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

2 Answers
May 7, 2017

(mathbf a - mathbf b) * (mathbf a - mathbf b) = mathbf a * mathbf a + mathbf b * mathbf b - 2 mathbfa * mathbf b

= abs(mathbf a)^2 + abs(mathbf b)^2 - 2 abs mathbf a * abs mathbf b cos theta

If abs mathbf a = abs mathbf b, then:

= 2 abs(mathbf a)^2 (1- cos theta)

Now: cos A = 1 - 2 sin^2 (A/2)

So putting it all together:

(mathbf a - mathbf b) * (mathbf a - mathbf b) = 2 abs(mathbf a)^2 ( 2 sin^2 (theta/2))

4 abs(mathbf a)^2 sin^2 (theta/2) = abs(mathbf a - mathbf b)^2

sin (theta/2) = (abs(mathbf a - mathbf b))/( 2 abs(mathbf a) )

May 7, 2017

sin(theta/2) = +-sqrt(1 - (((veca+vecb)*veca)/(|(veca+vecb)||veca|))^2)

Explanation:

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

The reference tells us that the vector, veca|vecb|+vecb|veca|, bisects the angle between the two vectors, veca and vecb.

Therefore, the dot-product between the bisector and the either vector must contain the cosine of theta/2.

The following is the dot-product with veca:

(veca|vecb|+vecb|veca|)*veca=|(veca|vecb|+vecb|veca|)||veca|cos(theta/2)

Because veca and vecb are the same magnitude we can write this as:

(veca|veca|+vecb|veca|)*veca=|(veca+vecb)||veca|^2cos(theta/2)

Divide both sides by |veca|:

(veca+vecb)*veca=|(veca+vecb)||veca|cos(theta/2)

Solve for cos(theta/2):

cos(theta/2) = ((veca+vecb)*veca)/(|(veca+vecb)||veca|)

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

sin(theta/2) = +-sqrt(1 - (((veca+vecb)*veca)/(|(veca+vecb)||veca|))^2)