Sink Cost Tank Cots

First, let's do some transformations:

$\displaystyle \frac{2 \ tan \ x}{1+tan^2 \ x} \times \frac{cos^2 \ x}{cos^2 \ x}=2 \ sin \ x \ cos \ x=sin \ 2x$

$\displaystyle \frac{sin^{2y} \ x}{cos^{\frac{y^2}{2}} \ x}+\frac{cos^{2y} \ x}{sin^{\frac{y^2}{2}} \ x}=\frac{sin^{2y} \ x}{cos^{\frac{y^2}{2}} \ x} \times \frac{sin^{\frac{y^2}{2}} \ x}{sin^{\frac{y^2}{2}} \ x} +\frac{cos^{2y} \ x}{sin^{\frac{y^2}{2}} \ x} \times \frac{cos^{\frac{y^2}{2}} \ x}{cos^{\frac{y^2}{2}} \ x}\implies$

$\implies \displaystyle \frac{sin^{2y} \ x}{cos^{\frac{y^2}{2}} \ x}+\frac{cos^{2y} \ x}{sin^{\frac{y^2}{2}} \ x}= \frac{sin^{ \frac{y^2}{2}+2y} \ x+cos^{\frac{y^2}{2}+2y} \ x}{ \left(sin \ x \ cos \ x \right)^{\frac{y^2}{2}}}\implies$

$\implies \displaystyle \frac{sin^{2y} \ x}{cos^{\frac{y^2}{2}} \ x}+\frac{cos^{2y} \ x}{sin^{\frac{y^2}{2}} \ x}= \frac{2^{\frac{y^2}{2}}\left(sin^{ \frac{y^2}{2}+2y} \ x+cos^{\frac{y^2}{2}+2y} \ x\right)}{ \left(2 \ sin \ x \ cos \ x \right)^{\frac{y^2}{2}}}\implies$

$\implies \displaystyle \frac{sin^{2y} \ x}{cos^{\frac{y^2}{2}} \ x}+\frac{cos^{2y} \ x}{sin^{\frac{y^2}{2}} \ x}= \frac{2^{\frac{y^2}{2}}\left(sin^{ \frac{y^2}{2}+2y} \ x+cos^{\frac{y^2}{2}+2y} \ x\right)}{ \left(sin \ 2x \right)^{\frac{y^2}{2}}}$

For equality to hold:

$\displaystyle\frac{2^{\frac{y^2}{2}}\left(sin^{ \frac{y^2}{2}+2y} \ x+cos^{\frac{y^2}{2}+2y} \ x\right)}{ \left(sin \ 2x \right)^{\frac{y^2}{2}}}=sin \ 2x \implies$

$\implies \displaystyle 2^{\frac{y^2}{2}} \ \left(sin^{ \frac{y^2}{2}+2y} \ x+cos^{\frac{y^2}{2}+2y} \ x \right)= \left(sin \ 2x \right)^{\frac{y^2}{2}+1}$

Now, define $f(x)$ and $g(x)$ as:

$\displaystyle f(x)=2^{\frac{y^2}{2}} \ \left(sin^{ \frac{y^2}{2}+2y} \ x+cos^{\frac{y^2}{2}+2y} \ x \right)$

$\displaystyle g(x)= \left(sin \ 2x \right)^{\frac{y^2}{2}+1}$

Derivating both $f(x)$ and $g(x)$ and equating to zero, one can find the critical points of the functions:

$\displaystyle f'(x)=2^{\frac{y^2}{2}} \ \left(\frac{y^2}{2}+2y\right) \ \left(sin^{\frac{y^2}{2}-2y-1} \ x \ cos \ x-cos^{\frac{y^2}{2}-2y-1} \ x \ sin \ x \right)$

$\displaystyle f'(x)=0 \implies sin^{\frac{y^2}{2}-2y-1} \ x \ cos \ x=cos^{\frac{y^2}{2}-2y-1} \ x \ sin \ x$

$\displaystyle f'(x)=0 \implies sin^{\frac{y^2}{2}-2y-2} \ x=cos^{\frac{y^2}{2}-2y-2} \ x \implies x=\frac{(2k+1) \pi}{4}, k \in \mathbb{Z}$

$\displaystyle g'(x)=\left(\frac{y^2}{2}+1\right) \ \left(sin \ 2x \right)^{\frac{y^2}{2}} \ 2 \ cos \ 2x$

$\displaystyle g'(x)=0 \implies \left(sin \ 2x \right)^{\frac{y^2}{2}}=0 \vee cos \ 2x=0\implies x=0 \vee x=\frac{(2k+1) \pi}{4}, k \in \mathbb{Z}$

Second derivative test shows us how do $f(x)$ and $g(x)$ increase or decrease, and that $x=\frac{\pi}{4}$ (since $x \in \left(0, \frac{\pi}{2}\right)$ ) is a maximum for $g(x)$ and a minimum for $g(x)$ ( $x=0$ is an inflection point).

Sketching both graphs, for any $y$ (in fact, the variation of $y$ defines a family of curves on the plane $xy$ ), $f(x)$ is greater than $g(x)$ everywhere, except for some $y$ , at minimum of $f(x)$ and maximum of $g(x)$ , where they're equal, verifying equality proposed. Substituting, gives:

$\displaystyle \frac{2 \ tan \ \frac{\pi}{4}}{1+tan^2 \ \frac{\pi}{4}}=\frac{sin^{2y} \ \frac{\pi}{4}}{cos^{\frac{y^2}{2}} \ \frac{\pi}{4}}+\frac{cos^{2y} \ \frac{\pi}{4}}{sin^{\frac{y^2}{2}} \ \frac{\pi}{4}}$

$\implies \displaystyle \frac{2.1}{1+1^2}=\frac{2^{-y}}{2^{-\frac{y^2}{4}}}+\frac{2^{-y}}{2^{-\frac{y^2}{4}}} \implies 2^0=2^{\frac{y^2}{4}-y+1} \implies y=2$

As there is only the solution $(\frac{\pi}{4}, 2)$ , $\alpha$ required for the problem is $\alpha=\frac{2 \pi}{3}$ . The given $(3n+1)\alpha$ angles happens to have the same trigonometric functions as each one is a complete number of $2 \pi$ rotations from another.

The sum has 100 terms ( $\frac{1-1}{3}=0, \frac{298-1}{3}=99, 99-0+1=100 \text{ terms}$ ) each of which is $sin \ 120^o=\frac{\sqrt{3}}{2}$ . Then, answer is $100 \ \frac{\sqrt{3}}{2}=50 \ \sqrt{3}$ . Thus, $a+b=50+3=\boxed{53}$