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Complex Numbers, EXT2 N2 2021 HSC 14c*

Using de Moivre’s theorem and the binomial expansion of `(cos theta + i sin theta)^5`, or otherwise, show that

      `cos5theta = 16cos^5theta-20cos^3 theta + 5cos theta`.   (3 marks)

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`text(See Worked Solution)`

Show Worked Solution

`(cos theta + i sin theta)^5 = cos5theta + i sin 5theta\ \ text{(by De Moivre)}`

`text(Using binomial expansion:)`

`(cos theta + i sin theta)^5`

`= cos^5theta + 5cos^4theta · isin theta + 10cos^3theta · i^2sin^2theta + 10 cos^2theta · i^3sin^3theta`

`+ 5costheta · i^4sin^4theta + i^5sin^5theta`

`= cos^5theta-10cos^3thetasin^2theta + 5costhetasin^4theta + i\ \ text{(imaginary part)}`
 

`text(Equating real parts:)`

`cos5theta` `= cos^5theta-10cos^3thetasin^2theta + 5costhetasin^4theta`
  `= cos^5theta-10cos^3theta(1-cos^2theta) + 5costheta(1-cos^2theta)sin^2theta`
  `= cos^5theta-10cos^3theta + 10cos^5theta + (5costheta-5cos^3theta)(1-cos^2theta)`
  `= 11cos^5theta-10cos^3theta + 5costheta-5cos^3theta-5cos^3theta + 5cos^5theta`
  `= 16cos^5theta-20cos^3theta + 5costheta`

Complex Numbers, EXT2 N2 2025 HSC 15c

  1. Show that
  2.     \(\dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \dfrac{\theta}{2}.\)   (3 marks)

  3. Use De Moivre's theorem to show that the sixth roots of \(-1\) are given by
  4.    \(\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right)\)  for  \(k=0,1,2,3,4,5\).   (2 marks)

  5. Hence, or otherwise, show the solutions to  \(\left(\dfrac{z-1}{z+1}\right)^6=-1\)  are
  6. \(z=i \cot \left(\dfrac{\pi}{12}\right), i \cot \left(\dfrac{3 \pi}{12}\right), i \cot \left(\dfrac{5 \pi}{12}\right), i \cot \left(\dfrac{7 \pi}{12}\right), i \cot \left(\dfrac{9 \pi}{12}\right)\), and \(i \cot \left(\dfrac{11 \pi}{12}\right)\).   (2 marks)

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i.     \(\text{Show} \ \ \dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \left(\frac{\theta}{2}\right)\)

\(\text{Method 1:}\)

\(\text{LHS}\) \(=\dfrac{2\cos^{2}\frac{\theta}{2}+2i\,\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^{2}\frac{\theta}{2}-2i\,\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \)  
  \(=\dfrac{2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2}+i\,\sin\frac{\theta}{2}\right)}{2\sin\frac{\theta}{2}\left(\sin\frac{\theta}{2}-i\,\cos\frac{\theta}{2}\right) } \)  
  \(=\dfrac{i\,\cos\frac{\theta}{2}\left(\sin\frac{\theta}{2}-i\,\cos\frac{\theta}{2}\right)}{\sin\frac{\theta}{2}\left( \sin\frac{\theta}{2}-i\,\cos\frac{\theta}{2}\right)} \)  
  \(=i\, \cot \left(\frac{\theta}{2}\right)\)  

 
\(\text{Method 2 (exponential form – ex-syllabus in 2027):}\)

\(\text{LHS}\) \(=\dfrac{1+e^{i \theta}}{1-e^{i \theta}} \times \dfrac{e^{-\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}}\)
  \(=\dfrac{e^{-\tfrac{i \theta}{2}}+e^{\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}-e^{\tfrac{i \theta}{2}}}\)
  \(=\dfrac{2 \cos \left(\frac{\theta}{2}\right)}{-2 i \sin \left(\frac{\theta}{2}\right)}\)
  \(=i \cot \left(\frac{\theta}{2}\right)\)

 
ii.   \(z=\cos \theta+i \sin \theta\)

\(\text{Find sixth roots of}\ -1 \ \text{(by De Moivre):}\)

\(z^6=\cos (6 \theta)+i \sin (6 \theta)=-1\)

\(\cos (6 \theta)\) \(=-1 \ \text{and} \ \ \sin (6 \theta)=0\)
\(6 \theta\) \(=\pi, 3 \pi, 5 \pi, \ldots\)
\(\theta\) \(=\dfrac{(2 k+1) \pi}{6}\ \ \text{for}\ \ k=0,1,2, \ldots, 5\)

\(\therefore \operatorname{Roots }=\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right) \ \  \text{for} \ \ k=0,1, \ldots, 5\)
 

iii.  \(\left(\dfrac{z-1}{z+1}\right)^6=-1\)

\(\text {Let} \ \ \alpha=\dfrac{z-1}{z+1} \ \Rightarrow \ \alpha^6=-1\)
 

\(\text {Using part (ii):}\)

\(\alpha=\operatorname{cis}\left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for}\ \ k=0,1,2,3,4,5\ \ldots\ (1)\)

\(\alpha=\dfrac{z-1}{z+1}\  \ \Rightarrow\ \ \alpha z+\alpha=z-1 \ \ \Rightarrow\ \ z=\dfrac{1+\alpha}{1-\alpha}\)
 

\(\text{Consider} \ \ \alpha=\operatorname{cis}\left(\frac{\pi}{6}\right) \ \text{(i.e. where}\ \ k=0 \ \ \text{from (1) above):}\)

\(z=\dfrac{1+\operatorname{cis}\left(\frac{\pi}{6}\right)}{1-\operatorname{cis}\left(\frac{\pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\frac{\pi}{12}\right) \quad \text{(using part (i))}\)

\(\text{Similarly}\ (k=1), \ z=\dfrac{1+\operatorname{cis}\left(\frac{3 \pi}{6}\right)}{1-\operatorname{cis}\left(\frac{3 \pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\frac{3 \pi}{12}\right)\)

\(\therefore z=i \cot \left(\frac{\pi}{12}\right), \, i \cot \left(\frac{3 \pi}{12}\right), \, i \cot \left(\frac{5 \pi}{12}\right), \ldots, i \cot \left(\frac{11 \pi}{12}\right)\)

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i.     \(\text{Show} \ \ \dfrac{1+\cos \theta+i \sin \theta}{1-\cos \theta-i \sin \theta}=i \cot \left(\frac{\theta}{2}\right)\)

\(\text{Method 1:}\)

\(\text{LHS}\) \(=\dfrac{2\cos^{2}\frac{\theta}{2}+2i\,\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^{2}\frac{\theta}{2}-2i\,\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \)  
  \(=\dfrac{2\cos\frac{\theta}{2}\left(\cos\frac{\theta}{2}+i\,\sin\frac{\theta}{2}\right)}{2\sin\frac{\theta}{2}\left(\sin\frac{\theta}{2}-i\,\cos\frac{\theta}{2}\right) } \)  
  \(=\dfrac{i\,\cos\frac{\theta}{2}\left(\sin\frac{\theta}{2}-i\,\cos\frac{\theta}{2}\right)}{\sin\frac{\theta}{2}\left( \sin\frac{\theta}{2}-i\,\cos\frac{\theta}{2}\right)} \)  
  \(=i\, \cot \left(\frac{\theta}{2}\right)\)  

 
\(\text{Method 2 (exponential form – ex-syllabus in 2027):}\)

\(\text{LHS}\) \(=\dfrac{1+e^{i \theta}}{1-e^{i \theta}} \times \dfrac{e^{-\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}}\)
  \(=\dfrac{e^{-\tfrac{i \theta}{2}}+e^{\tfrac{i \theta}{2}}}{e^{-\tfrac{i \theta}{2}}-e^{\tfrac{i \theta}{2}}}\)
  \(=\dfrac{2 \cos \left(\frac{\theta}{2}\right)}{-2 i \sin \left(\frac{\theta}{2}\right)}\)
  \(=i \cot \left(\frac{\theta}{2}\right)\)

 
ii.   \(z=\cos \theta+i \sin \theta\)

\(\text{Find sixth roots of}\ -1 \ \text{(by De Moivre):}\)

\(z^6=\cos (6 \theta)+i \sin (6 \theta)=-1\)

\(\cos (6 \theta)\) \(=-1 \ \text{and} \ \ \sin (6 \theta)=0\)
\(6 \theta\) \(=\pi, 3 \pi, 5 \pi, \ldots\)
\(\theta\) \(=\dfrac{(2 k+1) \pi}{6}\ \ \text{for}\ \ k=0,1,2, \ldots, 5\)

\(\therefore \operatorname{Roots }=\cos \left(\dfrac{(2 k+1) \pi}{6}\right)+i \sin \left(\dfrac{(2 k+1) \pi}{6}\right) \ \  \text{for} \ \ k=0,1, \ldots, 5\)
 

iii.  \(\left(\dfrac{z-1}{z+1}\right)^6=-1\)

\(\text {Let} \ \ \alpha=\dfrac{z-1}{z+1} \ \Rightarrow \ \alpha^6=-1\)

♦♦♦ Mean mark (iii) 23%.

\(\text {Using part (ii):}\)

\(\alpha=\operatorname{cis}\left(\dfrac{(2 k+1) \pi}{6}\right) \ \ \text{for}\ \ k=0,1,2,3,4,5\ \ldots\ (1)\)

\(\alpha=\dfrac{z-1}{z+1}\  \ \Rightarrow\ \ \alpha z+\alpha=z-1 \ \ \Rightarrow\ \ z=\dfrac{1+\alpha}{1-\alpha}\)
 

\(\text{Consider} \ \ \alpha=\operatorname{cis}\left(\frac{\pi}{6}\right) \ \text{(i.e. where}\ \ k=0 \ \ \text{from (1) above):}\)

\(z=\dfrac{1+\operatorname{cis}\left(\frac{\pi}{6}\right)}{1-\operatorname{cis}\left(\frac{\pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\frac{\pi}{12}\right) \quad \text{(using part (i))}\)

\(\text{Similarly}\ (k=1), \ z=\dfrac{1+\operatorname{cis}\left(\frac{3 \pi}{6}\right)}{1-\operatorname{cis}\left(\frac{3 \pi}{6}\right)} \ \Rightarrow \ z=i \cot \left(\frac{3 \pi}{12}\right)\)

\(\therefore z=i \cot \left(\frac{\pi}{12}\right), \, i \cot \left(\frac{3 \pi}{12}\right), \, i \cot \left(\frac{5 \pi}{12}\right), \ldots, i \cot \left(\frac{11 \pi}{12}\right)\)

Complex Numbers, EXT2 N2 2018 HSC 15b

  1. Use De Moivre's theorem and the expansion of `(costheta + isintheta)^8` to show that
  2. `sin8theta = ((8),(1)) cos^7thetasintheta-((8),(3)) cos^5thetasin^3theta`
  3.                  `+ ((8),(5)) cos^3thetasin^5theta-((8),(7)) costhetasin^7theta`   (2 marks)

  4. Hence, show that
  5. `(sin8theta)/(sin2theta) = 4(1-10sin^2theta + 24sin^4theta-16sin^6theta)`.   (3 marks)

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  1. `text(See Worked Solutions)`
  2. `text(See Worked Solutions)`
Show Worked Solution

i.   `text(By De Moivre)`

`costheta + isintheta^8 = cos8theta + isin8theta\ \ …\ (text{*})`
 

`text(Using Binomial Expansion)`

`(costheta + isintheta)^8`

`= cos^8theta + ((8),(1))cos^7theta * isintheta + ((8),(2)) cos^6theta *i^2sin^2theta`

`+ ((8),(3)) cos^5theta *i^3sin^3theta + ((8),(4)) cos^4theta *i^4sin^4theta + ((8),(5)) cos^3theta *i^5sin^5theta`

`+ ((8),(6)) cos^2theta *i^6sin^6theta + ((8),(7)) costheta *i^7sin^7theta + i^8sin^8theta`

 
`text(Equating imaginary parts of the expansion equation (*)):`

`isin8theta = ((8),(1)) cos^7theta* isintheta + ((8),(3)) icos^5theta* i^3sin^3theta`

`+ ((8),(5)) cos^3theta* i ^5sintheta + ((8),(7)) costheta *i^7sin^7theta`

`:. sin8theta = ((8),(1)) cos^7theta sintheta-((8),(3)) cos^5theta sin^3theta`

`+ ((8),(5)) cos^3theta sin^5theta-((8),(7)) costheta sin^7theta`
 

ii.    `sin8theta` `= 8cos^7theta sintheta-56cos^5 sin^3theta + 56cos^3theta sin^5theta-8costheta sin^7theta`
    `= 2sinthetacostheta (4cos^6theta-28cos^4theta sin^2theta + 28cos^2theta sin^4theta-4sin^6theta)`

 
`:. (sin8theta)/(sin2theta)`

  `= 4cos^6theta-28cos^4theta sin^2theta + 28cos^2theta sin^4theta-4sin^6theta`

  `= 4(1-sin^2theta)^3-28(1-sin^2theta)^2 sin^2theta + 28(1-sin^2theta) sin^4theta-4sin^6theta`

  `= 4(1-3sin^2theta + 3sin^4theta + sin^6theta)-28sin^2theta (1-2sin^2theta + sin^4theta)`

`+ 28sin^4theta (1-sin^2theta)-4sin^6theta`

  `= 4-40sin^2theta + 96sin^4theta-56sin^6theta`

  `= 4(1-10sin^2theta + 24sin^4theta-16sin^6theta)`

Complex Numbers, EXT2 N2 2016 HSC 12c

Let  `z = cos theta + i sin theta.`

  1. By considering the real part of `z^4`, show that `cos 4 theta` is
  2. `qquad cos^4 theta-6 cos^2 theta sin^2 theta + sin^4 theta.`   (2 marks)
  3. Hence, or otherwise, find an expression for  `cos 4 theta`  involving only powers of `cos theta.`   (1 mark)

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i.    `text(See Worked Solutions)`

ii.   `8cos^4theta-8cos^2theta + 1`

Show Worked Solution

i.   `z = costheta + isintheta`

`z^4` `= (costheta + isintheta)^4`
 

`= cos^4theta + 4cos^3theta*(isintheta) + 6cos^2theta*(isintheta)^2 +`

`4costheta*(isintheta)^3 + (isintheta)^4`
 

`= cos^4theta + 4icos^3thetasintheta-6cos^2thetasin^2theta -`

`4icosthetasin^3theta + sin^4theta`

 

`z^4 = cos4theta + isin4theta\ \ text{(by De Moivre)}`
 

`text(Equating real parts:)`

`cos4theta = cos^4theta-6cos^2thetasin^2theta + sin^4theta\ …\ text(as required)`

 

ii.    `cos4theta` `= cos^4theta-6cos^2theta(1-cos^2theta) + (1-cos^2theta)^2`
    `= cos^4theta-6cos^2theta + 6cos^4theta + 1-2cos^2theta + cos^4theta`
    `= 8cos^4theta-8cos^2theta + 1`

Complex Numbers, EXT2 N2 2007 HSC 8b

  1. Let `n` be a positive integer. Show that if  `z^2 != 1`  then
  2. `1 + z^2 + z^4 + … + z^(2n-2) = ((z^n-z^-n)/(z-z^-1)) z^(n-1)`.   (2 marks)

  3. By substituting  `z = cos theta + i sin theta`  where  `sin theta != 0`, into part (a), show that
  4. `1 + cos 2 theta + … + cos (2n-2) theta + i[sin 2 theta + … + sin (2n-2) theta]`
  5. `= (sin n theta)/(sin theta) [cos (n-1) theta + i sin (n-1) theta].`   (3 marks)

  6. Suppose  `theta = pi/(2n)`.  Using part (ii), show that
  7. `sin\ pi/n + sin\ (2 pi)/n + … + sin\ ((n-1) pi)/n = cot\ pi/(2n).`   (2 marks)

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a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `text(Proof)\ \ text{(See Worked Solutions)}`

c.    `text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

a.    `1 + z^2 + z^4 + … + z^(2n-2),\ z^2 != 1`

`text(GP where)\ a = 1,\ \ r = z^2,\ \ n\ text(terms):`

`S_n` `=(1((z^2)^n-1))/(z^2-1)`
  `=(z^(2n)-1)/(z^2-1)`
  `=((z^n-z^-n))/(z-z^-1) xx z^n/z`
  `=((z^n-z^-n)/(z-z^-1))z^(n-1)`

 

b.     `z` `= cos theta + i sin theta`
  `z^n` `= cos n theta + i sin n theta\ \ …\ text(etc)\ \ \ \ text{(De Moivre)}` 
  `z^-n` `= cos( -n theta) + i sin (-n theta)`
    `= cos n theta-i sin n theta`

 

`text(LHS)` `= 1 + (cos 2 theta + i sin 2 theta) + (cos 4 theta + i sin 4 theta) + `
  `… + (cos(2n-2) theta + i sin (2n-2) theta)`
  `= 1 + cos 2 theta + cos 4 theta + … + cos (2n-2) theta + `
  `i (sin 2 theta + sin 4 theta + … + sin (2n-2) theta)`
   

`text{Using part (a):}`

`text(LHS)` `=((cos n theta + i sin n theta-cos n theta + i sin n theta))/(cos theta + i sin theta-cos theta + i sin theta) xx`
  `[cos (n-1) theta + i sin (n-1) theta]`
  `=(2 i sin n theta)/(2 i sin theta) [cos (n-1) theta + i sin (n-1) theta]`
  `=(sin n theta)/(sin theta) [cos (n-1) theta + i sin (n-1) theta]\ \ text(… as required.)`

 

c.    `text{Equating the imaginary parts in part (b):}`

`sin 2 theta + sin 4 theta + … + sin 2 (n-1) theta = (sin (n theta) sin (n-1) theta)/(sin theta)`

`text(When)\ \ theta = pi/(2n):`

`sin\ (2 pi)/(2n) + sin\ (4 pi)/(2n) + … + sin\ (2(n-1) pi)/(2 n) = (sin\ (n pi)/(2n) sin\ ((n-1) pi)/(2n))/(sin\ pi/(2n))`

`:. sin\ pi/n + sin\ (2 pi)/n + … + sin\ ((n-1) pi)/n`

`=(sin\ pi/2)/(sin\ pi/(2n)) xx sin\ ((n-1) pi)/(2n)`

`=1/(sin\ pi/(2n)) sin (pi/2-pi/(2n))`

`=(cos\ pi/(2n))/(sin\ pi/(2n))`

`=cot\ pi/(2n)`

Complex Numbers, EXT2 N2 2009 HSC 7b

Let  `z = cos theta + i sin theta.`

  1. Show that  `z^n + z^-n = 2 cos(n theta)`, where `n` is a positive integer.   (2 marks)

  2. Let `m` be a positive integer. Show that
  3. `(2 cos theta)^(2m) = 2 [cos (2m theta) + ((2m), (1)) cos (2m-2) theta + ((2m), (2)) cos (2m-4) theta`
  4.           `+ … + ((2m), (m-1)) cos 2 theta] + ((2m), (m)).`   (3 marks)

  5. Hence, or otherwise, prove that
  6. `int_0^(pi/2) cos^(2m) theta\ d theta = pi/(2^(2m + 1)) ((2m), (m))`
  7. where `m` is a positive integer.   (2 marks)

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a.    `text(Proof)\ \ text{(See Worked Solutions)}`

b.    `text(Proof)\ \ text{(See Worked Solutions)}`

c.    `text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution
a.     `z` `= cos theta + i sin theta`
  `z^n` `= cos n theta + i sin n theta\ \ \ \ text{(De Moivre)}`
  `z^-n` `= cos (-n theta) + i sin (-n theta)\ \ \ \ text{(De Moivre)}`
    `= cos n theta-i sin n theta`
  `z^n + z^-n` `= cos n theta + i sin n theta + cos n theta-i sin n theta`
    `= 2 cos n theta,\ \ \ \ n > 0`

 

 

b.    `z + z^-1 = 2 cos theta`

`:.(2 cos theta)^(2m)`

`=(z + z^-1)^(2m)`

`=z^(2m) + ((2m), (1)) z^(2m-1) z^-1 + ((2m), (2)) z^(2m-2) z^-2+`

` … + ((2m), (2m-1)) z^1 z^-(2m-1) + z^-(2m)`

`=z^(2m) + ((2m), (1)) z^(2m-2) + ((2m), (2)) z^(2m-4)+`

` … + ((2m), (2m-1)) z^-(2m-2) + z^(-2m)`

`=z^(2m) + ((2m), (1)) z^(2m-2) + ((2m), (2)) z^(2m-4) + … + ((2m), (m)) z^(2m-2m) …`

`+ ((2m), (2)) z^-(2m-4) + ((2m), (1)) z^-(2m-2) + z^(-2m)`

`=(z^(2m) + z^(-2m)) + ((2m), (1)) (z^(2m-2) + z^-(2m-2)) + ((2m), (2))`

`(z^(2m-4) + z^-(2m-4)) + … + ((2m), (m-1)) (z + z^-1) + ((2m), (m))`

`=2 [cos 2 m theta + ((2m), (1)) cos (2m-2) theta + ((2m), (2)) cos (2m-4) theta`

`+ … + ((2m), (m-1)) cos 2 theta] + ((2m), (m))`

 

c.    `int_0^(pi/2) cos^(2m) d theta`

`=1/(2^(2m))  int_0^(pi/2) (2 cos theta)^(2m)`

`=1/(2^(2m)) int_0^(pi/2)[2(cos 2 m theta + ((2m), (1)) cos (2m-2) theta + ((2m), (2))`

`cos (2m-4) theta + … + ((2m), (m-1)) cos 2 theta) + ((2m), (m))] d theta`

`=1/(2^(2m)) [2((sin 2 m theta)/(2m) + ((2m), (1)) (sin (2m-2) theta)/(2m-2)`

`+ … + ((2m), (m-1)) (sin 2 theta)/2) + ((2m), (m)) theta]_0^(pi/2)`

`=1/(2^(2m)) [2(0 + 0 + … + 0) + ((2m), (m)) pi/2-(0)]`

`=pi/(2^(2m + 1)) ((2m), (m))`

Complex Numbers, EXT2 N2 2011 HSC 2d

  1. Use the binomial theorem to expand `(cos theta + i sin theta)^3.`   (1 mark)

  2. Use de Moivre’s theorem and your result from part (a) to prove that
  3.      `cos^3 theta = 1/4 cos 3 theta + 3/4 cos theta.`   (3 marks)

  4. Hence, or otherwise, find the smallest positive solution of
  5. `4 cos^3 theta-3 cos theta = 1.`   (2 marks)

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  1. `cos^3 theta + 3 i cos^2 theta sin theta-3 cos theta sin^2 theta-i sin^3 theta`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `theta = (2 pi)/3`
Show Worked Solution

a.    `(cos theta + i sin theta)^3`

`=sum_(k=0)^3 \ ^3C_k (cos theta)^(3-k) (i sin theta)^k`

`= cos^3 theta + 3 cos^2 theta (i sin theta)+ 3 cos theta (i sin theta)^2 + (i sin theta)^3`

`= cos^3 theta + 3 i cos^2 theta sin theta- 3 cos theta sin^2 theta-i sin^3 theta`

 

b.     `text(Using De Moivre’s Theorem)`

`(cos theta + i sin theta)^3 = cos 3 theta + i sin 3 theta`
 

`text(Equate real parts)`

`cos 3 theta` `= cos^3 theta-3 cos theta sin^2 theta`
`cos 3 theta` `= cos^3 theta-3 cos theta (1-cos^2 theta)`
`cos 3 theta` `= 4 cos^3 theta-3 cos theta`
`4 cos^3 theta`  `=cos 3 theta+3cos theta`
`:.cos^3 theta` `= 1/4 cos 3 theta + 3/4 cos theta\ \ \ text(… as required)`

 

c.    `text(If)\ \ \ 4 cos^3 theta-3 cos theta = 1`

`=>cos 3 theta = 1\ \ \ \ text{(from part (b))}`

`3 theta` `= 2 k pi`
`:. theta` `= (2 k pi)/3`

 

`:.\ text(Smallest positive solution occurs when)`

`theta = (2 pi)/3\ \ \ \ text{(i.e. when}\ k = 1 text{)}`