Question

Find each of the following in rectangular form x+iyand check your results by computer. Remember to save time by doing as much as you can in your head.

$\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathbf{}\mathit{\pi }\mathbf{-}\mathbf{2}\mathit{i}\mathbf{}\mathit{l}\mathit{n}\mathbf{3}\mathbf{)}\mathbf{.}$

Short answer

The rectangular form of the given question is, $\cos (\pi -2i\ln 3).=-\frac{41}{9}$.

Step-by-step solution

Given Information.

The given expression is$\cos (\pi -2i\ln 3).$ .

Meaning of rectangular form.

Representing the complex number in rectangular form means writing the given complex number in the form of x+iy, in which x is the real part and y is the imaginary part.

Step 3: Put the value in the formula.

Use the complex formula $cos\theta =\frac{e^i\theta +e^{-i\theta }}{2}$to rewrite the above expression.

And role="math" localid="1658746545257" $\cos (\pi -2i\ln 3)$can written as,

$$\begin{gathered} \cos (\pi -2i\ln 3)=\cos \pi .\cos (2i\ln 3)+\sin \pi .\sin (2i\ln 3) \\ \cos (\pi -2i\ln 3)=\cos (2i\ln 3) \end{gathered}$$

Now,

role="math" localid="1658747221542" $$\begin{gathered} -\cos (2i\ln 3)=-\frac{e^{\left(2i\ln 3\right)i}+e^{-\left(2i\ln 3\right)i}}{2} \\ =-\frac{e^{\left(-2i\ln 3\right)i}+e^{\left(2i\ln 3\right)i}}{2} \\ =-\frac{e^{\ln 3^{-2}}+e^{\ln 3^{-2}}}{2} \\ =-\frac{{\displaystyle \frac{1}{9}}+9}{2} \\ -\cos (2i\ln 3)=-\frac{1+81}{2\times 9} \\ -\cos (2i\ln 3)=-\frac{82}{2\times 9} \\ -\cos (2i\ln 3)=-\frac{41}{9} \\ so, \\ \cos (2i\ln 3)=-\cos (2i\ln 3) \end{gathered}$$

Substitute the value $-\cos (2i\ln 3)$of in the above equation.

$-\cos (2i\ln 3)=-\frac{41}{9}$

Therefore, the rectangular form of is$-\cos (2i\ln 3)=-\frac{41}{9}$