Question

If $cos\theta =\frac{1}{2}(a+\frac{1}{a})$ then $cos3\theta$ in terms of ${}^{\prime }{a}^{\prime }$=

• A. $\frac{1}{4}(a^3+\frac{1}{a^3})$
• B. $4(a^3+\frac{1}{a^3})$
• C. $\frac{1}{2}(a^3+\frac{1}{a^3})$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{2}(a^3+\frac{1}{a^3})$

Given

$\cos \theta =\frac{1}{2}(a+\frac{1}{a})$

$\cos 3\theta =4{\cos }^3\theta -3\cos \theta$

$⟹4{\left[\frac{1}{2}(a+\frac{1}{a})\right]}^3-3\left[\frac{1}{2}(a+\frac{1}{a})\right]$

$⟹\frac{4}{8}{(a+\frac{1}{a})}^3-\frac{3}{2}(a+\frac{1}{a})$

$⟹\frac{1}{2}(a^3+\frac{1}{a^3}+3a^2\frac{1}{a}+3a\frac{1}{a^2})-\frac{3}{2}(a+\frac{1}{a})$

$((a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2)$

$⟹\frac{1}{2}(a^3+\frac{1}{a^3}+3a+\frac{3}{a})-\frac{3}{2}(a-\frac{1}{a})$

$⟹\frac{1}{2}(a^3+\frac{1}{a^3}+3(a+\frac{1}{a})-3(a-\frac{1}{a}))$

$⟹\cos 3\theta =\frac{1}{2}(a^3+\frac{1}{a^3})$