If sin(theta/2) = 1/2(a+1/a), prove that: cos(theta) = -1/2(a^2 + 1/a^2)

Solution:

Given: $\sin\frac{\theta}{2} = \frac{1}{2}\left(a+\frac{1}{a}\right)$

To prove: $\cos\theta = -\frac{1}{2}\left(a^2 + \frac{1}{a^2}\right)$

We'll use the double angle formula for cosine:

$\cos(2\alpha) = 1 - 2\sin^2\alpha$

Let $\alpha = \frac{\theta}{2}$, then:

$\cos\theta = \cos\left(2 \times \frac{\theta}{2}\right) = 1 - 2\sin^2\frac{\theta}{2}$

First, let's find $\sin^2\frac{\theta}{2}$:

$\sin^2\frac{\theta}{2} = \left[\frac{1}{2}\left(a+\frac{1}{a}\right)\right]^2$

$= \frac{1}{4}\left(a+\frac{1}{a}\right)^2$

Now, let's expand $\left(a+\frac{1}{a}\right)^2$:

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

$= a^2 + 2 + \frac{1}{a^2}$

Therefore:

$\sin^2\frac{\theta}{2} = \frac{1}{4}\left(a^2 + 2 + \frac{1}{a^2}\right)$

$= \frac{1}{4}\left(a^2 + \frac{1}{a^2}\right) + \frac{1}{4} \times 2$

$= \frac{1}{4}\left(a^2 + \frac{1}{a^2}\right) + \frac{1}{2}$

Substituting back into the double angle formula:

$\cos\theta = 1 - 2\sin^2\frac{\theta}{2}$

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

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

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

Hence proved.