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

Question

2422 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-07T14:06:58

#(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) ) #

Answer 2 · 2017-05-07T14:19:54

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

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)#