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3.Trigonometrical Ratios, Functions and Identities
easy
यदि $\cos \theta = \frac{1}{2}\left( {a + \frac{1}{a}} \right),$ तब $\cos 3\theta $ का मान होगा
A
$\frac{1}{8}\left( {{a^3} + \frac{1}{{{a^3}}}} \right)$
B
$\frac{3}{2}\left( {a + \frac{1}{a}} \right)$
C
$\frac{1}{2}\left( {{a^3} + \frac{1}{{{a^3}}}} \right)$
D
$\frac{1}{3}\left( {{a^3} + \frac{1}{{{a^3}}}} \right)$
Solution
(c) $\because$ $\;\cos 3\theta = 4{\cos ^3}\theta – 3\cos \theta $
$\therefore \cos 3\theta = 4\frac{1}{{{2^3}}}{\left( {a + \frac{1}{a}} \right)^3} – 3\frac{1}{2}\left( {a + \frac{1}{a}} \right)$
$ \Rightarrow \cos 3\,\theta = \frac{1}{2}\left( {a + \frac{1}{a}} \right)\,\left[ {{{\left( {a + \frac{1}{a}} \right)}^2} – 3} \right]$
==> $\cos 3\theta = \frac{1}{2}\left( {{a^3} + \frac{1}{{{a^3}}}} \right)$.
Standard 11
Mathematics
