# If vec a and vec b are two equal vectors inclined at the angle theta then value of sin (theta/2)=?

May 7, 2017

$\left(m a t h b f a - m a t h b f b\right) \cdot \left(m a t h b f a - m a t h b f b\right) = m a t h b f a \cdot m a t h b f a + m a t h b f b \cdot m a t h b f b - 2 m a t h b f a \cdot m a t h b f b$

$= {\left\mid m a t h b f a \right\mid}^{2} + {\left\mid m a t h b f b \right\mid}^{2} - 2 \left\mid m \right\mid a t h b f a \cdot \left\mid m \right\mid a t h b f b \cos \theta$

If $\left\mid m \right\mid a t h b f a = \left\mid m \right\mid a t h b f b$, then:

$= 2 {\left\mid m a t h b f a \right\mid}^{2} \left(1 - \cos \theta\right)$

Now: $\cos A = 1 - 2 {\sin}^{2} \left(\frac{A}{2}\right)$

So putting it all together:

$\left(m a t h b f a - m a t h b f b\right) \cdot \left(m a t h b f a - m a t h b f b\right) = 2 {\left\mid m a t h b f a \right\mid}^{2} \left(2 {\sin}^{2} \left(\frac{\theta}{2}\right)\right)$

$4 {\left\mid m a t h b f a \right\mid}^{2} {\sin}^{2} \left(\frac{\theta}{2}\right) = {\left\mid m a t h b f a - m a t h b f b \right\mid}^{2}$

$\sin \left(\frac{\theta}{2}\right) = \frac{\left\mid m a t h b f a - m a t h b f b \right\mid}{2 \left\mid m a t h b f a \right\mid}$

May 7, 2017

$\sin \left(\frac{\theta}{2}\right) = \pm \sqrt{1 - {\left(\frac{\left(\vec{a} + \vec{b}\right) \cdot \vec{a}}{| \left(\vec{a} + \vec{b}\right) | | \vec{a} |}\right)}^{2}}$

#### Explanation:

Here is a reference for the vector that bisects the angle between two vectors .

The reference tells us that the vector, $\vec{a} | \vec{b} | + \vec{b} | \vec{a} |$, bisects the angle between the two vectors, $\vec{a} \mathmr{and} \vec{b}$.

Therefore, the dot-product between the bisector and the either vector must contain the cosine of $\frac{\theta}{2}$.

The following is the dot-product with $\vec{a}$:

$\left(\vec{a} | \vec{b} | + \vec{b} | \vec{a} |\right) \cdot \vec{a} = | \left(\vec{a} | \vec{b} | + \vec{b} | \vec{a} |\right) | | \vec{a} | \cos \left(\frac{\theta}{2}\right)$

Because $\vec{a}$ and $\vec{b}$ are the same magnitude we can write this as:

$\left(\vec{a} | \vec{a} | + \vec{b} | \vec{a} |\right) \cdot \vec{a} = | \left(\vec{a} + \vec{b}\right) | | \vec{a} {|}^{2} \cos \left(\frac{\theta}{2}\right)$

Divide both sides by $| \vec{a} |$:

$\left(\vec{a} + \vec{b}\right) \cdot \vec{a} = | \left(\vec{a} + \vec{b}\right) | | \vec{a} | \cos \left(\frac{\theta}{2}\right)$

Solve for $\cos \left(\frac{\theta}{2}\right)$:

$\cos \left(\frac{\theta}{2}\right) = \frac{\left(\vec{a} + \vec{b}\right) \cdot \vec{a}}{| \left(\vec{a} + \vec{b}\right) | | \vec{a} |}$

Use the identity, sin(theta/2) = +-sqrt(1 - cos^2(theta/2):

$\sin \left(\frac{\theta}{2}\right) = \pm \sqrt{1 - {\left(\frac{\left(\vec{a} + \vec{b}\right) \cdot \vec{a}}{| \left(\vec{a} + \vec{b}\right) | | \vec{a} |}\right)}^{2}}$