1. Regarding De Moivre’s Theorem

Can somebody help me to figure out the answer of that question?

Thanks in advanced...Really appreciate if have any help...

By Using De Moivre’s Theorem, show that :

(i) Cos 6theta = 32 cos^6theta - 48 cos^4theta + 18 cos^2theta -1

(ii) If z = costheta + i sintheta, nÎZ^+, proof that z^n + z^-n = 2 costheta, hence, Cos 4theta = 8 cos^4theta - 4cos2theta -3

(iii) Find the roots of z^3 = -1 + √3 i

2. Hello,
Originally Posted by ab.empire
Can somebody help me to figure out the answer of that question?

Thanks in advanced...Really appreciate if have any help...

By Using De Moivre’s Theorem, show that :

(i) Cos 6theta = 32 cos^6theta - 48 cos^4theta + 18 cos^2theta -1
We know that $\displaystyle (\cos \theta+i \sin \theta)^6=\cos(6 \theta)+i \sin(6 \theta)$, by de Moivre's theorem.

Expand the left hand side of the equation by using the binomial theorem, then group the real numbers together, and the imaginary numbers together.
The real part will be equal to $\displaystyle \cos(6\theta)$, which is what you want.

(ii) If z = costheta + i sintheta, nÎZ^+, proof that z^n + z^-n = 2 cos ntheta, hence, Cos 4theta = 8 cos^4theta - 4cos2theta -3
$\displaystyle z^n=(\cos \theta+i\sin \theta)^n=\cos(n\theta)+i \sin(n\theta)$
and $\displaystyle z^{-n}=(\cos \theta+i\sin \theta)^{-n}=\cos(-n\theta)+i \sin(-n\theta)=\cos(n\theta)-i\sin(n\theta)$ (because the cosine is even and the sine is odd)
add the two and you'll get what you want

as for the second part, just use let n=4 in the formula above. and use a similar method to 1)
(i'm looking for a simpler way...)

(iii) Find the roots of z^3 = -1 + √3 i
find the modulus of the RHS : $\displaystyle |-1+\sqrt{3} i|=\sqrt{1+3}=2$

so $\displaystyle -1+\sqrt{3} i=2 \left(-\frac 12+\frac{\sqrt{3}}{2} i\right)=2 (\cos(\theta)+i \sin(\theta))=2e^{i (\theta+2k\pi)}$

find the value of theta

,

,

,

,

,

,

,

âˆš1 cos6theta

Click on a term to search for related topics.