Imaginary numbers: express cos(5theta) in terms of....

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use the relationship eiθ=cosθ+isinθ\displaystyle e^{i\theta}=cos\theta+isin\theta
Express \(\displaystyle \L\cos5\theta\) in the terms of \(\displaystyle \L\cos\theta\)
And show that x=cos(110π)\displaystyle (\frac{1}{10}\pi) is a root of 16x4+20x2+5=0\displaystyle 16x^4+20x^2+5=0
Thanks :)
 
Re: Imaginary numbers

Hello, atomos!

Here's the first one . . .

Use the relationship:  eiθ = cos⁡θ + i⋅sin⁡θ\displaystyle \,e^{i\theta}\:=\:\cos\theta\,+\,i\cdot\sin\theta
Express cos⁡5θ\displaystyle \cos5\theta in the terms of cos⁡θ\displaystyle \cos\theta
Click to expand...
That relationshp says:  ei(5θ)  =  cos⁡(5θ) + i⋅sin⁡(5θ)  \displaystyle \,e^{i(5\theta)}\;=\;\cos(5\theta)\,+\,i\cdot\sin(5\theta)\; [1]


We also have:  ei5θ  =  (eiθ)5  =  (cos⁡θ + i⋅sin⁡θ)5\displaystyle \,e^{i5\theta}\;=\;\left(e^{i\theta}\right)^5\;=\;(\cos\theta\,+\,i\cdot\sin\theta)^5

    =  cos⁡5θ + 5i⋅cos⁡4θ⋅sin⁡θ − 10⋅cos⁡3θ⋅sin⁡2θ − 10i⋅cos⁡2θ⋅sin⁡3θ + 5⋅cos⁡θ⋅sin⁡4θ + i⋅sin⁡5θ\displaystyle \;\;= \;\cos^5\theta\,+\,5i\cdot\cos^4\theta\cdot\sin\theta\,-\,10\cdot\cos^3\theta\cdot\sin^2\theta\,-\,10i\cdot\cos^2\theta\cdot\sin^3\theta\,+\,5\cdot\cos\theta\cdot\sin^4\theta\,+\,i\cdot\sin^5\theta

    =  [cos⁡5θ − 10⋅cos⁡3θ⋅sin⁡2θ + 5⋅cos⁡θ⋅sin⁡4θ] + i[5⋅cos⁡4θ⋅sin⁡θ − 10⋅cos⁡2θ⋅sin⁡3θ + sin⁡5θ]  \displaystyle \;\;=\;[\cos^5\theta\,-\,10\cdot\cos^3\theta\cdot\sin^2\theta\,+\,5\cdot\cos\theta\cdot\sin^4\theta]\,+\,i[5\cdot\cos^4\theta\cdot\sin\theta\,-\,10\cdot\cos^2\theta\cdot\sin^3\theta\,+\,\sin^5\theta]\; [2]


Equate the real components of [1] and [2]:

cos⁡(5θ)  =  cos⁡5θ − 10⋅cos⁡3θ⋅sin⁡2θ + 5⋅cos⁡θ⋅sin⁡4θ\displaystyle \cos(5\theta) \;= \;\cos^5\theta\,-\,10\cdot\cos^3\theta\cdot\sin^2\theta\,+\,5\cdot\cos\theta\cdot\sin^4\theta

\(\displaystyle \;\;= \;\cos^5\theta\,-\,10\cdot\cos^3\theta\left(1\,-\,\cos^2\theta\right)\,+\,5\cdot\cos\theta\left(1\,-\,\cos^2\theta)^2\)

    =  cos⁡5θ − 10⋅cos⁡3θ + 10⋅cos⁡5θ + 5⋅cos⁡θ − 10⋅cos⁡3θ + 5⋅cos⁡4θ\displaystyle \;\;= \;\cos^5\theta\,-\,10\cdot\cos^3\theta\,+\,10\cdot\cos^5\theta\,+\,5\cdot\cos\theta\,-\,10\cdot\cos^3\theta\,+\,5\cdot\cos^4\theta


Therefore: cos⁡(5θ)  =  16⋅cos⁡5θ − 20⋅cos⁡3θ + 5⋅cos⁡θ\displaystyle \cos(5\theta) \;= \;16\cdot\cos^5\theta\,-\,20\cdot\cos^3\theta\,+\,5\cdot\cos\theta


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If we equate the imaginary components of [1] and [2], we have:

sin⁡(5θ)  =  5⋅cos⁡4θ⋅sin⁡θ − 10⋅cos⁡2θ⋅sin⁡3θ + sin⁡5θ\displaystyle \sin(5\theta) \;= \;5\cdot\cos^4\theta\cdot\sin\theta\,-\,10\cdot\cos^2\theta\cdot\sin^3\theta\,+\,\sin^5\theta

    =  5(1 − sin⁡2θ)2sin⁡θ − 10⋅(1 − sin⁡2θ)sin⁡3θ + sin⁡5θ\displaystyle \;\;=\;5(1\,-\,\sin^2\theta)^2\sin\theta \,- \,10\cdot(1\,-\,\sin^2\theta)\sin^3\theta\,+\,\sin^5\theta

    =  5⋅sin⁡θ − 10⋅sin⁡3θ + 5⋅sin⁡5θ − 10⋅sin⁡3θ + 10⋅sin⁡5θ + sin⁡5θ\displaystyle \;\;=\;5\cdot\sin\theta\,-\,10\cdot\sin^3\theta\,+\,5\cdot\sin^5\theta\,-\,10\cdot\sin^3\theta\,+\,10\cdot\sin^5\theta \,+\,\sin^5\theta


Therefore:  sin⁡(5θ)  =  16⋅sin⁡5θ − 20⋅sin⁡3θ + 5⋅sin⁡θ\displaystyle \,\sin(5\theta) \;= \;16\cdot\sin^5\theta\,-\,20\cdot\sin^3\theta\,+\,5\cdot\sin\theta
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