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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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)`
`(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` |