Hello ldx2
Welcome to Math Help Forum! Originally Posted by
ldx2 Can somebody solve this and give detailed description ?
I get stuck on this part 1/2 sin18..
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 $\displaystyle \sin^224^o-\sin^26^o = \frac{\sqrt5-1}{8}$
Grandad