Question

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

• A. $2\cos 6\theta$
• B. $2\cos 12\theta$
• C. $2\cos 3\theta$
• D. $2\sin 3\theta$

Answer 1

The correct option is B

$2\cos 12\theta$

Explanation for the correct option:

Step 1:Find the value of $x+\frac{1}{x}$using given relation

$\begin{array}{c}\Rightarrow x+\frac{1}{x}+2=4{\cos }^2\theta \\ \Rightarrow x+\frac{1}{x}=4{\cos }^2\theta -2\\ \Rightarrow x+\frac{1}{x}=2(2{\cos }^2\theta -1)\\ \Rightarrow x+\frac{1}{x}=2(\cos 2\theta )\mathrm{......}\left(ii\right)\end{array}$

Step 2: Find $x^2+\frac{1}{x^2}$, using equation $\left(ii\right)$

$\begin{array}{c}\Rightarrow {\left(x+\frac{1}{x}\right)}^2={\left(2\cos 2\theta \right)}^2\\ \Rightarrow x^2+\frac{1}{x^2}+2=4{\cos }^{2}2\theta \\ \Rightarrow x^2+\frac{1}{x^2}=4{\cos }^{2}2\theta -2\\ \Rightarrow x^2+\frac{1}{x^2}=2(2{\cos }^{2}2\theta -1)\\ \Rightarrow x^2+\frac{1}{x^2}=2\cos 2\left(2\theta \right)\\ \Rightarrow x^2+\frac{1}{x^2}=2\cos 4\theta .......\left(iii\right)\end{array}$

Step 3: Find $x^6+\frac{1}{x^6}$, using relation found in equation $\left(iii\right)$

$\begin{array}{c}\Rightarrow {\left(x^2+\frac{1}{x^2}\right)}^3={(2\cos 4\theta )}^3\\ \Rightarrow x^6+\frac{1}{x^6}+3·x^2\cdot \frac{1}{x^2}\left(x^2+\frac{1}{x^2}\right)=8{\cos }^{3}4\theta \end{array}$

From equation $\left(iii\right)$, $\begin{array}{c}{x}^2+\frac{1}{x^2}=2\cos 4\theta \end{array}$, then the above equation becomes,

$\begin{array}{c}\therefore x^6+\frac{1}{x^6}+3(2\cos 4\theta )=8{\cos }^{3}4\theta \\ \Rightarrow x^6+\frac{1}{x^6}=8{\cos }^{3}4\theta -6\cos 4\theta \\ =2(4{\cos }^{3}4\theta -3\cos 4\theta )\\ =2\cos 3\left(4\theta \right)\\ =2\cos 12\theta \end{array}$

Therefore, $x^6+\frac{1}{x^6}=2\cos 12\theta$.

Hence, the correct option is (B).