Question 1206981
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How do I find the trigonometric form of {{{6sqrt(6)-i }}} in angles with two decimal places?
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I want to exchange pleasantries with Edwin.


In his post 
https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.1206982.html 
Edwin completed the simplification which I omitted to carry out in my solution under that link.
Thanks to Edwin for this.


Here I wish to correct the annoying typo made by Edwin in his post for the current problem.


So, I copy here Edwin' solution and make the necessary corrections in it.


. . . . . . . . . . . . . . . . 


<pre>
{{{A + Bi}}}

{{{r=sqrt(A^2+B^2)}}}

{{{theta}}}{{{""=""}}}{{{matrix(1,9,angle,whose,tangent,is,B/A,in,quadrant,of,"(A,B)")}}}

{{{matrix(1,2,trig,form)}}}{{{""=""}}}{{{r(cos(theta)^""+i*sin(theta))}}}

{{{6sqrt(6) -i }}}, {{{A=6sqrt(6)}}}, {{{B=-1}}}   <<<---=== the correction is here: B= -1.
                                       From here, I support this change to the end without noticing.

{{{r=sqrt((6sqrt(6))^2+1^2)}}
{{{r=sqrt(36(6)+1)}}}
{{{r=sqrt(216+1)}}}
{{{r=sqrt(217)}}}

{{{theta}}}{{{""=""}}}{{{matrix(1,9,angle,whose,tangent,is,-1/(6sqrt(6)),in,quadrant,of,

(matrix(1,3,6sqrt(6),",",-1)))}}}

The quadrant in QIV since A is positive, while B is negative.

In degrees:

{{{theta}}}{{{""=""}}}{{{-3.892484484^o}}}


{{{matrix(1,2,trig,form)}}}{{{""=""}}}{{{sqrt(217)(cos(-3.892484484^o)^""+i*sin(-3.892484484^o))}}} = 

            {{{""=""}}}{{{sqrt(217)(cos(3.892484484^o)^""-i*sin(3.892484484^o))}}}

In radians:

{{{theta}}}{{{""=""}}}{{{-0.0679366703}}}

{{{matrix(1,2,trig,form)}}}{{{""=""}}}{{{sqrt(217)(cos(-0.0679366703)^""+i*sin(-0.0679366703))}}}

           {{{""=""}}}{{{sqrt(217)(cos(0.0679366703)^""-i*sin(0.0679366703))}}}.


Now round off as you were told.

ikleyn</pre>