Exact value for $\cos 36°$ Good morning! I'm having trouble with this problem... It's just taking me forever and I'm worn out and I'm lost on how to use a double angle identity for $72=2⋅36$
The problem reads as follows
An exact value for $\cos36°$ can be found using the following procedure. Begin by considering $\sin108°$. Note that $108=72+36$ and use the sine sum identity. Also note that $72=2⋅36$ and use double angle identities. If there are any common factors in each term, factor them out and cancel them if they are not equal to zero. You should eventually obtain a quadratic equation containing $\cos36°$. Use the quadratic formula to obtain the exact value for $\cos36°$. (Note that the quadratic formula should give two solutions. One can be disregarded - why?
Thank you in advance to anyone who can help. 
 A: \begin{align}
\sin 108^\circ&=\sin (72^\circ+36^\circ)\\
\sin(180^\circ-72^\circ)&=\sin 72^\circ\cos36^\circ+\cos 72^\circ\sin36^\circ\\
\sin72^\circ&=\sin (2\cdot36^\circ)\cos36^\circ+\cos (2\cdot36^\circ)\sin36^\circ\\
\sin (2\cdot36^\circ)&=2\sin36^\circ\cos36^\circ\cos36^\circ+(2\cos^236^\circ-1)\sin36^\circ\\
2\sin36^\circ\cos36^\circ&=\sin36^\circ(2\cos^236^\circ+2\cos^236^\circ-1)\\
2\cos36^\circ&=4\cos^236^\circ-1\\
4\cos^236^\circ-2\cos36^\circ-1&=0\tag1
\end{align}
Let $y=\cos36^\circ$, then $(1)$ becomes
$$
4y^2-2y-1=0\tag2
$$
Use complete square method or quadratic formula to obtain the roots of $(2)$ and take the positive value only because $\cos36^\circ>0$ since it lies on the first quadrant.
A: Hint: We have 
$$\sin(108^\circ)=\sin(72^\circ)=2\sin(36^\circ)\cos(36^\circ). \tag{1}$$
Also, 
$$\sin(108^\circ)=\sin(36^\circ +72^\circ)=\sin(36^\circ)\cos(72^\circ)+\cos(36^\circ)\sin(72^\circ). \tag{2}$$
As suggested in the OP, use double-angle identities on the right-hand side of (2).
A: Awesome proof : 

This is a regular pentagon. As every angle is $108^\circ$, $\angle CAD=36^\circ$ from symmetry.
Triangles $ABP$ and $AEP$ are similar.
$\frac{BE}{AB}=\frac{AE}{EP}$
$BE×EP=AB^2$
$(BP+EP)×EP=AB^2$
$(AB+EP)×EP=AB^2$
For the ratio $x=\frac{AB}{EP}$ we have the equation
$x+1=x^2$
with one positive solution $x=\Phi$, the 
golden ratio.
In $△AEP$, $AE=AB$ and $EP$ is one of the sides such that $\frac{AE}{EP}=\Phi$. Drop a perpendicular from $P$ to $AE$ to obtain two right triangles. Then say,
$cos(∠AEP)=(AE/2)/EP=(AE/EP)/2=ϕ/2$
But $∠AEP=36^\circ$ and we get the desired result.
Hence the answer is $$\frac{1+\sqrt 5}{4}$$ 
A: "Good Evening"
Use the following trig identities
s(x) = s(180 - x)
s(x + y) = s(x)c(y) + s(y)c(x) and in particular s(2x) = 2s(x)c(x)
c(2x) = $c^2(x) - s^2(x)$
when you get near the end note that $nc^2 - s^2 = (n+1)c^2 - 1$
It works out quite easily.
For the final part note the allowed range of the cosine function.
A: Using the identities $\sin 2t=2\sin t \cos t$, $\cos 2t=2\cos^t-1$ and $\sin t=\sin(180^{\circ}-t)$
\begin{align}
\sin(108^{\circ}) & =\sin(72^{\circ})\cos(36^{\circ})+\cos(72^{\circ})\sin(36^{\circ}) \\
\sin(72^{\circ})& =2\sin(36^{\circ})\cos^2(36^{\circ})+[2\cos^2(36^{\circ})-1]\sin(36^{\circ}) \\
2\sin(36^{\circ})\cos(36^{\circ})& =2\sin(36^{\circ})\cos^2(36^{\circ})+[2\cos^2(36^{\circ})-1]\sin(36^{\circ})\\
\end{align}
Note $\sin (36^{\circ})\neq 0$, so we can multiplicate by $\frac{1}{\sin 36^{\circ}}$, and we get
\begin{align}
2\cos(36^{\circ})& =2\cos^2(36^{\circ})+2\cos^2(36^{\circ})-1\\
4\cos^2(36^{\circ})-2\cos(36^{\circ})-1&=0 \;\;\quad\color{blue}{\text{(1)}}
\end{align}
Using the quadratic formula we get $\cos (36^{\circ})\in\{\frac{1+\sqrt{5}}{4},\frac{1-\sqrt{5}}{4}\}$. Since $\cos (36^{\circ})>0$ we conclude $\cos (36^{\circ})=\frac{1+\sqrt{5}}{4}$.
A: Using the diagram of a regular pentagon. 
Consider the length of the upper horizontal $CE$.  It follows that $$2\cos{36^{\circ}}=2\sin{18^{\circ}}+1$$ $Now \cos{36^{\circ}}=1-2\sin^2{18^{\circ}}$ so $s=\sin{18^{\circ}}$ satisfies $$4s^2+2s-1=0$$ Which gives $$s=\frac{-1+\sqrt{5}}{4}$$ And as $2\cos{36^{\circ}}=2s+1=\frac{1+\sqrt{5}}{2}$ therefore $$\cos{36^{\circ}}=\frac{1+\sqrt{5}}{4}.$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
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 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
We'll use the identity:
$\ds{\cos\pars{\bracks{n + 1}\theta} + \cos\pars{\bracks{n - 1}\theta}
     =2\cos\pars{n\theta}\cos\pars{\theta}}$

Set $\ds{\theta \equiv 36^{\circ}}$:
  $$
\begin{array}{rclcrcl}
n & = & 1 & \imp & \cos\pars{72^{\circ}} + 1 & = & 2\cos^{2}\pars{36^{\circ}}
\\[1mm]
n & = & 2 & \imp & \cos\pars{108^{\circ}} + \cos\pars{36^{\circ}} & = & 2\cos\pars{72^{\circ}}\cos\pars{36^{\circ}}
\end{array}
$$
  "Adding" both equations
  ( note that $\ds{\cos\pars{108^{\circ}} + \cos\pars{72^{\circ}} = 0}$ ):
  \begin{align}
&\color{#c00000}{1 + \cos\pars{36^{\circ}}}=
2\cos\pars{36^{\circ}}\bracks{\cos\pars{72^{\circ}} + \cos\pars{36^{\circ}}}
\\[3mm]&=2\cos\pars{36^{\circ}}\bracks{2\cos^{2}\pars{36^{\circ}} - 1 + \cos\pars{36^{\circ}}}
\\[3mm]&=2\cos\pars{36^{\circ}}\braces{2\bracks{\color{#c00000}{\cos\pars{36^{\circ}} + 1}}^{2}
-3\bracks{\color{#c00000}{\cos\pars{36^{\circ} + 1}}}}
\end{align}

Since $\ds{\color{#c00000}{\cos\pars{36^{\circ}} + 1} > 0}$, well'get
$$
1 = 2\cos\pars{36^{\circ}}\bracks{2\cos\pars{36^{\circ}} - 1}
$$
The positive root is given by:
$$
\color{#66f}{\large\cos\pars{36^{\circ}} = {1 + \root{5} \over 4}} \approx {\tt 0.8090}
$$
A: Last fall, I wrote up a proof that $\cos(36^\circ) = \varphi/2$ for an undergraduate trigonometry class.  As my proof is different from the others provided (while it is similar substantially similar to evilman999's argument, my argument is a bit more verbose, and I think that my picture is better ), and since this question has been asked again, I thought it reasonable to add my own two cents.  I will note that this isn't an exact answer for the question asked above (which outlines a specific strategy), but that this answer is appropriate for the more abstract question "How do we prove that $\cos(36^\circ) = \varphi/2$?"


Let $ABCDE$ be a regular pentagon, where it may assumed without loss of generality that the sides are of unit length.  The triangle $\triangle ABE$ is isosceles with $\measuredangle BAE = 108^\circ$ (this is the measure of each angle in a regular pentagon), which implies that
\begin{equation*}
 \measuredangle ABE = \measuredangle AEB = 36^\circ.
\end{equation*}
By similar arguments,
\begin{equation*}
 \measuredangle CBD
  = \measuredangle DEC
  = \measuredangle DCE
  = \measuredangle CDB
  = 36^\circ.
\end{equation*}
Next, observe that the measures of the three angles formed by the diagonals emanating from $B$ must sum to $108^\circ$.  This implies that
\begin{equation*}
 \measuredangle DBE
  = 108^\circ - 2\cdot 36^\circ
  = 36^\circ.
\end{equation*}
Similarly, $\measuredangle CEB = 36^\circ$.
Let $F$ denote the intersection of the diagonals $\overline{BD}$ and $\overline{EC}$.  By the angle-angle similarity theorem, $\triangle BEF \sim \triangle DCF$.  The goal now is to use this similarity to determine the length of $\overline{BE}$.  Note that
\begin{equation*}
 \measuredangle BCF + \measuredangle FCD = \measuredangle BCF + 36^\circ = 108^\circ
  \implies \measuredangle BCF = 72^\circ.
\end{equation*}
As the angle sum of a triangle is $180^\circ$, it follows that $\measuredangle BFC = 72^\circ$, and so $\triangle BCF$ is isosceles.  But then $\overline{BC} \cong \overline{BF}$.  In particular, this implies that $\overline{BF}$ is of unit length.
Let $x$ denote the length of $\overline{FD}$, so that the length of $\overline{BD}$ is $1+x$.  Then, as all of the diagonals in the regular pentagon $ABCDE$ are congruent, the length of $\overline{BE}$ must also be $1+x$.  By properties of similar triangles,
\begin{equation*}
 \frac{\ell(\overline{BE})}{\ell(\overline{BF})}
  = \frac{\ell(\overline{DC})}{\ell(\overline{DF})}
 \implies \frac{1+x}{1} = \frac{1}{x}
 \implies x^2 + x - 1 = 0
 \implies x = \frac{-1+\sqrt{5}}{2}.
\end{equation*}
Let $G$ denote the midpoint of $\overline{BE}$, and observe that $\overline{FG}$ is perpendicular to $\overline{BG}$.  Using properties of right-triangles (specifically, using thhe right-triangle definitions of the trigonometry functions, i.e. "SOH-CAH-TOA") and the earlier observation that $\measuredangle DBE = 36^\circ$, it is now possible to conclude that
\begin{equation*}
 \cos(36^\circ)
  = \frac{\ell(\overline{BG})}{\ell(\overline{BF})}
  = \frac{\frac{1}{2}(x+1)}{1}
  = \frac{1}{2}\left( \frac{-1+\sqrt{5}}{2} + 1 \right)
  = \frac{1+\sqrt{5}}{4} = \frac{\varphi}{2},
\end{equation*}
where $\varphi$ is the golden ratio.
