Algebra + trigo problems

Titu's leema,

$\dfrac{a_{1}^{2}}{1} + ......... \dfrac{a_{2015}^{2}}{1} \geq \dfrac{(a_{1} + .... a_{2015})^{2}}{1 + 1 + ...... 2015times}$

minmum value $x = \dfrac{1}{2015}$

$60450 = \dfrac{120900}{2} , \dfrac{60450}{2015} = 60$

$sin60^{\circ} = cos30^{\circ}$

$\text{The key technique here is to use Titu's Lemma.Titu's Lemma works this way}$ $:\dfrac{a_1^{2}}{b_1}+\dfrac{a_2^{2}}{b_2}+\dfrac{a_3^{2}}{b_n}+....\dfrac{a_n^{2}}{b_n}\geq \dfrac{(a_1+a_2+a_3+......a_n)^{2}}{b_1+b_2+b_3+.....b_n}.$ $\text{In this case,}$ $\dfrac{a_1^{2}}{1}+\dfrac{a_2^{2}}{1}+....\dfrac{a_{2015}^{2}}{1}\geq \dfrac{(a_1+a_2+.......a_{2015})^{2}}{2015(2015\ times\ 1)}.$ $\text{But it is given that}$ $a_1+a_2+.......a_{2015}=1$ $\text{substituting this value gives,}$ $\dfrac{a_1^{2}}{1}+\dfrac{a_2^{2}}{1}+....\dfrac{a_{2015}^{2}}{1}\geq \dfrac{1}{2015}.$ $\text{Thus,the minimum value of the given expression is,}$ $\dfrac{1}{2015}.$ $\text{Now,}$ $\ \dfrac{120900}{2015}=60\$ $\text{and}$ $\ \dfrac{60450}{2015}=30.\$ $\text{Thus,the required answer}$ $=\sin{60}-\cos{30}=0.$

Applying Cauchy-Schwarz Inequality:

$\quad (\sum{a_k^2})(\sum{1^2})\geq (\sum{a_k * 1})^2$

$\Rightarrow (\sum{a_k^2})(2015)\geq (1)$

$\Rightarrow (\sum{a_k^2})\geq (\frac { 1 }{ 2015 })$

$Therefore,$

$\quad E_{min} = 2015 \\$

$\sin { \frac { 120900 }{ 2015 } } -\cos { \frac { 60450 }{ 2015 } } = \quad \sin { 60 } -\cos { 30\quad = } \quad \boxed { 0 }$

If a(k)>0, then the average value of a(k) for any k∈[1,2015] is 1/2015. Thus E=2015*1/(2015²)=1/2015 and hence x=1/2015. sin(120900/2015)-cos(60450/2015)= sin(60)-cos(30)=√3/2-√3/2=0 QED.

It is possible to solve this by using the Cauchy Schwarz Inequality and making up another sequence which comprises of only 1.