SOLUTION: Hi, re-asking this question as my coding didn't work last time. Use De Moivres Theorem to simplify (cos(5pi/6)-sin(5pi/6))^7. I have tried a few examples and can't seem to figure o

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Hi, re-asking this question as my coding didn't work last time. Use De Moivres Theorem to simplify (cos(5pi/6)-sin(5pi/6))^7. I have tried a few examples and can't seem to figure o      Log On


   



Question 1205952: Hi, re-asking this question as my coding didn't work last time. Use De Moivres Theorem to simplify (cos(5pi/6)-sin(5pi/6))^7. I have tried a few examples and can't seem to figure out this one. Thank you
Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

De Moivre's Theorem:
If z = r*cis(theta), then z^n = r^n*cis(n*theta)
where cis(theta) = cos(theta)+i*sin(theta) and n is an integer.

I'll assume that your expression is (cos(5pi/6)-i*sin(5pi/6))^7
I introduced the imaginary number 'i' just before the "sin"

The minus sign between those trig functions means we cannot immediately use De Moivre's Theorem just yet. We need to replace the minus sign with a plus sign.

We can use the idea that sine is an odd function. Odd as in "not even", though I suppose someone could argue that it is a strange function.
Since sine is odd, we know that -sin(x) = sin(-x) for all real numbers x.

Cosine being even means cos(-x) = cos(x)
It has y-axis symmetry.

z = cos(5pi/6) - i*sin(5pi/6)
z = cos(5pi/6) + i*sin(-5pi/6) ... since sine is odd
z = cos(-5pi/6) + i*sin(-5pi/6) ... since cosine is even
z = cis(-5pi/6)

Now we can use De Moivre's Theorem
z = cis(-5pi/6)
z^7 = cis( 7*(-5pi/6) )
z^7 = cis( -35pi/6 )
z^7 = cis( pi/6 )

How did we go from -35pi/6 to pi/6?
I'm using the fact that both sine and cosine have period 2pi.
sin(x+2pi) = sin(x)
cos(x+2pi) = sin(x)
We can add and subtract multiples of 2pi to find coterminal angles.

Add 3 multiples of 2pi onto angle -35pi/6 radians to end up at pi/6
-35pi/6 + 3*2pi = pi/6
This is the same as rotating a full 360 degrees exactly three times. We end up pointing at the same direction we started at (somewhere in the northeast).

From here, we can use the unit circle to wrap things up.
z^7 = cis( pi/6 )
z^7 = cos( pi/6 ) + i*sin( pi/6 )
z^7 = sqrt(3)/2 + i*(1/2)
z^7 = sqrt(3)/2 + (1/2)*i
z^7 = (1/2)*( sqrt(3) + i )


Therefore,
(cos(5pi/6)-i*sin(5pi/6))^7 = sqrt(3)/2 + (1/2)*i
or
(cos(5pi/6)-i*sin(5pi/6))^7 = (1/2)*( sqrt(3) + i )
depending on how you prefer to write the answer.

Various computer algebra systems (CAS) can be used to verify the answer.
Some examples include: WolframAlpha and GeoGebra.