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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 \[\lambda =\]
Options
- \[\frac{1}{4}\]
- \[\frac{1}{2}\]
1
none of these
Solution
\[\text{ Given } : \]
\[\text{ cos } x = \frac{1}{2}\left( a + \frac{1}{a} \right) \]
\[\cos3x = \lambda\left( a^3 + \frac{1}{a^3} \right)\]
\[\text{ Now } , \]
\[ \cos^3 x = \frac{1}{8}\left[ a^3 + \frac{1}{a^3} + 3a\frac{1}{a}\left( a + \frac{1}{a} \right) \right]\]
\[ \Rightarrow \cos^3 x = \frac{1}{8}\left( a^3 + \frac{1}{a^3} + 3 \times 2\text { cos } x \right) \left[ \because \text { cos } x = \frac{1}{2}\left( a + \frac{1}{a} \right) \right]\]
\[ \Rightarrow \cos^3 x = \frac{1}{8}\left( \frac{\cos3x}{\lambda} + 6\text{ cos } x \right)\]
\[ \Rightarrow \cos^3 x = \frac{1}{8}\left( \frac{4 \cos^3 x - 3\text{ cos } x}{\lambda} + 6\text{ cos } x \right)\]
\[ \Rightarrow \cos^3 x = \frac{4 \cos^3 x}{8\lambda} - \frac{3\text{ cos } x}{8\lambda} + \frac{6\text{ cos } x}{8}\]
\[\text{ On comparing the powers of } \cos^3 x \text{ on both sides, we get} \]
\[1 = \frac{4}{8\lambda}\]
\[ \Rightarrow \lambda = \frac{1}{2}\]
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