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Mathematics Question on Complex Numbers and Quadratic Equations

If x+1x=2 cos⁡θ\sqrt{x} + \frac{1}{\sqrt{x}} = 2 \, \cos \theta, then x6+x6=x^6 + x^6 =

A

2 cos⁡12θ2 \, \cos 12 \theta

B

2 cos⁡6θ2 \, \cos 6\theta

C

2 sin⁡3θ2 \, \sin 3 \theta

D

2 cos⁡3θ2 \, \cos 3 \theta

Answer

2 cos⁡12θ2 \, \cos 12 \theta

Explanation

Solution

x+1x=2 cos⁡θ\sqrt{x} + \frac{1}{\sqrt{x}} = 2 \, \cos \theta ...(i)
Squaring (i) both sides, we get
x+1x+2=4cos⁡2θx + \frac{1}{x} + 2 = 4 \cos^{2} \theta
⇒x+1x=4cos⁡2θ−2=2(2cos⁡2θ−1)\Rightarrow x + \frac{1}{x} = 4 \cos ^{2} \theta - 2 = 2\left(2 \cos ^{2} \theta-1\right)
⇒x+1x=2(cos⁡2θ)\Rightarrow x + \frac{1}{x} = 2\left( \cos 2\theta\right) ...(ii)
Cubing (ii) both sides, we get
⇒(x+1x)3=8cos⁡32θ\Rightarrow \left(x +\frac{1}{x}\right)^{3} = 8 \cos^{3} 2\theta
⇒x3+1x3+3(x+1x)=8cos⁡32θ\Rightarrow x^{3} + \frac{1}{x^{3}} + 3\left(x +\frac{1}{x}\right) = 8 \cos^{3} 2 \theta
⇒x3+1x3=8cos⁡32θ−6cos⁡2θ=2(4cos⁡32θ−3cos⁡2θ)\Rightarrow x^{3} + \frac{1}{x^{3}} = 8 \cos ^{3} 2\theta - 6 \cos 2\theta = 2\left(4 \cos ^{3} 2\theta - 3 \cos 2\theta\right)
⇒x3+1x3=2(cos⁡6θ)\Rightarrow x^{3} + \frac{1}{x^{3}} = 2\left(\cos 6\theta\right)...(iii)
Squaring (iii) both sides we get
⇒(x3+1x3)2=4cos⁡26θ\Rightarrow \left(x^{3} + \frac{1}{x^{3}}\right)^{2} = 4 \cos ^{2} 6 \theta
x6+1x6+2=4cos⁡26θx^{6} + \frac{1}{x^{6}} + 2 = 4 \cos ^{2} 6 \theta
=4cos⁡26θ−2=2(2cos⁡26θ−1)= 4 \cos ^{2} 6 \theta - 2 = 2\left(2 \cos ^{2} 6\theta - 1\right)
⇒x6+1x6=2cos⁡12θ\Rightarrow x^{6} + \frac{1}{x^{6} } = 2 \cos 12 \theta