Given that
[tex]x=6\tan\theta \\ \\ \Rightarrow\tan\theta= \frac{x}{6} [/tex]
Consider a right triangle with angle θ, the side opposite the angle θ is x and that side adjacent angle x is 6.
Then the hypothenus is given by [tex] \sqrt{x^2+36} [/tex]
[tex]\sin{ \frac{2\theta}{4}} =\sin{2\left( \frac{\theta}{2} \right)} \\ \\ =2\sin{\frac{\theta}{2}}\cos{\frac{\theta}{2}}=2\cdot \sqrt{\frac{1-\cos{\theta}}{2}} \cdot \sqrt{\frac{1+\cos{\theta}}{2}} \\ \\ =\sqrt{1-\cos^2\theta}[/tex]
Recall that [tex]\cos\theta= \frac{adjacent}{hypothenuse} = \frac{6}{\sqrt{x^2+36}} [/tex]
Therefore,
[tex] \frac{\theta}{2} -\sin \frac{2\theta}{4} = \frac{1}{2} \tan^{-1}
\frac{x}{6} -\sqrt{1-\cos^2\theta} \\ \\ =\frac{1}{2} \tan^{-1}
\frac{x}{6}-\sqrt{1-\left(\frac{6}{\sqrt{x^2+36}}\right)^2}=\frac{1}{2}
\tan^{-1} \frac{x}{6}-\sqrt{1-\frac{36}{x^2+36}} \\ \\ =\frac{1}{2}
\tan^{-1} \frac{x}{6}-\sqrt{\frac{x^2}{x^2+36}}=\bold{\frac{1}{2}
\tan^{-1} \frac{x}{6}-\frac{x}{\sqrt{x^2+36}}}[/tex]
