How do you write the trigonometric form into a complex number in standard form #6(cos((5pi)/12)+isin((5pi)/12))#?

1 Answer
Aviv S.
Mar 25, 2018

The complex number in standard form is #(3sqrt6-3sqrt2)/2+(3sqrt6+3sqrt2)/2i#.

Explanation:

To convert from trig form to standard form, simply compute the trig functions' values and expand the multiplication.

First, let's find the #sin# and #cos# of #(5pi)/12# using the respective angle-sum identities:

#sin(A+B)=sinAcosB+cosAsinB#

#cos(A+B)=cosAcosB-sinAsinB#

We can figure out that #(5pi)/12# is the sum of #pi/6# and #pi/4#:

#color(white)=pi/6+pi/4#

#=(2pi)/12+pi/4#

#=(2pi)/12+(3pi)/12#

#=(2pi+3pi)/12#

#=(5pi)/12#

Now we can use those angle sum formulae.

Here is a unit circle to remind us of some trig values:

enter image source here

Here's computing #sin((5pi)/12)#:

#color(white)=sin((5pi)/12)#

#=sin((2pi)/12+(3pi)/12)#

#=sin(pi/6+pi/4)#

#=sin(pi/6)cos(pi/4)+cos(pi/6)sin(pi/4)#

#=1/2*sqrt2/2+sqrt3/2*sqrt2/2#

#=sqrt2/4+sqrt6/4#

#=(sqrt6+sqrt2)/4#

Here's computing #cos((5pi)/12)#:

#color(white)=cos((5pi)/12)#

#=cos((2pi)/12+(3pi)/12)#

#=cos(pi/6+pi/4)#

#=cos(pi/6)cos(pi/4)-sin(pi/6)sin(pi/4)#

#=sqrt3/2*sqrt2/2-1/2*sqrt2/2#

#=sqrt6/4-sqrt2/4#

#=(sqrt6-sqrt2)/4#

Now we can plug in the values to the trig form of the complex number:

#color(white)=6(cos((5pi)/12)+isin((5pi)/12))#

#=6((sqrt6-sqrt2)/4+i*(sqrt6+sqrt2)/4)#

#=6*(sqrt6-sqrt2)/4+6*i*(sqrt6+sqrt2)/4#

#=3*(sqrt6-sqrt2)/2+3*i*(sqrt6+sqrt2)/2#

#=(3sqrt6-3sqrt2)/2+(3sqrt6+3sqrt2)/2i#

That's it. Hope this helped!

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