Question

If $\cos x=\frac{1}{2}\left(a+\frac{1}{a}\right),$ and $\cos 3x=\lambda \left(a^3+\frac{1}{a^3}\right)$, then $\lambda =$
(a) $\frac{1}{4}$

(b) $\frac{1}{2}$

(c) 1

(d) none of these

Answer 1

(b) $\frac{1}{2}$
$$\begin{gathered} \mathrm{Given}: \\ \cos x=\frac{1}{2}\left(a+\frac{1}{a}\right) \\ \cos 3x=\lambda \left(a^3+\frac{1}{a^3}\right) \\ \mathrm{Now}, \\ {\cos }^{3}x=\frac{1}{8}\left[a^3+\frac{1}{a^3}+3a\frac{1}{a}\left(a+\frac{1}{a}\right)\right] \\ \Rightarrow {\cos }^{3}x=\frac{1}{8}\left(a^3+\frac{1}{a^3}+3\times 2\cos x\right)\left[\because \cos x=\frac{1}{2}\left(a+\frac{1}{a}\right)\right] \\ \Rightarrow {\cos }^{3}x=\frac{1}{8}\left(\frac{\cos 3x}{\lambda }+6\cos x\right) \\ \Rightarrow {\cos }^{3}x=\frac{1}{8}\left(\frac{4{\cos }^{3}x-3\cos x}{\lambda }+6\cos x\right) \\ \Rightarrow {\cos }^{3}x=\frac{4{\cos }^{3}x}{8\lambda }-\frac{3\cos x}{8\lambda }+\frac{6\cos x}{8} \\ \mathrm{On}\mathrm{comparing}\mathrm{the}\mathrm{powers}\mathrm{of}{\cos }^{3}x\mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ 1=\frac{4}{8\mathrm{\lambda }} \\ \Rightarrow \mathrm{\lambda }=\frac{1}{2} \end{gathered}$$