Question

If $\cos \theta =\left(\frac{1}{2}\right)\left(a+\frac{1}{a}\right)$, then the value of $\cos 3\theta$ is

• A. $\left(\frac{1}{8}\right)\left(a^3+\frac{1}{a^3}\right)$
• B. $\left(\frac{3}{2}\right)\left(a+\frac{1}{a}\right)$
• C. $\left(\frac{1}{2}\right)\left(a^3+\frac{1}{a^3}\right)$
• D. $\left(\frac{1}{3}\right)\left(a^3+\frac{1}{a^3}\right)$

Answer 1

The correct option is C

$\left(\frac{1}{2}\right)\left(a^3+\frac{1}{a^3}\right)$

Find the value of $\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{3}\mathit{\theta }$:

Given, $\cos \theta =\left(\frac{1}{2}\right)\left(a+\frac{1}{a}\right)$

As we know,

$\begin{array}{rcl}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{3}\mathit{\theta }\mathbf{}& \mathbf{=}& \mathbf{}\mathbf{4}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{3}}\mathit{\theta }\mathbf{}\mathbf{–}\mathbf{}\mathbf{3}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }\\ & =& cos\theta [4co{s}^2\theta –3]\\ & =& \left(\frac{1}{2}\right)\left(a+\frac{1}{a}\right)\left(4{\left[\left(\frac{1}{2}\right)\left(a+\frac{1}{a}\right)\right]}^2–3\right)\\ & =& \left(\frac{1}{2}\right)\left(a+\frac{1}{a}\right)\left[\left(\frac{4}{4}\right)\left(a^2+\frac{1}{a^2}+2\right)–3\right]\\ & =& \left(\frac{1}{2}\right)\left(a+\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}-1\right)\end{array}$

$\mathbf{\therefore }\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{3}\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\right)}\mathbf{(}{\mathbf{a}}^{\mathbf{3}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{a}}^{\mathbf{3}}}\mathbf{)}$ ; $\left[\mathbf{\because }{\mathit{x}}^{\mathbf{3}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{y}}^{\mathbf{3}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{(}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathit{y}\mathbf{)}\mathbf{(}{\mathit{x}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{y}}^{\mathbf{2}}\mathbf{}\mathbf{–}\mathbf{}\mathit{x}\mathit{y}\mathbf{)}\right]$

Hence, Option ‘C’ is Correct.