The Whimsical Cosine

BY ALGEBRAIC MANIPULATION :

$\displaystyle \sum_{r=1}^{10} sec^{4r}\theta= sec^4\theta \cdot \dfrac{(1-sec^{40}\theta)}{(1-sec^4\theta)}$ ---(1) Using G.M. sum formula

Given : $cos^8\theta +cos^4\theta=1$

$\implies sec^8\theta=1+sec^4\theta$

$sec^{16}\theta=(sec^8\theta)^2=(1+sec^4\theta)^2=2+3sec^4\theta$

$sec^{32}\theta=(sec^{16}\theta)^2=13+21sec^4\theta$

$sec^{40}\theta=sec^{32}\theta \cdot sec^8\theta=34+55sec^4\theta$

Using above values in (1)

$\displaystyle \sum_{r=1}^{10} sec^{4r}\theta= sec^4\theta \cdot \dfrac{(1-(34+55sec^\theta))}{(1-sec^4\theta)}$

$\quad =\dfrac{(-33-55sec^4\theta) \cdot sec^4\theta}{1-sec^4\theta}$

$\quad =\dfrac{-55-88sec^4\theta}{1-sec^4\theta}$

Adding and subtracting 88 in the numerator :

$\quad =88+\dfrac{143}{sec^4\theta-1}$

$\quad =88+\dfrac{143}{cos^4\theta}$

So $\large \boxed{a=143}$

enjoy !

Given that $\cos^8 {\theta} + \cos^4 {\theta} = 1$ . Dividing throughout by $\cos^8 {\theta}$ :

$\Rightarrow 1 + \sec^4 {\theta} = sec^8 {\theta} \quad \Rightarrow \sec^8 {\theta} - sec^4 {\theta} - 1 = 0 \quad \Rightarrow \varphi^2 - \varphi -1 = 0$

$\Rightarrow \sec^4 {\theta} = \varphi = \dfrac {1+\sqrt{5}}{2} = \text{golden ratio}$

We note that:

\(\begin{array} {} \varphi &= \varphi & & = 0 + \varphi \\ \varphi^2 & = 1 + \varphi & & = 1 + \varphi \\ \varphi^3 & = \varphi + \varphi^2 & = \varphi + 1 + \varphi & = 1 + 2 \varphi \\ \varphi^4 & = \varphi + 2\varphi^2 & = \varphi + 2(1 + \varphi) & = 2 + 3 \varphi \\ \varphi^5 & = 2\varphi + 3\varphi^2 & = 2\varphi + 3(1 + \varphi) & = 3 + 5 \varphi \\ \varphi^6 & & & = 5 + 8\varphi \\ \varphi^7 & & & = 8 + 13 \varphi \\ \varphi^8 & & & = 13 + 21 \varphi \\ \varphi^9 & & & = 21 + 34 \varphi \\ \varphi^{10} & & & = 34 + 55 \varphi \end{array} \)

We note that: $\varphi^n = F_{n-1} + F_n \varphi$ , where $F_n$ is the $n^{th}$ Fibonacci number.

Therefore, $\displaystyle \sum_{r=1}^{10} {\sec^{4r}{\theta}} = \sum_{r=0}^{9} {F_r} + \sec^{4}{\theta} \sum_{r=1}^{10} {F_r} = 88 + \dfrac {143}{\cos^{4}{\theta}}$

$\Rightarrow a = \boxed{143}$

Golden Approach:

$\cos^8 \theta+\cos^4 \theta=1.$ Clearly $\cos \theta \ne 0.$ Now putting $x=\frac{1}{\cos^4 \theta}=\sec^4 \theta$ we get, $x^2-x-1=0$ $\Rightarrow x=\phi$ . Now $\sum_{r=1}^{10} \sec^{4r} \theta=\sum_{n=1}^{10} \phi^n$ Using the equation $\phi^n=F_n \phi+F_{n-1}$ where $F_n$ is the nth number of the Fibonacci sequence $1,1,2,3,5,8...$ we find that $\sum_{r=1}^{10} \sec^{4r} \theta=143 \sec^4 \theta+88$ $\Rightarrow a=\boxed{143}$

golden ratio and its property that g+1=g^2 {g-golden ratio} here g-1=cos^4(theta)