Intereting Posts

Easy example why complex numbers are cool
Parallel lines divide a circle's area into thirds
Why not write the solutions of a cubic this way?
How exactly are the beta and gamma distributions related?
A mathematical phenomenon regarding perfect squares…
Half iteration of exponential function
Given the sequence of functions, $f_1(x):=\sin(x)$ and $f_{n+1}(x):=\sin(f_n(x))$, why $|f_n(x)|\leq f_n(1)$?
The product of a subgroup and a normal subgroup is a subgroup
What do modern-day analysts actually do?
Bounds on off-diagonal entries of a correlation matrix
Question on relative homology
Mental estimate for tangent of an angle (from $0$ to $90$ degrees)
The Cantor set is homeomorphic to infinite product of $\{0,1\}$ with itself – cylinder basis – and it topology
Cauchy's Theorem vs. Fundamental Theorem of Contour Integration.
Limits and exponents and e exponent form

The problem reads as follows:

Noting that $t=\frac{\pi}{5}$ satisfies $3t=\pi-2t$, find the exact value of

$$\cos(36^\circ)$$

it says that you may find useful the following identities:

$$\cos^2 t+\sin^2 t = 1,\\

\sin 2t = 2\sin t\cos t,\\

\sin 3t = 3\sin t – 4\sin^3 t.

$$

- Distance between a point and a m-dimensional space in n-dimensional space ($m<n$)
- Prove the formula for the inverse of a matrix
- Is perspective transform affine? If it is, why it's impossible to perspective a square by an affine transform, given by matrix and shift vector?
- How can I characterize the type of solution vector that comes out of a matrix?
- Shortest and most elementary proof that the product of an $n$-column and an $n$-row has determinant $0$
- What does this theorem in linear algebra actually mean?

Do I have to do a system of linear equations in function of ..what? $t$? $\cos$?

Thanks in advance ðŸ™‚

- If matrix A is invertible, is it diagonalizable as well?
- What are some applications of elementary linear algebra outside of math?
- Angular alignment of points on two concentric circles
- A matrix is similar to its transpose
- for any ring $A$ the matrix ring $M_n(A)$ is simple if and only if $A$ is simple
- Number of solutions to the equations $x + 2y + 4z = 9\\4yz + 2xz + xy = 13\\ xyz = 3$
- Simultaneous diagonalization of quadratic forms
- Is the product of three positive semidefinite matrices positive semidefinite
- Matrix multiplication using Galois field
- There is a subspace $W$ of $V$ such that $V = U \oplus W$

Let $t=\frac{\pi}{5}$ (so $t$ is $36^\circ$). Since $108=180-72$, we have $3t=\pi-2t$ and therefore

$$\sin(3t)=\sin(\pi-2t).$$

But $\sin(\pi-2t)=\sin(2t)=2\sin t\cos t$.

Also, by the identity you were given, $\sin(3t)=3\sin t-4\sin^3 t$. Thus

$$3\sin t-4\sin^3 t=2\sin t\cos t.$$

But $\sin t\ne 0$, so we can cancel a $\sin t$ and obtain

$$3-4\sin^2 t=2\cos t.$$

Substitute $1-\cos^2 t$ for $\sin^2 t$ and simplify a bit. We get

$$4\cos^2 t-2\cos t-1=0.$$

Use the Quadratic Formula to solve this quadratic equation for $\cos t$, rejecting the negative root. We get

$$\cos t=\frac{2+\sqrt{20}}{8}.$$

We can simplify this to $\dfrac{1+\sqrt{5}}{4}.$

To find $\cos{\pi/5}$, note that

$$\sin{(3 \pi/5)} = \sin{(2 \pi/5)}$$

and

$$\sin{(3 \pi/5)} = 3 \sin{(\pi/5)} – 4 \sin^3{(\pi/5)} = 2 \sin{(\pi/5)} \cos{(\pi/5)}$$

Thus

$$2 \cos{(\pi/5)} = 3 – 4 \sin^2{(\pi/5)} = 4 \cos^2{(\pi/5)} – 1$$

Let $y=\cos{(\pi/5)}$. Then

$$4 y^2-2 y-1=0 \implies y = \frac{1 + \sqrt{5}}{4}$$

because $y>0$. Thus, $\cos{(\pi/5)} = (1+\sqrt{5})/4$.

Since rlgordonma has given the answer suggested by the supplied hint, here is another method of computing $\cos(\pi/5)$.

In this answer, it is shown that

$$

\tan(5\theta)=\frac{5\tan(\theta)-10\tan^3(\theta)+\tan^5(\theta)}{1-10\tan^2(\theta)+5\tan^4(\theta)}

$$

which, since $\tan(\pi/2)=\infty$, implies that

$$

5\tan^4(\pi/10)-10\tan^2(\pi/10)+1=0

$$

which, by the quadratic formula, yields

$$

\tan^2(\pi/10)=\frac{5-2\sqrt{5}}{5}

$$

Adding $1$ and taking the reciprocal yields

$$

\frac{1+\cos(\pi/5)}{2}=\cos^2(\pi/10)=\frac{5+\sqrt5}{8}

$$

Therefore,

$$

\cos(\pi/5)=\frac{1+\sqrt5}{4}=\phi/2

$$

where $\phi$ is the Golden Ratio.

Here is a way using roots of unity.

We have $\cos 36 = \frac{\omega + \omega^{-1}}{2}$ where $\omega = \text{exp} \left ( \frac{2 \pi}{10} \right )$. We have $-w$ is a primitive $5^{th}$ root of unity, so it follows $\omega^4 – \omega^3 + \omega^2 – \omega + 1 = 0$. Now, $x^4 – x^3 + x^2 – x + 1 = x^2(x^2 – x + 1 – \frac{1}{x} + \frac{1}{x^2}) = x^2\left ( \left (x + \frac{1}{x} \right )^2 – \left (x + \frac{1}{x} \right ) -1 \right )$. Thus $0 = \omega^2 \cdot ((2 \cos 36)^2 – (2 \cos 36) – 1)$. Therefore $4 \cos^3 36 – 2 \cos 36 – 1 = 0$. Using the quadratic formula we then arrive at $$\cos 36 = \frac{1 + \sqrt{5}}{4}$$

http://mathuprising.comlu.com/images/pentagon.png

This is a graphic that I created for my website about 2 days ago. Using $\Delta ABC$ invoke the Law of Cosines ($\angle BAC = \angle BCA = 36^\circ$).

BEGIN QUOTE: Ptolemy used geometric reasoning based on Proposition 10 of Book XIII of Euclid’s Elements to find the chords of $72^\circ$ and $36^\circ$. That Proposition states that if an equilateral pentagon is inscribed in a circle, then the area of the square on the side of the pentagon equals the sum of the areas of the squares on the sides of the hexagon and the decagon inscribed in the same circle. END QUOTE

http://en.wikipedia.org/wiki/Ptolemy%27s_table_of_chords#How_Ptolemy_computed_chords

Finding the chord of $72^\circ$ amounts to finding the sine of $36^\circ$.

Here is Proposition 10 of Book VIII.

Another with roots of unity …

Let $\omega:=\exp\left(\frac{\pi i}{5}\right)$ and $\phi$ is the golden ratio defined via relation $$\phi^2:=\phi+1\qquad|\quad\phi>0\tag{a}$$i.e., explicitly : $\phi=\frac{1+\sqrt{5}}{2}$ (but only (a) is sufficient to proceed further through this text, without radicals of 5’s at start), then following holds true :

Via Euler’s formula: $$\Re\{\omega\}=\Re\bigg\{ \exp\left(\frac{\pi i}{5}\right) \bigg\} = \cos \left(\frac{\pi}{5}\right)$$

Similarly

$$\Im\{\omega\}=\sin \left(\frac{\pi}{5}\right)$$

Moreover, it’s clear, by properities of the function cosine (resp. sine) on the interval $(0,\frac{\pi}{2})$, that $$\Re\{\omega\}>0\qquad\mathrm{resp.}\qquad\Im\{\omega\}>0\tag{0}$$ By Moivre’s formula also $$\omega^5 = e^{i\pi}=-1$$

I.e.:

$$\omega^5+1=0$$

Since $\omega\neq -1$ we divide this by factor $\omega+1$, hence

$$\omega^4-\omega^3+\omega^2-\omega+1=0\tag{1}$$

This could be wieved as a polynomial degree 4 in $\omega$, so we may guess, it can be factorised more suggesting :

$$\omega^4-\omega^3+\omega^2-\omega+1 = (\omega^2-\alpha\omega+1)(\omega^2-\beta\omega+1)\tag{2}$$

Expanding out we get a system for $\alpha$ and $\beta$ :

$$\alpha+\beta=1$$

$$\alpha\beta=-1$$

Subctracting $\phi$ in the equation for golden ratio and dividing through $\phi$ we get an equivalent relation $$\phi-\frac{1}{\phi}=1\tag{b}$$

So the conditions for $\alpha$ and $\beta$ are met when :

$$\alpha=\phi$$ $$\beta=-\frac{1}{\phi}$$

Now we have from $(1)$ and $(2)$ two quadratic equations for $\omega$ :

$$\omega=\frac{\alpha}{2}\pm\frac{1}{2}\sqrt{\alpha^2-4}\qquad \mathrm{or}\qquad \omega=\frac{\beta}{2}\pm\frac{1}{2}\sqrt{\beta^2-4}\tag{3}$$

Dividing $(b)$ by $\phi$ and rearanging we get

$$\frac{1}{\phi}=1-\frac{1}{\phi^2}\tag{c}$$

So then $$\alpha^2-4=\phi^2-4\overset{(a)}{=}\phi-3\overset{(b)}{=}\frac{1}{\phi}-2\overset{(c)}{=}-\frac{1}{\phi^2}-1<0 \quad\mathrm{since}\quad \phi>0$$

… and also … $$\beta^2-4=\frac{1}{\phi^2}-4\overset{(c)}{=}-\frac{1}{\phi}-3<0 \quad\mathrm{since}\quad \phi>0$$

Taking square roots make us some imaginary numbers, from (3) then, taking real part

$$\Re\{\omega\}=\frac{\alpha}{2}\qquad \mathrm{or}\qquad \Re\{\omega\}=\frac{\beta}{2}$$

However, since $\beta=-\frac{1}{\phi}<0$ is by $(0)$ and $(3)$ nessesarily and uniquely :

$$\omega=\frac{\alpha}{2}+\frac{i}{2}\sqrt{4-\alpha^2}=\frac{\phi}{2}+\frac{i}{2}\sqrt{3-\phi}$$

Ergo

$$\cos\left(\frac{\pi}{5}\right)=\Re\{\omega\}=\frac{\phi}{2}=\frac{1+\sqrt{5}}{4}$$

and, as a bonus :

$$\sin\left(\frac{\pi}{5}\right)=\Im\{\omega\}=\frac{1}{2}\sqrt{3-\phi}=\frac{\sqrt{5-\sqrt{5}}}{4}$$

- How can an ordered pair be expressed as a set?
- Three tangents meet opposide edges in collinear points
- Does $x_{n+2} = (x_{n+1} + x_{n})/2$ converge?
- Free boolean algebra
- Proving connectedness of the $n$-sphere
- Please correct my answer (Probability)
- Does Riemann integrable imply Lebesgue integrable?
- What are cohomology rings good for?
- Why is it called a series?
- When is the Lie algebra of the center of Lie group the center of its Lie algebra
- Question about solvable groups
- Proof of Wilson's Theorem using concept of group.
- Are cyclic groups always abelian?
- How to prove $\sum^n_{i=1} \frac{1}{i(i+1)} = \frac{n}{n+1}$?
- Proper notation for distinct sets