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What is trigonometric form of complex z = (sin a+cos a)+i(sin a-cos a)?

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sparangello | Student, College Freshman | eNoter

Posted July 5, 2013 at 5:20 PM via web

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What is trigonometric form of complex z = (sin a+cos a)+i(sin a-cos a)?

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aruv | High School Teacher | Valedictorian

Posted July 5, 2013 at 6:13 PM (Answer #1)

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`Z=(sin(a)+cos(a))+i(sin(a)-cos(a))` 

Let

`Z=rcos(theta)+irsin(theta)`

`=(sin(a)+cos(a))+i(sin(a)-cos(a))`

`therefore`

`sin(a)+cos(a)=rcos(theta)`

`sin(a)-cos(a)=rsin(theta)`

squaring and adding

`r^2=(sin(a)+cos(a))^2+(sin(a)-cos(a))^2`

`r^2=2`

`r=sqrt(2)`

Divide `rsin(theta) by rcos(theta)`

so

`tan(theta)=(sin(a)-cos(a))/(sin(a)+cos(a))`

`tan(theta)=(tan(a)-1)/(tan(a)+1)`

`tan(theta)=tan(a-pi/4)`

`theta=a-pi/4`

`Thus```

`Z=sqrt(2)(cos(a-pi/4)+isin(a-pi/4))`

 

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