Respuesta :
first of all let's recall that sin(2θ) => 2sin(θ)cos(θ).
let's also recall that our angle is between 0 < θ < π/2, or namely is in the First Quadrant, meaning that the adjacent(x) and opposite(y) sides are both positive.
well, we know that the sec(θ) = 9/4, let's recall that secant is the reciprocal of cosine, namely if secant is 9/4, then cosine is the same thing upside-down, namely cos(θ) = 4/9.
well, what's the sine anyway?
[tex]sec(\theta ) = \cfrac{\stackrel{hypotenuse}{9}}{\stackrel{adjacent}{4}}\impliedby \textit{let's find the opposite side}\\\\\\\textit{using the pythagorean theorem}\\\\c^2=a^2+b^2\implies c^2-a^2=b^2\implies \pm\sqrt{c^2-a^2}=b\qquad\begin{cases}c=hypotenuse\\a=adjacent\\b=opposite\\\end{cases}\\\\\\b = \pm\sqrt{9^2-4^2}\implies b = \pm\sqrt{65}\implies \stackrel{\textit{1st Quadrant}}{b = +\sqrt{65}}\\\\[-0.35em]\rule{34em}{0.25pt}[/tex]
[tex]sin(\theta ) = \cfrac{\stackrel{opposite}{\sqrt{65}}}{\stackrel{hypotenuse}{9}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\sin(2\theta )\implies 2sin(\theta )cos(\theta )\implies 2\cdot \cfrac{\sqrt{65}}{9}\cdot \cfrac{4}{9}\implies \cfrac{8\sqrt{65}}{81}[/tex]
