Polynomials Syllabus Example Question (1 Viewer)

wagig · Mar 15, 2014

i) Use De Moivre's theorem to express cos(5x), sin(5x) in powers of sin(x) and cos(x). [I can do this part]
Hence express tan(5x) as a rational function of t, where t = tan(x) [I can't do this part without resulting to double-angle results]
ii) Deduce that: tan(pi/5)tan(2pi/5)tan(3pi/5)tan(4pi/5) = 5

Kurosaki · Mar 15, 2014

wagig said:
i) Use De Moivre's theorem to express cos(5x), sin(5x) in powers of sin(x) and cos(x). [I can do this part]
Hence express tan(5x) as a rational function of t, where t = tan(x) [I can't do this part without resulting to double-angle results]
ii) Deduce that: tan(pi/5)tan(2pi/5)tan(3pi/5)tan(4pi/5) = 5

I think you do $tan(5x)=\frac{sin(5x)}{cos(5x)}$ , then divide by something for both numerator and denominator (some power of cos(x)) for the first part

Sy123 · Mar 15, 2014

(a)

$\\(\cos x + i\sin x)^5 = \cos^5 x -10\cos^3x \sin^2x+5\cos x \sin^4 x + i(5\cos^4 x \sin x - 10 \cos^2 x \sin^3 x + \sin^5 x) \\ \\ $Therefore$ \ \cos(5x) = \cos^5 x - 10\cos^3x \sin^2x + 5\cos x \sin^4x \ \ \ \ (1) \\ \sin(5x) = 5\cos^4x \sin x - 10\cos^2x \sin^3 x + \sin^5x \ \ \ \ \ (2) \\ \\ $By using De Moivre's and equating real and imaginary$ \\ \\ $Therefore$ \ \ \frac{(1)}{(2)} = \tan (5x) = \frac{5\cos^4x \sin x - 10\cos^2x \sin^3 x + \sin^5x}{\cos^5 x - 10\cos^3x \sin^2x + 5\cos x \sin^4x} = \frac{5\cos^4x \sin x - 10\cos^2x \sin^3 x + \sin^5x}{\cos^5 x - 10\cos^3x \sin^2x + 5\cos x \sin^4x} \times \frac{\frac{1}{\cos^5x}}{\frac{1}{\cos^5 x}} \\ \\ \Rightarrow \ \tan(5x) = \frac{5 \tan(x) - 10\tan^3x + \tan^5x}{1 - 10\tan^2x + 5\tan^4x} = \frac{5t+10t^3 + t^5}{1-10t^2 +5t^4} \ \ $Where$ \ t=\tan x$

(b)

$\\ $In the above rational function, let$ \ \ \tan(5x) = 0 \ \ $we find that$ \\ \\ t^5 + 10t^3 + 5t = 0 \ \ \ (1-10t^2+5t^4 \neq 0 \ \ (*)) \Rightarrow \ \ t^4+10t^2 + 5 = 0 \ \ (t\neq 0) \\ \\ $The solutions to this equation are the solutions to$ \ \tan(5x) = 0 \ $Hence the solutions to the polynomial are$ \ \ x=\frac{k\pi}{5} \ \ k \ $is integer$ \\ \\ $Since$ \ t=\tan x \\ \\ $Then the solutions are$ \ \ t=\tan \frac{k\pi}{5} \ \ $we take 4 solutions, being$ \ k=1,2,3,4 \\ \\ $Since$ \ t\neq 0 \ $is not a solution and$ \ \ \tan \frac{k\pi}{5} = \tan \left(\frac{k\pi}{5} + \pi \right) \\ \\ $Take the quartic again$ \\ \\ t^4 + 10t^2 +5 \\ \\ $Product of roots therefore is$ \ 5 \\ \\ \therefore \ \ \tan \frac{\pi}{5} \tan \frac{2\pi}{5} \tan \frac{3\pi}{5} \tan \frac{4\pi}{5} = 5$

Polynomials Syllabus Example Question (1 Viewer)

wagig

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Kurosaki

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Sy123

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