Exact value for $\cos 36°$

Question

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.

Answer 1

"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.

Answer 2

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).

Answer 3

\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.

Answer 4

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 $\cos (36^{\circ})=\frac{1+\sqrt{5}}{4}$.

Answer 5

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 $$\frac{1+\sqrt 5}{4}$$

Answer 6

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 $$\cos{36^{\circ}}=\frac{1+\sqrt{5}}{4}.$$

Answer 7

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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: $$ \color{#66f}{\large\cos\pars{36^{\circ}} = {1 + \root{5} \over 4}} \approx {\tt 0.8090} $$