MHB Find tan 15° and cos 72° (using geometry)

[Albert1]

find :
(1)$ tan \,\, 15^o$
(2)$cos\,\,72^o$
(using geometry)

 

[Prove It]

find :
(1)$ tan \,\, 15^o$
(2)$cos\,\,72^o$
(using geometry)

Spoiler

$\displaystyle \begin{align*} \tan{ \left( 2\,\theta \right) } &\equiv \frac{ 2 \tan{ \left( \theta \right) } }{1 - \tan^2{ \left( \theta \right) } } \\ \tan{ \left( 30^{\circ} \right) } &= \frac{ 2\tan{ \left( 15^{\circ} \right) } }{ 1 - \tan^2{ \left( 15^{\circ} \right) } } \\ \frac{1}{\sqrt{3}} &= \frac{2\tan{ \left( 15^{\circ} \right) }}{1 - \tan^2{ \left( 15^{\circ} \right) } } \\ 1 - \tan^2{ \left( 15^{\circ} \right) } &= 2\,\sqrt{3}\,\tan{ \left( 15^{\circ} \right) } \\ 0 &= \tan^2{ \left( 15^{\circ} \right) } + 2\,\sqrt{3}\,\tan{ \left( 15^{\circ} \right) } - 1 \\ \tan{ \left( 15^{\circ} \right) } &= \frac{-2\,\sqrt{3} \pm \sqrt{ \left( 2\,\sqrt{3} \right) ^2 - 4 \left( 1 \right) \left( -1 \right) } }{2\left( 1 \right) } \\ \tan{ \left( 15^{\circ} \right) } &= \frac{-2\,\sqrt{3} \pm \sqrt{ 16 }}{2} \\ \tan{ \left( 15^{\circ} \right) } &= \frac{-2\,\sqrt{3} \pm 4}{2} \\ \tan{ \left( 15^{\circ} \right) } &= -\sqrt{3} \pm 2 \end{align*}$

and as $\displaystyle \begin{align*} 15^{\circ} \end{align*}$ is in the first quadrant, the amount needs to be positive, thus it must be that $\displaystyle \begin{align*} \tan{ \left( 15^{\circ} \right) } = 2 - \sqrt{3} \end{align*}$.

 

[Albert1]

Spoiler

$\displaystyle \begin{align*} \tan{ \left( 2\,\theta \right) } &\equiv \frac{ 2 \tan{ \left( \theta \right) } }{1 - \tan^2{ \left( \theta \right) } } \\ \tan{ \left( 30^{\circ} \right) } &= \frac{ 2\tan{ \left( 15^{\circ} \right) } }{ 1 - \tan^2{ \left( 15^{\circ} \right) } } \\ \frac{1}{\sqrt{3}} &= \frac{2\tan{ \left( 15^{\circ} \right) }}{1 - \tan^2{ \left( 15^{\circ} \right) } } \\ 1 - \tan^2{ \left( 15^{\circ} \right) } &= 2\,\sqrt{3}\,\tan{ \left( 15^{\circ} \right) } \\ 0 &= \tan^2{ \left( 15^{\circ} \right) } + 2\,\sqrt{3}\,\tan{ \left( 15^{\circ} \right) } - 1 \\ \tan{ \left( 15^{\circ} \right) } &= \frac{-2\,\sqrt{3} \pm \sqrt{ \left( 2\,\sqrt{3} \right) ^2 - 4 \left( 1 \right) \left( -1 \right) } }{2\left( 1 \right) } \\ \tan{ \left( 15^{\circ} \right) } &= \frac{-2\,\sqrt{3} \pm \sqrt{ 16 }}{2} \\ \tan{ \left( 15^{\circ} \right) } &= \frac{-2\,\sqrt{3} \pm 4}{2} \\ \tan{ \left( 15^{\circ} \right) } &= -\sqrt{3} \pm 2 \end{align*}$

and as $\displaystyle \begin{align*} 15^{\circ} \end{align*}$ is in the first quadrant, the amount needs to be positive, thus it must be that $\displaystyle \begin{align*} \tan{ \left( 15^{\circ} \right) } = 2 - \sqrt{3} \end{align*}$.

Spoiler

the answer is correct ,can you prove it using "geometry" ?

 

[Prove It]

Spoiler

the answer is correct ,can you prove it using "geometry" ?

Trigonometry IS geometry!

 

[kaliprasad]

find :
(1)$ tan \,\, 15^o$
(2)$cos\,\,72^o$
(using geometry)

kindly excuse me as I cannot draw a figure as I am not familiar but my solution is as below(for tan 15) ( I even not using tan 30)

Spoiler

Draw an equilateral triangle ABC side 2 and draw AD perpendiculaor to BC

AB = 2 , BD = 1 and $AD=sqrt{3}$

Now extend BD to E such that BE = AB = 2

join AE
AEB isn isosceles so $\angle AE = 30^\circ$ and $DE = DB + BE = 3$

so $AE = \sqrt{12}$

Extend DE to F such that EF = AE and join AF. So $DF = 3+ \sqrt{12}$

AEF is isosceles

so $\angle AFD = 15^\circ$ and

hence $\tan \, 15^\circ = \frac{AD}{DF}= \frac{\sqrt{3}}{3+\sqrt{12}} =\frac{\sqrt{3}(\sqrt{12}-3)}{12-9}$ = $\frac{6-3\sqrt{3}}{3}=2-\sqrt{3}$

 

[Opalg]

[sp]

[TIKZ][scale=3]
\coordinate [label=above left: $A$] (A) at (0,2) ;
\coordinate [label=above right: $B$] (B) at (2,2) ;
\coordinate [label=below right: $C$] (C) at (2,0) ;
\coordinate [label=below left: $D$] (D) at (0,0) ;
\coordinate [label=right: $E$] (E) at (1,1.732) ;
\coordinate [label=above : $F$] (F) at (1,2) ;
\coordinate [label=below : $G$] (G) at (1,0) ;
\draw (A) -- (B) -- (C) -- (D) -- (A) -- (E) ;
\draw (D) -- (E) -- (C) ;
\draw (F) -- (G) ;[/TIKZ]​

In the diagram, $ABCD$ is a square with side $2$, and $CDE$ is an equilateral triangle whose height $EG$ is $\sqrt3$.

In the isosceles triangle $ADE$ the angle at $D$ is $30^\circ$, so the angles at $A$ and $E$ are $75^\circ$. Therefore the angle $EAF$ is $15^\circ$. You can then see from the triangle $AFE$ that $\tan15^\circ = \dfrac{EF}{AF} = \dfrac{FG - EG}{1} = 2-\sqrt3.$[/sp]
Edit: kaliprasad just beat me to it with what looks like a very similar solution.

 

[Prove It]

Spoiler

As for evaluating $\displaystyle \begin{align*} \cos{ \left( 72^{\circ} \right) } \end{align*}$, we need to look at a regular pentagram.

View attachment 6433

If we have the side lengths as 1 unit long, and if we call the unknown length "x" then in the middle triangle from the cosine rule we can write

$\displaystyle \begin{align*} x^2 &= x^2 + 1^2 - 2\left( x \right) \left( 1 \right) \cos{ \left( 72^{\circ} \right) } \\ 0 &= 1 - 2\,x\cos{ \left( 72^{\circ} \right) } \\ 2\,x \cos{ \left( 72^{\circ} \right) } &= 1 \\ x &= \frac{1}{2\cos{ \left( 72^{\circ} \right) } } \end{align*}$

If we look at the left hand triangle, we can relate the three sides by the cosine rule

$\displaystyle \begin{align*} x^2 &= 1^2 + 1^2 - 2\left( 1 \right) \left( 1 \right) \cos{ \left( 108 ^{\circ} \right) } \\ x^2 &= 2 - 2\cos{ \left( 108^{\circ} \right) } \\ x^2 &= 2 - 2\cos{ \left( 180^{\circ} - 72^{\circ} \right) } \\ x^2 &= 2 + 2\cos{ \left( 72^{\circ} \right) } \\ \left[ \frac{1}{2\cos{ \left( 72^{\circ} \right) } } \right] ^2 &= 2 + 2\cos{ \left( 72^{\circ} \right) } \\ \frac{1}{4\cos^2{\left( 72^{\circ} \right) } } &= 2 + 2\cos{ \left( 72^{\circ} \right) } \\ 1 &= 4\cos^2{ \left( 72^{\circ} \right) } \left[ 2 + 2\cos{ \left( 72^{\circ} \right) } \right] \\ 1 &= 8\cos^2{ \left( 72^{\circ} \right) } + 8\cos^3{ \left( 72^{\circ} \right) } \\ 0 &= 8\cos^3{ \left( 72^{\circ} \right) } + 8\cos^2{ \left( 72^{\circ} \right) } - 1 \end{align*}$

If we let $\displaystyle \begin{align*} y = \cos{ \left( 72^{\circ} \right) } \end{align*}$ then we have the polynomial equation $\displaystyle \begin{align*} 8\,y^3 + 8\,y^2 - 1 &= 0 \end{align*}$.

We can see that $\displaystyle \begin{align*} y = -\frac{1}{2} \end{align*}$ is a solution, and so $\displaystyle \begin{align*} \left( 2\,y + 1 \right) \end{align*}$ is a factor. Thus

$\displaystyle \begin{align*} 8\,y^3 + 8\,y^2 - 1 &= 0 \\ 8\,y^3 + 4\,y^2 + 4\,y^2 - 1 &= 0 \\ 8\,y^3 + 4\,y^2 + 4\,y^2 + 2\,y - 2\,y - 1 &= 0 \\ 4\,y^2 \left( 2\,y + 1 \right) + 2\,y \left( 2\,y + 1 \right) - 1 \left( 2\,y + 1 \right) &= 0 \\ \left( 2\,y + 1 \right) \left( 4\,y^2 + 2\,y - 1 \right) &= 0 \end{align*}$

It's pretty obvious that as $\displaystyle \begin{align*} 0 < 72 < 90 \end{align*}$ that $\displaystyle \begin{align*} \cos{ \left( 72^{\circ} \right) } > 0 \end{align*}$, so $\displaystyle \begin{align*} -\frac{1}{2} \end{align*}$ can't be the solution, thus

$\displaystyle \begin{align*} 4\,y^2 + 2\,y - 1 &= 0 \\ y &= \frac{-2 \pm \sqrt{2^2 - 4\left( 4 \right) \left( -1 \right) }}{2\left( 4 \right) } \\ y &= \frac{-2 \pm \sqrt{ 20 }}{8} \\ y &= \frac{-2 \pm 2\,\sqrt{5}}{8} \\ y &= \frac{-1 \pm \sqrt{5}}{4} \end{align*}$

Again, as we know $\displaystyle \begin{align*} \cos{ \left( 72^{\circ} \right) } > 0 \end{align*}$ that means that $\displaystyle \begin{align*} \frac{-1 - \sqrt{5}}{4} \end{align*}$ can't be the solution, and thus we can say for certain, as it's the only other possible solution to this polynomial equation, that $\displaystyle \begin{align*} \cos{ \left( 72^{\circ} \right) } = \frac{\sqrt{5} - 1}{4} \end{align*}$.

 

[Opalg]

$\cos72^\circ$, "using geometry":
[sp]

[TIKZ]\coordinate [label=above: $A$] (A) at (90:3) ;
\coordinate [label=right: $B$] (B) at (18:3) ;
\coordinate [label=below right: $C$] (C) at (306:3) ;
\coordinate [label=below left: $D$] (D) at (234:3) ;
\coordinate [label=left: $E$] (E) at (162:3) ;
\coordinate [label=below: $G$] (G) at (0,-2.43) ;
\draw (A) -- (B) -- (C) -- (D) -- (E) -- cycle;
\draw (A) -- (C) -- (E) -- (B) -- (D) -- cycle ;
\draw (A) -- (G) ;
\node at (-1.5,-0.5) {$F$} ;
[/TIKZ]​

In the diagram, $ABCDE$ is a regular pentagon with side $2$. Let $d$ be the length of a diagonal such as $AD$.

The quadrilateral $ABCF$ is a rhombus, and so $AF = FC = AB = 2$. Thus the triangle $DFC$ is isosceles. It has the same angles ($36^\circ,72^\circ,72^\circ$) as the triangle $DAC$. Comparing lengths of sides in these two triangles, you see that $\dfrac{DF}{DC} = \dfrac{DC}{DA}$, so that $\dfrac{DF}{2} = \dfrac2d.$

Therefore $d = DA = DF + FA = \dfrac4d + 2$, from which $d^2 - 2d - 4 = 0$. The positive solution of that quadratic is $d = \sqrt5+1$. Hence $\cos72^\circ = \dfrac{DG}{DA} = \dfrac1{\sqrt5+1}$. You can tidy this up by rationalising the denominator, getting $\cos72^\circ = \frac14(\sqrt5-1)$.
[/sp]