Question

What is the product of 3√2cis(π/12) and 2√5cis(4π/3) ?


___cis(___)

Answer 1

Answer:

Did you mean $cos$ instead of $cis$ ?

If that's the case, then the answer is:

$3\sqrt{2} cos(\frac{\pi}{12})*2\sqrt{5} cos(\frac{4 \pi}{3}) =6\sqrt{10} cos(0.66043941112\pi) \approx -9.16$

Step-by-step explanation:

As you may know:

$\sqrt{ab} =\sqrt{a}* \sqrt{b}$

So:

$3\sqrt{2} *2\sqrt{5} =3*2*\sqrt{2*5} =6\sqrt{10}$

Now:

$cos(\frac{\pi}{4}) =\frac{\sqrt{6} +\sqrt{2} }{4} \\\\and\\\\cos(\frac{4\pi}{3} )=-\frac{1}{2}$

So:

$cos(\frac{\pi}{12} )*cos(\frac{4\pi}{3} )=(\frac{\sqrt{6}+\sqrt{2} }{4} )*(-\frac{1}{2} ) =\frac{-\sqrt{6}-\sqrt{2} }{8}$

Now, using the inverse cosine:

$arccos(\frac{-\sqrt{6}-\sqrt{2} }{8})=2.074831602=0.66043941112\pi$

Therefore, the equivalent expression for the  product of 3√2cis(π/12) and 2√5cis(4π/3) is:

$3\sqrt{2} cos(\frac{\pi}{12})*2\sqrt{5} cos(\frac{4 \pi}{3}) =6\sqrt{10} cos(0.66043941112\pi) \approx -9.16$