rewrite \\( \\sin \\left( 2 \\tan ^{-1} \\frac{u}{3} \\right) \\) as an algebraic expression in \\( u…

Explanation:

Step1: Use the double - angle formula

The double - angle formula for sine is (\sin(2\theta)=2\sin\theta\cos\theta). Let (\theta = \tan^{- 1}\frac{u}{3}), then (\sin\left(2\tan^{-1}\frac{u}{3}\right)=2\sin\left(\tan^{-1}\frac{u}{3}\right)\cos\left(\tan^{-1}\frac{u}{3}\right)).

Step2: Use the right - triangle relationship

If (\theta=\tan^{-1}\frac{u}{3}), then (\tan\theta=\frac{u}{3}). In a right - triangle, if (\tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{u}{3}), by the Pythagorean theorem, the hypotenuse (r=\sqrt{u^{2}+9}). So (\sin\theta=\frac{u}{\sqrt{u^{2}+9}}) and (\cos\theta=\frac{3}{\sqrt{u^{2}+9}}).

Step3: Substitute into the double - angle formula

Substitute (\sin\theta=\frac{u}{\sqrt{u^{2}+9}}) and (\cos\theta=\frac{3}{\sqrt{u^{2}+9}}) into (2\sin\theta\cos\theta). We get (2\times\frac{u}{\sqrt{u^{2}+9}}\times\frac{3}{\sqrt{u^{2}+9}}=\frac{6u}{u^{2}+9}).

Answer:

(\frac{6u}{u^{2}+9})