'Di'gonometry

Relevant wiki: Fundamental Trigonometric Identities - Problem Solving - Medium

$\begin{aligned} \frac {\sin^4 \theta}5 + \frac {\cos^4 \theta}7 & = \frac 1{12} \\ 84 \sin^4 \theta + 60 \cos^4 \theta & = 35 \\ 84 \sin^4 \theta + 60 (1-\sin^2 \theta)^2 & = 35 \\ 84 \sin^4 \theta + 60 (1-2\sin^2 \theta + \sin^4 \theta) & = 35 \\ 144 \sin^4 \theta - 120\sin^2 \theta + 25 & = 0 \\ (12 \sin^2 \theta - 5)^2 & = 0 \\ \implies \sin^2 \theta & = \frac 5{12} \\ \implies \cos^2 \theta & = 1 - \frac 5{12} = \frac 7{12} \end{aligned}$

Now, we have:

$\begin{aligned} \frac {\sin^{2016} \theta}{5^{1007}} + \frac {\cos^{2016} \theta}{7^{1007}} & = \frac {\left(\sin^2 \theta\right)^{1008}}{5^{1007}} + \frac {\left(\cos^2 \theta\right)^{1008}}{7^{1007}} \\ & = \frac {\left(\frac 5{12} \right)^{1008}}{5^{1007}} + \frac {\left(\frac 7{12} \right)^{1008}}{7^{1007}} \\ & = \frac 5{12^{1008}} + \frac 7{12^{1008}} \\ & = 12^{-1007} \end{aligned}$

$\implies a = \boxed{-1007}$

Relevant wiki: Titu's Lemma

Applying Titu 's lemma :

$\dfrac{(\sin^2\theta)^2}{5}+\dfrac{(\cos^2\theta)^2}{7}\ge \dfrac{\left(\overbrace{\sin^2\theta+\cos^2\theta}^{\color{#D61F06}{1}}\right)^2}{5+7}=\dfrac1{12}$ So we see according to question equality holds in above inequality which is the case when $\dfrac{\sin^2\theta}5=\dfrac{\cos^2\theta}7$ or $\sin^2\theta =\dfrac{5}{12},\cos^2\theta=\dfrac7{12}$ . Direct substitution in required expression gives: $\dfrac{\left(\frac{5}{12}\right)^{1008}}{5^{1007}}+\dfrac{\left(\frac{7}{12}\right)^{1008}}{5^{1007}}=12^{-1008}(5+7)=12^{-1007}$

Relevant wiki: Fundamental Trigonometric Identities - Problem Solving - Medium

The given expression is of the form $\dfrac{{\sin}^4 \theta}{x} + \dfrac{{\cos}^4 \theta}{y} = \dfrac{1}{x + y}$ .

$\begin{aligned} \dfrac{{\sin}^4 \theta}{x} + \dfrac{{\cos}^4 \theta}{y} & = \dfrac{1}{x + y}\\ \\ (x + y)\left(\dfrac{{\sin}^4 \theta}{x} + \dfrac{{\cos}^4 \theta}{y}\right) & = 1\\ \\ {\sin}^4 \theta + {\cos}^4 \theta + \dfrac{y}{x}{\sin}^4 \theta + \dfrac{x}{y}{\cos}^4 \theta & = {\sin}^4 \theta + {\cos}^4 \theta + 2{\sin}^2 \theta {\cos}^2 \theta\\ \\ \dfrac{y}{x}{\sin}^4 \theta + \dfrac{x}{y}{\cos}^4 \theta - 2{\sin}^2 \theta {\cos}^2 \theta & = 0\\ \\ {\left(\sqrt{\dfrac{y}{x}} {\sin}^2 \theta - \sqrt{\dfrac{x}{y}} {\cos}^2 \theta\right)}^2 & = 0\\ \\ \sqrt{\dfrac{y}{x}} {\sin}^2 \theta - \sqrt{\dfrac{x}{y}} {\cos}^2 \theta & = 0\\ \\ \sqrt{\dfrac{y}{x}} {\sin}^2 \theta & = \sqrt{\dfrac{x}{y}} {\cos}^2 \theta\\ \\ \dfrac{{\sin}^2 \theta}{{\cos}^2 \theta} & = \dfrac{x}{y}\\ \\ \dfrac{{\sin}^2 \theta}{x} & = \dfrac{{\cos}^2 \theta}{y}\\ \\ \dfrac{{\sin}^2 \theta}{x} & = \dfrac{{\cos}^2 \theta}{y} = \dfrac{{\sin}^2 \theta + {\cos}^2 \theta}{a + b} \left(\text{Rule of proportions}\right)\\ \\ \dfrac{{\sin}^2 \theta}{x} & = \dfrac{{\cos}^2 \theta}{y} = \dfrac{1}{a + b}\\ \\ {\sin}^2 \theta = \dfrac{x}{x+y} & \left(\text{and}\right) {\cos}^2 \theta = \dfrac{y}{x+y} \longrightarrow \boxed{1} \end{aligned}$

Now, we see that $2016 = 4×504$ and $1007 = 2×504 - 1$ . So, the expression we have to evaluate can be written as $\dfrac{{\sin}^{4n} \theta}{x^{2n - 1}} + \dfrac{{\cos}^{4n} \theta}{y^{2n - 1}}$ .

$\begin{aligned} \dfrac{{\sin}^{4n} \theta}{x^{2n - 1}} + \dfrac{{\cos}^{4n} \theta}{y^{2n - 1}} & = \dfrac{{\left(\sin^2 \theta\right)}^{2n}}{x^{2n - 1}} + \dfrac{{\left(\cos^2 \theta\right)}^{2n}}{y^{2n - 1}}\\ \\ \text{Plugging in the values from} \ \boxed{1}:-\\ \\ & = \dfrac{1}{x^{2n - 1}}{\left(\dfrac{x}{x+y}\right)}^{2n} + \dfrac{1}{y^{2n - 1}}{\left(\dfrac{y}{x+y}\right)}^{2n}\\ \\ & = \dfrac{x}{{(x+y)}^{2n}} + \dfrac{y}{{(x+y)}^{2n}}\\ \\ & = \dfrac{x+y}{{(x+y)}^{2n}}\\ \\ & = \dfrac{1}{{(x+y)}^{2n-1}} \end{aligned}$ ]

So, plugging in $x=5$ and $y=7$ and $n=504$ , we get:-
$\begin{aligned} \dfrac{{\sin}^{2016} \theta}{5^{1007}} + \dfrac{{\cos}^{2016} \theta}{7^{1007}} & = \dfrac{1}{{(5+7)}^{2×504 - 1}}\\ \\ & = \dfrac{1}{{12}^{1007}}\\ \\ & = {12}^{-1007}\\ \\ \therefore a & = \color{#3D99F6}{\boxed{-1007}} \end{aligned}$

Almost same as Mr. Chew-Seong Cheong .
$\sin^2 \theta+\cos^2 \theta=1, so~we~have,\\ \begin{aligned} \frac {\sin^4 \theta}5 + \frac {\cos^4 \theta}7 & = \frac 1{12} \\ 7 \sin^4 \theta + 5 (1-\cos^2 \theta)^2 & = \frac{35}{12} \\ 144 \sin^4 \theta - 120 \sin^2 \theta +60& = 35 \\ (12 \sin^2 \theta)^2 - 10(12\sin^2 \theta) + 25 & = 0 \\ (12 \sin^2 \theta - 5)^2 & = 0 \\ \implies \sin^2 \theta & = \frac 5{12} \\ \implies \cos^2 \theta & = 1 - \frac 5{12} = \frac 7{12} \\ \frac {\sin^{2016} \theta}{5^{1007}} + \frac {\cos^{2016} \theta}{7^{1007}} & = \frac {\left(\sin^2 \theta\right)^{1008}}{5^{1007}} + \frac {\left(\cos^2 \theta\right)^{1008}}{7^{1007}} \\ & = \frac {\left(\frac 5{12} \right)^{1008}}{5^{1007}} + \frac {\left(\frac 7{12} \right)^{1008}}{7^{1007}} \\ & = \frac 5{12^{1008}} + \frac 7{12^{1008}} \\ & = 12^{-1007} \end{aligned} \\ \therefore~ a=\Large \color{#D61F06}{-1007} .$