Trigonometric form of [2/3(cos(π/2)+isin(π/2))][3(cos(2π/3)+isin(2π/3))]

[CantEven]

Trigonometric form of [2/3(cos(π/2)+isin(π/2))][3(cos(2π/3)+isin(2π/3))]

Perform the operation and leave the result in trigonometric form. (Let 0≤θ<2π).

[2/3(cos(π/2)+isin(π/2))][3(cos(2π/3)+isin(2π/3))]

What is the trig form of this? I really don't get it.

 

[Deleted member 4993]

Perform the operation and leave the result in trigonometric form. (Let 0≤θ<2π).

[2/3(cos(π/2)+isin(π/2))][3(cos(2π/3)+isin(2π/3))]

What is the trig form of this? I really don't get it.

Use:

r[cos(Θ) + i sin(Θ)] = r * e^iΘ

Then use a^m * aⁿ = a^m+n

 

[CantEven]

Use:

r[cos(Θ) + i sin(Θ)] = r * e^iΘ

Then use a^m * aⁿ = a^m+n

This is the answer I ended up with:

=(-2)[cos(-pi/6)+isin(-pi/6)]

Is this right?

 

[stapel]

This is the answer I ended up with:

=(-2)[cos(-pi/6)+isin(-pi/6)]

Is this right?

Does it evaluate to the same decimal (or exact) form?

. . . . .\(\displaystyle \left[\, \dfrac{2}{3}\, \left(\cos\left(\dfrac{\pi}{2}\right)\, +\, i\, \sin\left(\dfrac{\pi}{2}\right)\right)\, \right]\, \left[\, 3\, \left(\cos\left(\dfrac{2\pi}{3}\right)\, +\, i\, \sin\left(\dfrac{2\pi}{3}\right)\right)\, \right]\, \)

. . . . . . . . . .\(\displaystyle =\, \left[\, \dfrac{2}{3}\, (0\, +\, 1i)\, \right]\, \left[\, 3\, \left(-\dfrac{1}{2}\, +\, \dfrac{\sqrt{\strut 3\,}}{2}\, i\right)\, \right]\)

. . . . . . . . . .\(\displaystyle =\, \left[\, \dfrac{2}{3}\, i\, \right]\, \left[\, -\dfrac{3}{2}\, +\, \dfrac{3\, \sqrt{\strut 3\,}}{2}\, i\, \right]\)

. . . . . . . . . .\(\displaystyle =\,-i\, -\, \sqrt{\strut 3\, }\, =\, -\sqrt{\strut 3\,}\, -\, i\)

Your proposed solution, on the other hand:

. . . . .\(\displaystyle (-2)\, \left[\cos\left(-\dfrac{\pi}{6}\right)\, +\, i\, \sin\left(-\dfrac{\pi}{6}\right)\right]\)

. . . . . . . . . .\(\displaystyle =\, (-2)\, \left[\dfrac{\sqrt{\strut 3\,}}{2}\, -\, \dfrac{1}{2}\, i\right]\)

. . . . . . . . . .\(\displaystyle =\, -\sqrt{\strut 3\,}\, +\, i\)

How did you obtain your solution? Please show all of your steps. Thank you!

 

[pka]

Perform the operation and leave the result in trigonometric form. (Let 0≤θ<2π).
[2/3(cos(π/2)+isin(π/2))][3(cos(2π/3)+isin(2π/3))]

This is the answer I ended up with:
=(-2)[cos(-pi/6)+isin(-pi/6)] Is this right?

Note that the requested range is \(\displaystyle 0\le\theta\le2\pi \).
POLAR FORM: \(\displaystyle x+y\bf{i}=r(\cos(\theta)+\bf{i}\sin(\theta) \) where \(\displaystyle r=\sqrt{x^2+y^2}~\&~\theta=\text{Arg}(x+y\bf{i}) \)
Because \(\displaystyle r \) is a distance it is not negative. Yet you have \(\displaystyle -2 \). WHY?
\(\displaystyle z=r(\cos(\theta)+\bf{i}\sin(\theta)~\&~w=s(\cos( \zeta)+\bf{i}\sin(\zeta)) \)
Then \(\displaystyle z\cdot w= r\cdot s(\cos(\theta+\zeta)+\bf{i}\sin(\theta+\zeta) \)
Be sure that you have \(\displaystyle 0\le\theta+\zeta \le 2\pi \)