SOLUTION: How do I find the trigonometric form of {{{6sqrt(6)-i }}} in angles with two decimal places? Thank you.

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Question 1206981: How do I find the trigonometric form of 6sqrt%286%29-i+ in angles with two decimal places?
Thank you.
Found 2 solutions by Edwin McCravy, ikleyn:
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
A+%2B+Bi

r=sqrt%28A%5E2%2BB%5E2%29

theta%22%22=%22%22

matrix%281%2C2%2Ctrig%2Cform%29%22%22=%22%22r%28cos%28theta%29%5E%22%22%2Bi%2Asin%28theta%29%29

6sqrt%286%29+-i+, A=6sqrt%286%29, B=1

r=sqrt%28%286sqrt%286%29%29%5E2%2B1%5E2%29%7D%7D%0D%0A%7B%7B%7Br=sqrt%2836%286%29%2B1%29
r=sqrt%28216%2B1%29
r=sqrt%28217%29

theta%22%22=%22%22

The quadrant in QI since A and B are both positive.

In degrees:

theta%22%22=%22%223.892484484%5Eo

matrix%281%2C2%2Ctrig%2Cform%29%22%22=%22%22sqrt%28217%29%28cos%283.892484484%5Eo%29%5E%22%22%2Bi%2Asin%283.892484484%5Eo%29%29

In radians:

theta%22%22=%22%220.0679366703

matrix%281%2C2%2Ctrig%2Cform%29%22%22=%22%22sqrt%28217%29%28cos%280.0679366703%29%5E%22%22%2Bi%2Asin%280.0679366703%29%29

Now round off as you were told.

Edwin

Answer by ikleyn(52855) About Me  (Show Source):
You can put this solution on YOUR website!
.
How do I find the trigonometric form of 6sqrt%286%29-i+ in angles with two decimal places?
~~~~~~~~~~~~~~~~~~~~~~~~~~

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.

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

A+%2B+Bi

r=sqrt%28A%5E2%2BB%5E2%29

theta%22%22=%22%22

matrix%281%2C2%2Ctrig%2Cform%29%22%22=%22%22r%28cos%28theta%29%5E%22%22%2Bi%2Asin%28theta%29%29

6sqrt%286%29+-i+, A=6sqrt%286%29, B=-1   <<<---=== the correction is here: B= -1.
                                       From here, I support this change to the end without noticing.

r=sqrt%28%286sqrt%286%29%29%5E2%2B1%5E2%29%7D%7D%0D%0A%7B%7B%7Br=sqrt%2836%286%29%2B1%29
r=sqrt%28216%2B1%29
r=sqrt%28217%29

theta%22%22=%22%22

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

In degrees:

theta%22%22=%22%22-3.892484484%5Eo


matrix%281%2C2%2Ctrig%2Cform%29%22%22=%22%22sqrt%28217%29%28cos%28-3.892484484%5Eo%29%5E%22%22%2Bi%2Asin%28-3.892484484%5Eo%29%29 = 

            %22%22=%22%22sqrt%28217%29%28cos%283.892484484%5Eo%29%5E%22%22-i%2Asin%283.892484484%5Eo%29%29

In radians:

theta%22%22=%22%22-0.0679366703

matrix%281%2C2%2Ctrig%2Cform%29%22%22=%22%22sqrt%28217%29%28cos%28-0.0679366703%29%5E%22%22%2Bi%2Asin%28-0.0679366703%29%29

           %22%22=%22%22sqrt%28217%29%28cos%280.0679366703%29%5E%22%22-i%2Asin%280.0679366703%29%29.


Now round off as you were told.

ikleyn