Everything Just Disappear?

Relevant wiki: Sum and Difference Trigonometric Formulas - Problem Solving

The question is of the form $\dfrac{\sin {(3^0x)}^°}{\cos {(3^1x)}^°} + \dfrac{\sin {(3^1x)}^°}{\cos {(3^2x)}^°} + \dfrac{\sin {(3^2x)}^°}{\cos {(3^3)x}^°} + \cdots + \dfrac{\sin {(3^{n-1}x)}^°}{\cos {(3^nx)}^°}$ where $x=2$ .

$\begin{aligned} \dfrac{\sin {(3^0x)}^°}{\cos {(3^1x)}^°} + \dfrac{\sin {(3^1x)}^°}{\cos {(3^2x)}^°} + \dfrac{\sin {(3^2x)}^°}{\cos {(3^3x)}^°} + \cdots + \dfrac{\sin {(3^{n-1}x)}^°}{\cos {(3^nx)}^°} & = \dfrac{1}{2} × 2\left(\dfrac{\sin {(3^0x)}^°}{\cos {(3^1x)}^°} + \dfrac{\sin {(3^1x)}^°}{\cos {(3^2x)}^°} + \dfrac{\sin {(3^2x)}^°}{\cos {(3^3x)}^°} + \cdots + \dfrac{\sin {(3^{n-1}x)}^°}{\cos {(3^nx)}^°}\right)\\ \\ & = \dfrac{1}{2} \left(\dfrac{2\sin {(3^0x)}^°}{\cos {(3^1x)}^°} + \dfrac{2\sin {(3^1x)}^°}{\cos {(3^2x)}^°} + \dfrac{2\sin {(3^2x)}^°}{\cos {(3^3x)}^°} + \cdots + \dfrac{2\sin {(3^{n-1}x)}^°}{\cos {(3^nx)}^°}\right)\\ \\ & = \dfrac{1}{2} \left(\dfrac{2×\sin {(3^0x)}^° \cos {(3^0x)}^°}{\cos {(3^1x)}^° \cos {(3^0x)}^°} + \dfrac{2\sin {(3^1x)}^° \cos{(3^1x)}^°}{\cos {(3^2x)}^° \cos {(3^1x)}^°} + \dfrac{2\sin {(3^2)}^° \cos {(3^2x)}^°}{\cos {(3^3x)}^° \cos{(3^2x)}^°} + \cdots + \dfrac{2\sin {(3^{n-1}x)}^° \cos {(3^{n-1}x)}^°}{\cos {(3^nx)}^° \cos {(3^{n-1}x)}^°}\right)\\ \\ & = \dfrac{1}{2} \left(\dfrac{\sin {(2×3^0x)}^°}{\cos {(3^1x)}^° \cos {(3^2x)}^°} + \dfrac{\sin {(2×3^1x)}^°}{\cos {(3^2x)}^° \cos {(3^1x)}^°} + \dfrac{\sin {(2×3^2x)}^°}{\cos {(3^3x)}^° \cos {(3^2x)}^°} \cdots \dfrac{\sin {(2×3^{n-1}x)}^°}{\cos {(3^nx)}^° \cos {(3^{n-1}x)}^°}\right)\\ \\ & = \dfrac{1}{2} \left(\dfrac{\sin {(3^1x - 3^0x)}^°}{\cos {(3^1x)}^° \cos {(3^2x)}^°} + \dfrac{\sin {(3^2x - 3^1x)}^°}{\cos {(3^2x)}^° \cos {(3^1x)}^°} + \dfrac{\sin {(3^3x - 3^2x)}^°}{\cos {(3^3x)}^° \cos {(3^2x)}^°} \cdots \dfrac{\sin {(3^nx - 3^{n-1}x)}^°}{\cos {(3^nx)}^° \cos {(3^{n-1}x)}^°}\right)\\ \\ & = \dfrac{1}{2} \left(\dfrac{\sin {(3^1x)}^° \cos {(3^0x)}^° - \sin{(3^0x)}^°\cos{(3^1x)}^°}{\cos {(3^1x)}^° \cos {(3^0x)}^°} + \dfrac{\sin {(3^2x)}^° \cos {(3^1x)}^° - \sin{(3^1x)}^°\cos{(3^1x)}^°}{\cos {(3^2x)}^° \cos {(3^1x)}^°} + \dfrac{\sin {(3^3x)}^° \cos {(3^2x)}^° - \sin{(3^2x)}^°\cos{(3^3x)}^°}{\cos {(3^3x)}^° \cos {(3^2x)}^°} + \cdots + \dfrac{\sin {(3^nx)}^° \cos {(3^{n-1}x)}^° - \sin{(3^{n-1}x)}^°\cos{(3^nx)}^°}{\cos {(3^nx)}^° \cos {(3^{n-1}x)}^°}\right)\\ \\ & = \dfrac{1}{2} \left(\tan {(3^1x)}^° - \tan {(3^0x)}^° + \tan {(3^2x)}^° - \tan {(3^1x)}^° + \tan {(3^3x)}^° - \tan {(3^2x)}^° + \cdots + \tan {(3^nx)}^° - \tan {(3^{n-1}x)}^°\right)\\ \\ & = \dfrac{1}{2} \left(\tan {(3^nx)}^° - \tan{(3^0x)}^°\right)\\ \\ & = \dfrac{1}{2} \left(\tan {(3^nx)}^° - \tan x^°\right) \end{aligned}$

So, plugging in the $x=2$ and substituting $3^n = 59049$ , we get:-

$\begin{aligned} \dfrac{\sin 2^°}{\cos 6^°} + \dfrac{\sin 6^°}{\cos {18}^°} + \dfrac{\sin {18}^°}{\cos {54}^°} + \dfrac{\sin {54}^°}{\cos {162}^°} + \cdots \dfrac{\sin {39366}^°}{\cos {118098}^°} & = \dfrac{1}{2} \left(\tan {118098}^° - \tan 2^°\right)\\ \\ & = \dfrac{1}{2} \left(\tan {(18)}^° - \tan {(2)}^°\right)\\ \\ \implies \alpha = {18}^° & \left(\text{and}\right) \beta = 2^°\\ \\ \therefore \alpha + \beta & = 18 + 2\\ \\ & = \color{#3D99F6}{\boxed{20}} \end{aligned}$

We're essentially calculating the sum $\displaystyle S:= \sum_{n=0}^9 \dfrac{ \sin(3^n \cdot y^\circ)}{\cos (3^{n+1} \cdot y^\circ)}$ , where $y=2$ . Since it is hinted that this sum has a closed form of $\dfrac12 (\tan \alpha - \tan \beta)$ , this suggests that $S$ might be expressed as $f(m) - f(m- k) = (f(m) - f(m-1)) + (f(m-1) + f(m-2)) + (f(m-2) + f(m-3)) + \cdots + ( f(m-k+1) - f(m-k)) \; ,$ so $S$ could possibly be expressed as a telescoping sum and we wish to determine the function $f(m)$ .

In other words, we want to find $A$ and $B$ satisfying the trigonometric identity $\dfrac{\sin x}{\cos 3x} = \dfrac12 (\tan A - \tan B)$ . If we substitute $x$ as some common values, like say $45^\circ, 60^\circ, 120^\circ$ , we can see that the trigonometric identity holds true when $A = 3x, B = x$ , this suggests that

$\dfrac{\sin x}{\cos 3x} = \dfrac12 ( \tan 3x - \tan x )$

is a trigonometric identity. To make sure our assumption is correct, we need to prove that it is true first.

Claim : $\dfrac{\sin x}{\cos 3x} = \dfrac12 ( \tan 3x - \tan x )$ is a trigonometric identity.

Proof : Proving this identity is equivalent to proving that $2 \sin x =\cos 3x (\tan 3x - \tan x)$ is true.

Looking at the RHS, we have $\cos 3x (\tan 3x - \tan x) = \sin 3x - \tan x \cos 3x$ . By triple angle identities , we know that

$\sin(3x) = -4\sin^3 x + 3\sin x \qquad \qquad \cos(3x) =4\cos^3 x - 3\cos x \;$

Then, $\begin{aligned} RHS &=& \sin 3x - \tan x \cos 3x \\ &=& (-4\sin^3 + 3\sin x) + \dfrac{\sin x}{\cos x}(4\cos^3 x - 3\cos x) \\ &=& -4\sin^3 + 3\sin x + 4\cos^3 x \sin x + 3\sin x \\ &=& 6\sin x - 4\sin x ( \cos^2 x + \sin^2 x)\\ &=& 6\sin x - 4\sin x (1) \\ &= &2\sin x = LHS \quad \blacksquare \end{aligned}$

Now back to the sum, we now know that $S$ can also be expressed as $\displaystyle \sum_{n=0}^9 \dfrac12 \left [\tan(3^{n+1} y^\circ) - \tan(3^n y ^\circ) \right ] = \dfrac12 \sum_{n=0}^9 [ f(m+1) - f(m) ]$ , where $f(m) = \tan(3^m \cdot y^\circ)$ . This clearly shows that $S$ telescopes and is equal to:

$S \; = \; \dfrac12 [ f(10) - f(0) ] \; =\; \dfrac12 [ \tan(3^{10} \cdot y^\circ) - \tan (y^\circ) ]\; =\; \dfrac12 [ \tan(3^{10} \cdot 2^\circ) - \tan 2^\circ] \; .$

Since we want to minimize the sum of angles $\alpha$ and $\beta$ , then we need to find the minimum positive integers of $\alpha$ and $\beta$ such that $\tan \alpha = \tan(3^{10} \cdot 2^\circ)$ and $\tan \beta = \tan 2$ .

It is obvious that $\min(\beta) = 2$ . And since $g(x) = \tan x$ is a periodic function with a fundamental period of $180^\circ$ , then $\min( \alpha) = (3^{10} \cdot 2 ) \bmod{180} = 18$ .

Our answer is $\min (\alpha ) + \min(\beta ) = 18 + 2 = \boxed{20}$ .