Question

if x+1/x=2cos π/5 then show,x5+1/x5= -2​

Answer 1

Given info : x + 1/x = 2cosπ/5.

To show : x⁵ + 1/x⁵ = -2

proof : here we have to use, x⁵ + 1/x⁵ = (x² + 1/x²)(x³ + 1/x³) - (x + 1/x)

so we have to find the values of (x² + 1/x²) and (x³ + 1/x³) at first.

x² + 1/x² = (x + 1/x)² - 2 = [2cos²π/5]² - 2 = 2[2cos²π/5 - 1] = 2cos(2π/5)

[ ∵ 2cos²Ф - 1 = cos2Ф ]

now, x³ + 1/x³ = (x + 1/x)³ - 3(x + 1/x) = [2cosπ/5]³ - 3[2cosπ/5] = 2[4cos³(π/5) - 3cos(π/5) ] = 2cos(3π/5)

[∵ 4cos³Ф - 3cosФ = cos3Ф ]

now, LHS =  x⁵ + 1/x⁵ = 2cos(2π/5) × 2cos(3π/5) - 2cos(π/5)

                    = 2{2cos(2π/5) cos(3π/5)} - 2cos(π/5)

                    = 2[cos(2π + 3π)/5 + cos(3π - 2π)/5] - 2cos(π/5)

[∵ 2cosAcosB = cos(A + B) + cos(A - B) ]

                    = 2cosπ = 2cos(π/5) - 2cos(π/5) = 2cosπ = 2 × -1 = -2 = RHS

hence proved.

Answer 2

SOLUTION

GIVEN

$\displaystyle \sf \: x + \frac{1}{x} = 2 \cos \frac{\pi}{5}$

TO PROVE

$\displaystyle \sf \: {x}^{5} + \frac{1}{ {x}^{5} } = - 2$

EVALUATION

Here it is given that

$\displaystyle \sf \: x + \frac{1}{x} = 2 \cos \frac{\pi}{5}$

$\displaystyle \sf \implies \: \frac{ {x}^{2} + 1}{x} = 2 \cos \frac{\pi}{5}$

$\displaystyle \sf \implies \: {x}^{2} + 1= 2x \cos \frac{\pi}{5}$

$\displaystyle \sf \implies \: {x}^{2} - 2x \cos \frac{\pi}{5} + 1 = 0$

$\displaystyle \sf \implies \: x = \frac{ + 2\cos \frac{\pi}{5} \pm \: \sqrt{4{\cos}^{2} \frac{\pi}{5} - 4.1.1 } }{2.1}$

$\displaystyle \sf \implies \: x = \frac{ + 2\cos \frac{\pi}{5} \pm \: \sqrt{4{\cos}^{2} \frac{\pi}{5} - 4 } }{2}$

$\displaystyle \sf \implies \: x = \frac{ 2\cos \frac{\pi}{5} \pm \: 2i \sqrt{1 - {\cos}^{2} \frac{\pi}{5} } }{2}$

$\displaystyle \sf \implies \: x = \frac{ 2\cos \frac{\pi}{5} \pm \: 2i \sin \frac{\pi}{5} }{2}$

$\displaystyle \sf \implies \: x = \cos \frac{\pi}{5} \pm \: i \sin \frac{\pi}{5}$

Case : I

$\boxed{ \: \displaystyle \sf \: x = \cos \frac{\pi}{5} + \: i \sin \frac{\pi}{5} \: }$

Using De Moivre's theorem on complex number we get

$\displaystyle \sf \: {x}^{5} = \cos \frac{5\pi}{5} \pm \: i \sin \frac{5\pi}{5}$

$\displaystyle \sf \implies \: {x}^{5} = \cos \pi \pm \: i \sin \pi$

$\displaystyle \sf \implies \: {x}^{5} = - 1$

$\displaystyle \sf \implies \: \frac{1}{ {x}^{5} } = - 1$

Thus we get

$\displaystyle \sf \: {x}^{5} + \frac{1}{ {x}^{5} }$

$\displaystyle \sf \: = - 1 - 1$

$\displaystyle \sf \: = - 2$

Case : II

$\boxed{ \: \displaystyle \sf \: x = \cos \frac{\pi}{5} - \: i \sin \frac{\pi}{5} \: }$

Similarly it can be shown that

$\displaystyle \sf \: {x}^{5} + \frac{1}{ {x}^{5} } = - 2$

Hence proved

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