Question

Determine the values of sin(2a) and tan(2a), given tan(a) = -7/24 and Pi over 2 ≤ 0 ≤ Pi

Answer 1

$\displaystyle \tan{a} = \frac{-7}{24}\\ \tan(\pi - a) = \frac{7}{24}\\ a = \pi - \arctan\left(\frac{7}{24}\right)\\ \begin{aligned} \sin(2a) &= 2\cos{a}\sin{a} \\&= 2\cos\left(\pi - \arctan\left(\frac{7}{24}\right)\right)\sin\left(\pi - \arctan\left(\frac{7}{24}\right)\right) \\&= -2\cos\left(\arctan\left(\frac{7}{24}\right)\right) \sin\left(\arctan\left(\frac{7}{24}\right)\right) \end{aligned}\\ \textsf{Let}\,y = \cos(\arctan(x))\\ y = \cos{u}, x = \tan{u}\\ 1 + x^2 = \sec^2{u}\\ \cos{u} = \frac{1}{\sqrt{1 + x^2}}\\ y = \cos{u} = \frac{1}{\sqrt{1 + x^2}}\\ \begin{aligned} \sin{u} &= \sqrt{1 - \cos^2{u}} \\&= \sqrt{1 - \left(\frac{1}{\sqrt{1 + x^2}}\right)^2} = \frac{x}{\sqrt{1 + x^2}} = \sin(\arctan(x)) \end{aligned}\\ \begin{aligned} \sin(2a) &= -2 \times \frac{1}{\sqrt{1 + \frac{7^2}{24^2}}} \times \frac{7/24}{\sqrt{1 + \frac{7^2}{24^2}}} \\&= \frac{-14 \times 24}{(\sqrt{7^2 + 24^2})(\sqrt{7^2 + 24^2})} \\&= \frac{-14 \times 24}{7^2 + 24^2} = -\frac{336}{625} \end{aligned}\\ \begin{aligned} \tan(2a) &= \frac{2\tan{a}}{1 - \tan^2{a}} \\&= \frac{2 \times -7/24}{1 - \frac{7^2}{24^2}} = -\frac{7}{12} \times \frac{24^2}{24^2 - 7^2} \\&= \frac{-14 \times 24}{24^2 - 7^2} = \frac{-336}{527} \end{aligned}\\$