Question

Write the given expression as an algebraic expression in x. tan(2 cos^-1(x))

Answer 1

$\bf tan\left[ 2cos^{-1}(x) \right]\implies tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right] \\\\\\ \textit{if we say }cos^{-1}\left( \frac{x}{1} \right)=\theta\textit{ that means }tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right]\iff tan(2\theta)\\\\ -----------------------------\\\\ cos^{-1}\left( \frac{x}{1} \right)=\theta\implies cos(\theta)=\cfrac{x}{1}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}\\\\ -----------------------------$

$\bf \textit{again, using the pythagorean theorem to get the opposite side} \\\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{1^2-x^2}=b \\\\\\ \pm\sqrt{1-x^2}=b\\\\ -----------------------------\\\\ tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\implies tan(2\theta)=\cfrac{2\cdot \frac{\pm\sqrt{1-x^2}}{x}}{1-\left( \frac{\pm\sqrt{1-x^2}}{x} \right)^2}$

$\bf tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{1-\frac{1-x}{x}}\implies tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{\frac{x-1+x}{x}} \\\\\\ tan(2\theta)=\cfrac{\pm2\sqrt{1-x^2}}{x}\cdot \cfrac{x}{2x-1} \implies tan(2\theta)=\cfrac{\pm 2\sqrt{1-x^2}}{2x-1}$