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3.Trigonometrical Ratios, Functions and Identities
easy
यदि $x + \frac{1}{x} = 2\,\cos \theta ,$ तो ${x^3} + \frac{1}{{{x^3}}} = $
A
$\cos \,\,3\theta $
B
$2\,\cos \,3\theta $
C
$\frac{1}{2}\cos \,3\theta $
D
$\frac{1}{3}\cos \,3\theta $
Solution
(b) यहाँ $x + \frac{1}{x} = 2\cos \theta $,
अब ${x^3} + \frac{1}{{{x^3}}} = {\left( {x + \frac{1}{x}} \right)^3} – 3x\frac{1}{x}\left( {x + \frac{1}{x}} \right)$
$= {(2\cos \theta )^3} – 3(2\cos \theta ) = 8{\cos ^3}\theta – 6\cos \theta $
$= 2(4{\cos ^3}\theta – 3\cos \theta ) = 2\cos 3\theta $.
ट्रिक : $x = 1$ $ \Rightarrow $ $\theta = {0^\circ }$.
तब ${x^3} + \frac{1}{{{x^3}}} = 2 = 2\cos 3\theta $.
Standard 11
Mathematics
