If cos(B/3) = 1/2(m+1/m) then, prove that: cosB = 1/2(m^3 + 1/m^3)

Solution:

Given: $\cos\frac{B}{3} = \frac{1}{2}\left(m+\frac{1}{m}\right)$

To prove: $\cos B = \frac{1}{2}\left(m^3 + \frac{1}{m^3}\right)$

We'll use the triple angle formula for cosine:

$\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$

Let $\theta = \frac{B}{3}$, then:

$\cos B = \cos\left(3 \times \frac{B}{3}\right) = 4\cos^3\frac{B}{3} - 3\cos\frac{B}{3}$

Substituting the given value $\cos\frac{B}{3} = \frac{1}{2}\left(m+\frac{1}{m}\right)$:

$\cos B = 4 \times \left[\frac{1}{2}\left(m+\frac{1}{m}\right)\right]^3 - 3 \times \frac{1}{2}\left(m+\frac{1}{m}\right)$

$= 4 \times \frac{1}{8}\left(m+\frac{1}{m}\right)^3 - \frac{3}{2}\left(m+\frac{1}{m}\right)$

$= \frac{1}{2}\left(m+\frac{1}{m}\right)^3 - \frac{3}{2}\left(m+\frac{1}{m}\right)$

Now, let's expand $\left(m+\frac{1}{m}\right)^3$:

$\left(m+\frac{1}{m}\right)^3 = m^3 + 3m^2 \times \frac{1}{m} + 3m \times \frac{1}{m^2} + \frac{1}{m^3}$

$= m^3 + 3m + \frac{3}{m} + \frac{1}{m^3}$

$= m^3 + \frac{1}{m^3} + 3\left(m + \frac{1}{m}\right)$

Substituting back:

$\cos B = \frac{1}{2}\left[m^3 + \frac{1}{m^3} + 3\left(m + \frac{1}{m}\right)\right] - \frac{3}{2}\left(m+\frac{1}{m}\right)$

$= \frac{1}{2}\left(m^3 + \frac{1}{m^3}\right) + \frac{3}{2}\left(m + \frac{1}{m}\right) - \frac{3}{2}\left(m+\frac{1}{m}\right)$

$= \frac{1}{2}\left(m^3 + \frac{1}{m^3}\right) + \frac{3}{2}\left(m + \frac{1}{m}\right) - \frac{3}{2}\left(m + \frac{1}{m}\right)$

$= \frac{1}{2}\left(m^3 + \frac{1}{m^3}\right)$

Hence proved.