If \( \cos x=\frac12\left(a+\frac1a\right) \), Find \( \lambda \)
Question
If
\[ \cos x=\frac12\left(a+\frac1a\right) \]
and
\[ \cos 3x=\lambda\left(a^3+\frac1{a^3}\right), \]
then \( \lambda \) is
(a) \( \frac14 \)
(b) \( \frac12 \)
(c) \( 1 \)
(d) none of these
Solution
Use the triple-angle identity:
\[ \cos 3x=4\cos^3x-3\cos x \]
Given
\[ \cos x=\frac12\left(a+\frac1a\right) \]
Therefore,
\[ \cos 3x = 4\left[\frac12\left(a+\frac1a\right)\right]^3 – 3\left[\frac12\left(a+\frac1a\right)\right] \]
\[ = \frac12\left(a+\frac1a\right)^3 -\frac32\left(a+\frac1a\right) \]
Using
\[ \left(a+\frac1a\right)^3 = a^3+\frac1{a^3} + 3\left(a+\frac1a\right) \]
Substituting,
\[ \cos 3x = \frac12\left[ a^3+\frac1{a^3} + 3\left(a+\frac1a\right) \right] -\frac32\left(a+\frac1a\right) \]
\[ = \frac12\left(a^3+\frac1{a^3}\right) \]
Comparing with
\[ \cos 3x = \lambda\left(a^3+\frac1{a^3}\right), \]
we get
\[ \lambda=\frac12 \]
Final Answer
\[ \boxed{\lambda=\frac12} \]
Hence, the correct option is (b) \( \frac12 \).