyle x = 2(1-2x^2)^2-1$
for which $\displaystyle 0<x<1$.
$\displaystyle \Rightarrow x=2(1-4x^2+4x^4)-1$

$\displaystyle \Rightarrow 8x^4-8x^2-x+1=0$
Noting that $\displaystyle x = 1$ satisfies this equation, we can factorise:
$\displaystyle (x-1)(8x^3+8x^2-1)=0$
and then discard the factor $\displaystyle (x-1)$, since $\displaystyle 0<x<1$.

We then note that $\displaystyle x = -\tfrac12$ is also a root. Hence $\displaystyle (2x+1)$ is a factor, which again we discard for the same reason; leaving us with the remaining factor:
$\displaystyle 4x^2+2x-1=0$
the positive root of which is:
$\displaystyle x = \frac{\sqrt5-1}{4}=\sin18^o$
Hence $\displaystyle \sin^224^o-\sin^26^o = \frac{\sqrt5-1}{8}$

• Dec 14th 2009, 07:49 AM
ldx2
Thanks for the quick reply and solving it. EDIT: I understood until the part where you say x=-1/2 is also a root and how you found out (2x+1) is also a factor. Could you explain that part in more detail ?
• Dec 15th 2009, 12:50 PM
By inspection (which is a posh word for 'trial and error'), when $\displaystyle x = -\tfrac12, 8x^3+8x^2-1= -1+2-1=0$. So, using the remainder theorem, $\displaystyle (2x+1)$ is a factor.