Question

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

 

 

विकल्प

• \[\frac{1}{4}\]

   

• \[\frac{1}{2}\]

   

• 1

• none of these

Answer 1

\[\frac{1}{2}\]

\[\text{ Given } : \]

\[\text{ cos } x = \frac{1}{2}\left( a + \frac{1}{a} \right) \]

\[\cos3x = \lambda\left( a^3 + \frac{1}{a^3} \right)\]

\[\text{ Now } , \]

\[ \cos^3 x = \frac{1}{8}\left[ a^3 + \frac{1}{a^3} + 3a\frac{1}{a}\left( a + \frac{1}{a} \right) \right]\]

\[ \Rightarrow \cos^3 x = \frac{1}{8}\left( a^3 + \frac{1}{a^3} + 3 \times 2\text { cos } x \right) \left[ \because \text { cos } x = \frac{1}{2}\left( a + \frac{1}{a} \right) \right]\]

\[ \Rightarrow \cos^3 x = \frac{1}{8}\left( \frac{\cos3x}{\lambda} + 6\text{ cos } x \right)\]

\[ \Rightarrow \cos^3 x = \frac{1}{8}\left( \frac{4 \cos^3 x - 3\text{ cos } x}{\lambda} + 6\text{ cos } x \right)\]

\[ \Rightarrow \cos^3 x = \frac{4 \cos^3 x}{8\lambda} - \frac{3\text{ cos } x}{8\lambda} + \frac{6\text{ cos } x}{8}\]

\[\text{ On comparing the powers of } \cos^3 x \text{ on both sides, we get} \]

\[1 = \frac{4}{8\lambda}\]

\[ \Rightarrow \lambda = \frac{1}{2}\]