How can you use trigonometric functions to simplify # 8 e^( ( 19 pi)/12 i ) # into a non-exponential complex number?

Question

How can you use trigonometric functions to simplify #  8 e^(  (  19   pi)/12   i    )     # into a non-exponential complex number?

Violet Anderson · Accepted Answer

The answer is #=2(sqrt6-sqrt2)-2i(sqrt6+sqrt2)#

Apply Euler's identity

#e^(itheta)=costheta+ isin theta#

#19/12pi=5/4pi+1/3pi#

#cos(19/12pi)=cos(5/4pi+1/3pi)#

#=cos(5/4pi)cos(1/3pi)-sin(5/4pi)sin(1/3pi)#

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

#=(sqrt6-sqrt2)/4#

#sin(19/12pi)=sin(5/4pi+1/3pi)#

#=sin(5/4pi)cos(1/3pi)+sin(1/3pi)cos(5/4pi)#

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

#=-(sqrt2+sqrt6)/4#

Therefore,

#8e^(19/12pi)=8cos(19/12pi)+i8sin(19/12pi)#

#=8*((sqrt6-sqrt2)/4)-8i(-(sqrt2+sqrt6)/4)#

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

Caroline Adams · Answer

undefined You can use Euler's formula, which states that $ e^{ix} = \cos(x) + i\sin(x) $, to express the complex number in terms of trigonometric functions.

Given $ 8 e^{\frac{19\pi}{12}i} $, we have:

$$ e^{\frac{19\pi}{12}i} = \cos\left(\frac{19\pi}{12}\right) + i\sin\left(\frac{19\pi}{12}\right) $$

Since $ \frac{19\pi}{12} $ is in the third quadrant, where both sine and cosine are negative, we have:

$$ \cos\left(\frac{19\pi}{12}\right) = -\cos\left(\frac{\pi}{12}\right) $$
$$ \sin\left(\frac{19\pi}{12}\right) = -\sin\left(\frac{\pi}{12}\right) $$

Substituting these values into the expression, we get:

$$ 8 e^{\frac{19\pi}{12}i} = 8 \left( -\cos\left(\frac{\pi}{12}\right) - i\sin\left(\frac{\pi}{12}\right) \right) $$

Therefore, $ 8 e^{\frac{19\pi}{12}i} $ simplifies to $ -8\cos\left(\frac{\pi}{12}\right) - 8i\sin\left(\frac{\pi}{12}\right) $ as a non-exponential complex number.