What is the Cartesian form of #( 16,(pi )/12#?

To convert the given polar coordinates (16, π/12) to Cartesian coordinates, you can use the following formulas:

[ x = r \cos(\theta) ] [ y = r \sin(\theta) ]

Substituting the given values:

[ x = 16 \cos\left(\frac{\pi}{12}\right) ] [ y = 16 \sin\left(\frac{\pi}{12}\right) ]

Using the values of cosine and sine of π/12 (which is 15 degrees):

[ \cos\left(\frac{\pi}{12}\right) \approx \frac{\sqrt{6}+\sqrt{2}}{4} ] [ \sin\left(\frac{\pi}{12}\right) \approx \frac{\sqrt{6}-\sqrt{2}}{4} ]

Substituting these values into the equations:

[ x \approx 16 \times \frac{\sqrt{6}+\sqrt{2}}{4} ] [ y \approx 16 \times \frac{\sqrt{6}-\sqrt{2}}{4} ]

Simplify:

[ x \approx 4\sqrt{6} + 4\sqrt{2} ] [ y \approx 4\sqrt{6} - 4\sqrt{2} ]

So, the Cartesian coordinates are approximately:

[ \left(4\sqrt{6} + 4\sqrt{2}, 4\sqrt{6} - 4\sqrt{2}\right) ]