How do you write the trigonometric form into a complex number in standard form $6 (cos (\frac{5 π}{12}) + i sin (\frac{5 π}{12}))$ $6(cos((5pi)/12)+isin((5pi)/12))$ ?

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How do you write the trigonometric form into a complex number in standard form $6 (cos (\frac{5 π}{12}) + i sin (\frac{5 π}{12}))$ $6(cos((5pi)/12)+isin((5pi)/12))$ ?

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Answer 1 · 2018-03-25T03:09:07

The complex number in standard form is $\frac{3 \sqrt{6} - 3 \sqrt{2}}{2} + \frac{3 \sqrt{6} + 3 \sqrt{2}}{2} i$ $(3sqrt6-3sqrt2)/2+(3sqrt6+3sqrt2)/2i$ .

Explanation:

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

First, let's find the $sin$ $sin$ and $cos$ $cos$ of $\frac{5 π}{12}$ $(5pi)/12$ using the respective angle-sum identities:

$sin (A + B) = sin A cos B + cos A sin B$ $sin(A+B)=sinAcosB+cosAsinB$

$cos (A + B) = cos A cos B - sin A sin B$ $cos(A+B)=cosAcosB-sinAsinB$

We can figure out that $\frac{5 π}{12}$ $(5pi)/12$ is the sum of $\frac{π}{6}$ $pi/6$ and $\frac{π}{4}$ $pi/4$ :

$= \frac{π}{6} + \frac{π}{4}$ $color(white)=pi/6+pi/4$

$= \frac{2 π}{12} + \frac{π}{4}$ $=(2pi)/12+pi/4$

$= \frac{2 π}{12} + \frac{3 π}{12}$ $=(2pi)/12+(3pi)/12$

$= \frac{2 π + 3 π}{12}$ $=(2pi+3pi)/12$

$= \frac{5 π}{12}$ $=(5pi)/12$

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 $sin (\frac{5 π}{12})$ $sin((5pi)/12)$ :

$= sin (\frac{5 π}{12})$ $color(white)=sin((5pi)/12)$

$= sin (\frac{2 π}{12} + \frac{3 π}{12})$ $=sin((2pi)/12+(3pi)/12)$

$= sin (\frac{π}{6} + \frac{π}{4})$ $=sin(pi/6+pi/4)$

$= sin (\frac{π}{6}) cos (\frac{π}{4}) + cos (\frac{π}{6}) sin (\frac{π}{4})$ $=sin(pi/6)cos(pi/4)+cos(pi/6)sin(pi/4)$

$= \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2}$ $=1/2*sqrt2/2+sqrt3/2*sqrt2/2$

$= \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}$ $=sqrt2/4+sqrt6/4$

$= \frac{\sqrt{6} + \sqrt{2}}{4}$ $=(sqrt6+sqrt2)/4$

Here's computing $cos (\frac{5 π}{12})$ $cos((5pi)/12)$ :

$= cos (\frac{5 π}{12})$ $color(white)=cos((5pi)/12)$

$= cos (\frac{2 π}{12} + \frac{3 π}{12})$ $=cos((2pi)/12+(3pi)/12)$

$= cos (\frac{π}{6} + \frac{π}{4})$ $=cos(pi/6+pi/4)$

$= cos (\frac{π}{6}) cos (\frac{π}{4}) - sin (\frac{π}{6}) sin (\frac{π}{4})$ $=cos(pi/6)cos(pi/4)-sin(pi/6)sin(pi/4)$

$= \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2}}{2}$ $=sqrt3/2*sqrt2/2-1/2*sqrt2/2$

$= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$ $=sqrt6/4-sqrt2/4$

$= \frac{\sqrt{6} - \sqrt{2}}{4}$ $=(sqrt6-sqrt2)/4$

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

$= 6 (cos (\frac{5 π}{12}) + i sin (\frac{5 π}{12}))$ $color(white)=6(cos((5pi)/12)+isin((5pi)/12))$

$= 6 (\frac{\sqrt{6} - \sqrt{2}}{4} + i \cdot \frac{\sqrt{6} + \sqrt{2}}{4})$ $=6((sqrt6-sqrt2)/4+i*(sqrt6+sqrt2)/4)$

$= 6 \cdot \frac{\sqrt{6} - \sqrt{2}}{4} + 6 \cdot i \cdot \frac{\sqrt{6} + \sqrt{2}}{4}$ $=6*(sqrt6-sqrt2)/4+6*i*(sqrt6+sqrt2)/4$

$= 3 \cdot \frac{\sqrt{6} - \sqrt{2}}{2} + 3 \cdot i \cdot \frac{\sqrt{6} + \sqrt{2}}{2}$ $=3*(sqrt6-sqrt2)/2+3*i*(sqrt6+sqrt2)/2$

$= \frac{3 \sqrt{6} - 3 \sqrt{2}}{2} + \frac{3 \sqrt{6} + 3 \sqrt{2}}{2} i$ $=(3sqrt6-3sqrt2)/2+(3sqrt6+3sqrt2)/2i$

That's it. Hope this helped!

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How do you write the trigonometric form into a complex number in standard form $6 (cos (\frac{5 π}{12}) + i sin (\frac{5 π}{12}))$ $6(cos((5pi)/12)+isin((5pi)/12))$ ?

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How do you write the trigonometric form into a complex number in standard form 6(cos(5π12)+isin(5π12))6(cos((5pi)/12)+isin((5pi)/12))?

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How do you write the trigonometric form into a complex number in standard form $6 (cos (\frac{5 π}{12}) + i sin (\frac{5 π}{12}))$ $6(cos((5pi)/12)+isin((5pi)/12))$ ?