Question

rewrite \\( \cos \left( \tan ^ { - 1 } \frac { w } { \sqrt { 9 - w ^ { 2 } } } \
ight) \\) as an algebraic expression in \\( w \\).
\\( \cos \left( \tan ^ { - 1 } \frac { w } { \sqrt { 9 - w ^ { 2 } } } \
ight) = \\)

Explanation:

Step1: Let $\theta=\tan^{-1}\frac{w}{\sqrt{9 - w^{2}}}$

By the definition of the inverse - tangent function, $\tan\theta=\frac{w}{\sqrt{9 - w^{2}}}$, where $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$.

Step2: Use the right - triangle relationship

If $\tan\theta=\frac{w}{\sqrt{9 - w^{2}}}=\frac{\text{opposite}}{\text{adjacent}}$, then by the Pythagorean theorem, the hypotenuse $r=\sqrt{w^{2}+(\sqrt{9 - w^{2}})^{2}}=\sqrt{w^{2}+9 - w^{2}} = 3$.

Step3: Find $\cos\theta$

Since $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, and $\theta=\tan^{-1}\frac{w}{\sqrt{9 - w^{2}}}$, we have $\cos(\tan^{-1}\frac{w}{\sqrt{9 - w^{2}}})=\frac{\sqrt{9 - w^{2}}}{3}$.

Answer:

$\frac{\sqrt{9 - w^{2}}}{3}$