Question

Find the exact value of sin (2θ) given sec θ = 9/4 and 0 < θ < π/2

Answer 1

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?

`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√(c^2-a^2)=b\qquad\begin{cases}c=hypotenuse\\a=adjacent\\b=opposite\\\end{cases}\\\\\\b = \pm√(9^2-4^2)\implies b = \pm√(65)\implies \stackrel{\textit{1st Quadrant}}{b = +√(65)}\\\\[-0.35em]\rule{34em}{0.25pt}`

`sin(\theta ) = \cfrac{\stackrel{opposite}{√(65)}}{\stackrel{hypotenuse}{9}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\sin(2\theta )\implies 2sin(\theta )cos(\theta )\implies 2\cdot \cfrac{√(65)}{9}\cdot \cfrac{4}{9}\implies \cfrac{8√(65)}{81}`