Minimize Cosine

For positive integers $k$ , $(\cos\theta+i\sin\theta)^k = e^{ki\theta} = \cos{k\theta}+i\sin{k\theta}$ , so equating real parts and using $\sin^2\theta = 1-\cos^2\theta$ gives us an expression for $\cos{k\theta}$ in terms of $\cos\theta$ . For convenience, let $x = \cos\theta \in [-1,1]$ . Then $\cos{4\theta} = 8x^4 - 8x^2 + 1$ , $\cos{3\theta} = 4x^3-3x$ , and $\cos{2\theta} = 2x^2 -1$ [Note: Or you can obtain this by the sum and difference formulas - Calvin]

So, $2\cos{4\theta}+8\cos{3\theta}+32\cos{2\theta}+56\cos\theta+100$ can be rewritten as $2(8x^4-8x^2+1)+8(4x^3-3x)+32(2x^2-1)+56x+100$ which is equal to $16x^4+32x^3+48x^2+32x+ 70 = 16(x^2+x+1)^2 + 54$ where we use the factorization $x^4+2x^3+3x^2+2x+1 = (x^2+x+1)^2 .$

But $x^2+x+1 = (x+\frac{1}{2})^2+\frac{3}{4}$ is clearly minimized over the interval $[-1,1]$ at $x=-\frac{1}{2}$ , so the desired minimum is $16(\frac{3}{4})^2 + 54 = 63$ .

Let $f(\theta)=2 \cos 4\theta + 8 \cos 3\theta + 32 \cos 2\theta + 56 \cos \theta + 100$ . We want to find the minimum value of this function.

Using the trig identities $\cos 4\theta = 8\cos^4 \theta - 8\cos^2 \theta + 1$ , $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$ , and $\cos 2\theta = 2\cos^2 \theta - 1$ , we substitute and get

$f(\theta)=16\cos^4 \theta + 32\cos^3 \theta + 48\cos^2 \theta + 32\cos \theta + 70$ .

Since this function is differentiable everywhere, to find its minimum value we can examine where its derivative is zero. So we differentiate:

$f'(\theta)=-64\cos^3 \theta \sin \theta - 96\cos^2 \theta \sin \theta - 96\cos \theta \sin \theta + 32$

And factor:

$f'(\theta)=-32 \sin \theta (2\cos \theta + 1)(\cos^2 \theta + \cos \theta + 1)$

There are no real values of $\cos \theta$ for which $\cos^2 \theta + \cos \theta + 1 = 0$ . Therefore $f'(\theta)=0$ only if either $\sin \theta = 0$ or $\cos\theta = -\frac{1}{2}$ . In other words, only if $\cos\theta = -1$ , $-\frac{1}{2}$ , or $1$ . The respective values of $f(\theta)$ are $70$ , $63$ , and $198$ . Therefore the minimum value of $f(\theta)$ is 63.

The first part of the solution is to expand all the cosines into powers of $\cos(x)$ . This is a very boring part of the solution so i will omit it. In the end, we can replace $\cos(x)=t$ to get:

$70 + 32 t + 48 t^2 + 32 t^3 + 16 t^4$

We will now try to minimize this polynomial, respecting $-1 \leq t \leq 1$ :

If there is an extreme value the derivative of this function will have to be equal to 0 in that point. The derivative is:

$32 + 96 t + 96 t^2 + 64 t^3$

We have to solve $32 + 96 t + 96 t^2 + 64 t^3 = 0$

We factor out 32 to get

$1 + 3 t + 3 t^2 + 2 t^3 = 0$

We will write this in a slightly different way to get

$1 + 3 t + 3 t^2 + t^3 = -t^3$

That is,

$(t+1)^3 = (-t)^3$

Since the function $x^3$ is a monotonic function, this implies

$t+1 = -t$

Solving this we get $t = -\frac{1}{2}$

Replacing this into the starting polynomial we get 63. Note that it is a good idea to also check the endpoints of the interval $[-1,1]$ to make sure that the minimum is not there.

It is well known that, for any positive integer $n$ , $\cos n\theta$ is a polynomial of degree $n$ in $\cos\theta$ (it is the Chebyshev polynomial of the first kind). Therefore, the strategy which should be used for solving this problem is to express everything in terms of $x=\cos\theta$ and then find the minimum of a quartic function obtained.

By using the multiple angle formulae, we easily infer $\cos 2\theta = 2x^2-1 , \cos 3\theta = 4x^3-3x, \cos 4\theta = 8x^4-8x^2+1$ . Now, after some manipulation, the original expression becomes $f(x) = 16x^4+32x^3+48x^2+32x+70$ , with derivative $f'(x) = 64x^3+96x^2+96x+32 = 32(2x+1)(x^2+x+1)$ .

Note that as $x^2+x+1 = (x+\frac12)^2 + \frac12 > 0$ for all real $x$ , the first derivative of $f$ has only one zero, namely $-\frac12$ . It is well-known that the extremes of $f$ can be obtained only at 3 points - at $-\frac 12$ , and at the boundaries of the intervals at which $f$ is defined, in this case, these are -1 and 1, as $-1\leq \cos \theta \leq 1$ . By hand-checking, the minimum value of the expression is 63.

Using the sum and product formulas, we get that

$\begin{aligned} 2 \cos 4 \theta + 8 \cos 3\theta + 32 \cos 2 \theta + 56 \cos \theta + 46 &= 16\cos^4 \theta + 32 \cos ^3 \theta + 48 \cos^2 \theta + 32 \cos \theta + 16 \\ &= 16 ( \cos^2 \theta + \cos \theta + 1)^2 \\ &= 16 \left(\left(\cos\theta + \frac{1}{2}\right)^2 +\frac{3}{4} \right)^2 \end{aligned}$

This achieves it's minimum when $\cos \theta = \frac {-1} {2}$ , which has value $16 \left(\frac {3}{4}\right) ^2 = 9$ . Hence, the minimum of the expression is $100 - 46 + 9 = 63$ .

Let $\cos(\theta) = x$ . \begin{align} \cos(2 \theta) & =2 \cos(\theta) - 1 & = 2x^2-1\ \cos(3 \theta) & = 4 \cos^3(\theta) - 3 \cos(\theta) & = 4x^3 -3x\ \cos(4 \theta) & = 2 \cos^2(2 \theta) - 1 = 2(2x^2-1)^2-1 & = 8x^4 - 8x^2+1 \end{align} We want to find the minimum of $2 \cos(4 \theta)+8 \cos(3 \theta)+32 \cos(2 \theta)+56 \cos(\theta)+100$ . Equivalently, we can find the minimum of $\cos(4 \theta) + 4 \cos(3 \theta) + 16 \cos(2 \theta) + 28 \cos( \theta) + 50$ and multiply by $2$ . Hence, \begin{align} \cos(4 \theta) + 4 \cos(3 \theta) + 16 \cos(2 \theta) + 28 \cos( \theta) + 50 & = (8x^4 - 8x^2+1) + 4(4x^3-3x) + 16(2x^2-1) + 28x + 50\ & = 8x^4 + 16x^3 + 24x^2 + 16x + 35 \end{align} Since $x = \cos( \theta)$ , we have $x \in [-1,1]$ and hence we want to find the minimum of $$f(x) = 8x^4 + 16x^3 + 24x^2 + 16x + 35$$ in the domain $x \in [-1,1]$ .

To do this, let us first find $f'(x)$ . We get that $$f'(x) = 32x^3 + 48x^2 + 48x + 16 = 16(2x^3 + 3x^2 + 3x + 1) = 16(2x+1)(x^2 + x + 1)$$ Setting $f'(x) = 0$ , we get that $x=-\dfrac12$ , which also lies in the domain $[-1,1]$ . Hence, the global minimum of $f(x)$ is $$f(-1/2) = 8 \times \dfrac1{16} - 16 \times \dfrac18 + 24 \times \dfrac14 - 16 \times \dfrac12 + 35 = \dfrac12 - 2 + 6 - 8 + 35 = 31 \dfrac12$$

Hence, the minimum of $2 \cos(4 \theta)+8 \cos(3 \theta)+32 \cos(2 \theta)+56 \cos(\theta)+100$ is $$2 \times 31 \dfrac12 = 63$$

2\cos4?+8\cos3?+32\cos2?+56\cos?+100 2\cos 4\theta + 8\cos 3\theta + 32\cos 2\theta + 56 \cos \theta + 100 we substitue all \cos with more than 1 \theta : \cos 4\theta = \cos 2(2\theta) = 2 (\cos 2\theta)^2 -1 = 2 (2\cos^2 \theta -1)^2 -1 = 2 (4\cos^4 \theta - 4\cos^2 \theta +1) -1 = 8\cos^4 \theta - 8\cos^2 \theta + 1

\cos 3\theta = 4\cos^3 \theta - 3 \cos \theta

\cos 2\theta = 2\cos^2 \theta - 1

so we get a new formula :

2\cos 4\theta + 8\cos 3\theta + 32\cos 2\theta + 56 \cos \theta + 100 =16 \cos^4 \theta - 16\cos^2 \theta + 2 + 32 \cos^3 \theta - 24 \cos \theta + 64 \cos^2 \theta - 32 + 56 \cos \theta + 100 =16 \cos^4 \theta + 32 \cos^3 \theta + 48 \cos^2 \theta + 32 \cos \theta + 70

now we are looking for the minimum possible value, so the value of \cos \theta must be negative, and we have choice of -0,5 (\theta = 120^\circ) or -1 (\theta= 180^\circ). because we have \cos^4 \theta and \cos^2 \theta, we will get a positive number from inserting the negative one, so we must minimize both, where we can minimize it by inserting the \theta by 120^\circ so we will get these number : 16 \times 0,0625 + 32 \times (-0,125) + 48 \times 0,25 + 32 \times (-0,5) + 70 = 1 - 4 + 12 - 16 + 70 = \underline{63}

2cos4θ+8cos3θ+32cos2θ+56cosθ+100=2(2(2 cos^(2)θ-1)^(2)-1)+8(4 cos^(3)θ-3 cosθ)+32(2 cos^(2)θ-1)+56 cosθ+100. let cosθ=x ,then equation becomes 16 x^4 +32 x^3 + 48 x^2 +32 x+70 , differentiating and equating to zero to find minima,we get . 64 x^3+96 x^2 +96 x +32=0, x=-1/2 is a valid result (between -1 and 1). Again differentiating and substituting x=-1/2 we get positive value(48), so we confirm that x=-1/2 gives the minimum value of this equation.Thus substituting cosθ=-1/2 we get the value of equation as 67. 16 (-1/2)^4+32 (-1/2)^3 +48 (-1/2)^2 +32 (-1/2)-30+100=67.

[;2cos(4\theta )+8cos(3\theta )+32cos(2\theta )+56cos(\theta )+100 ;] Let [;cos(\theta )=u;] therefore: [;cos(2\theta )=2{ u }^{ 2 }-1;] [;cos(3\theta )={ 4u }^{ 3 }-{ 3u };] ;[cos(4\theta )=2{ ({ 2u }^{ 2 }-1) }^{ 2 }-1={ 8u }^{ 4 }-8{ u }^{ 2 }+1;] Substituting in the above equation we have: [;{ 16u }^{ 4 }+{ 32u }^{ 3 }+48{ u }^{ 2 }+32{ u }+100;] Deriving the expression we find: [;2{ u }^{ 3 }+3{ u }^{ 2 }+3u+1=0;] Now we note that as [;-1\le u\le 1;] as solution of the cubic equation is [;\frac { -1 }{ 2 };] and verification, we see that it is the value that minimizes the expression. Sorry for my bad english.

$f(\theta) = 2 \cos4\theta + 8 \cos3\theta + 32 \cos2\theta + 56 \cos\theta +100$

Simplifying this function to contain only $\cos\theta$ terms:

$f(\theta) = 16 (\cos^4\theta + 2 \cos^3\theta + 3 \cos^2\theta + 2 \cos\theta) + 70$

$f(\theta) = 16 (A) + 70$

Now function will have minimum value if and only if $A \leq 0$

Let $\cos\theta = x$

So, $x^4 + 2 x^3 + 3 x^2 + 2 x \leq 0$

Factorizing, we get $x ( x + 1) (x^2 + x + 2) \leq 0$

But $(x^2 + x + 2) > 0$ for all x, so now

$x (x + 1) \leq 0$

Case 1:

$x (x + 1) = 0$ ; which gives either $x = 0$ or $x + 1 = 0$

If $x = 0$ then $\theta = (\frac {\pi}{2} + n\pi)$ which gives $A > 0$ and hence $f(\theta) > 70$

if $x = -1$ then $\theta = (2n + 1)\pi$ which gives $A = 0$ and hence $f(\theta) = 70$

Case 2: $x (x + 1) < 0$ ; where either $x > 0$ and $x + 1 < 0$ and that is not possible. So $x < 0$ and $x + 1 > 0$ Simplifying, $0 > x > -1$

$0 > \cos\theta > -1$ that gives

$(2n + 1)\frac {\pi}{2} > \theta > (2n + 1)\pi$

Now we see that among $\theta = (\frac {\pi}{2} + \frac {\pi}{3})$ ; $\theta = (\frac {\pi}{2} + \frac {\pi}{4})$ and $\theta = (\frac {\pi}{2} + \frac {\pi}{6})$

$A < 0$ only for $\theta = (\frac {\pi}{2} + \frac {\pi}{6})$

So evaluating we get $f(\theta) = 63$

We want to minimize the function $f(x)= 2cos4x+8cos3x+32cos2x+56cosx$ . Now using the sum-to-product identity of cosine we can rewrite f(x) as : $f(x)=4cos(\frac{7}{2}x)cos(\frac{1}{2}x)+64cos(\frac{3}{2}x)cos(\frac{1}{2}x)+24cosx+6cos3x$ . Now, since 64 is greater than the sum of the other "coefficients" (which is 34) and $-1\leq cosx\leq 1$ , f(x) is minimized when $64cos(\frac{3}{2}x)cos(\frac{1}{2}x)$ is minimized. But it is easy to see that $64cos(\frac{3}{2}x)cos(\frac{1}{2}x)$ is minimized when $x=\frac{2}{3} \pi$ . Now substituting $x=\frac{2}{3} \pi$ in f(x) we obtain f(x)=-37. Now to obtain the desired result we must add 100 to -37 which makes 63.

The derivative is $-8(sin \theta + 3 sin 2\theta + 8 sin 3\theta + 7 sin 4\theta)$ .

For $\theta = \frac {2 \pi}{3}$ it is null: $sin \theta + 3 sin 2\theta + 8 sin 3\theta + 7 sin 4\theta = \frac{1}{2} ( -1 + 8 +0 -7) = 0$ .

I can't demonstrate that this local min is global, I guessed it from the plot :(

$cos 4\theta = 8cos^4 \theta - 8cos^2 \theta + 1$ , $cos 3\theta = 4cos^3 \theta - 3cos \theta$ , $cos 2\theta = 2 cos^2 \theta -1$ . Substitute for these into the equation, and let $x= cos \theta$ . The statement now translates to finding minimum of, say $y=16x^4 + 32x^3+48x^2+32x+70$ . Taking $\frac{dy}{dx} = 0$ , we get $2x^3+3x^2+3x+1 =0$ .

ie, $x^3 + (1+x)^3 = 0$ for $-1 \le x \le 1$ ,

$-x^3 = (1+x)^3$ for $-1 \le x \le 1$ .

Only $x=\frac{-1}{2}$ satisfies this equality by taking cube root on both sides. Also, $\frac{d^2y}{dx^2}$ is $> 0$ at $x= \frac{-1}{2}$ . Therefore we have a minimum at $x= \frac{-1}{2}$ . Substituting this value, we get minimum value of $y = 63$ .