Trigometric identity conversion within an integral

[Ritorufon]

Homework Statement

The problem is finding the average value of momentum in an infinite potential well but the theory I understand, its the mathematical execution I'm having trouble with.

Homework Equations

The expectation value for the momentum is found using the conjugate formula

For odd solutions
\psi_n=Bcos(\frac{n\pi}{2a}x)

then the expectation value is

<p_x> =\int^{a}_{-a} B cos(\frac{n\pi}{2a}x) (-i\hbar \frac {d}{dx}) B cos(\frac{n\pi}{2a}x)

which is equivalent to

<p_x>= B^2(+{i}{\hbar})\frac{n\pi}{2a}\int^{a}_{-a}cos\frac{n\pi}{2a}xsin\frac{n\pi}{2a}x dx

heres what I don't get, in my notes it multipies the above equation by a half which yields

<p_x>=B^2(+{i}{\hbar})\frac{n\pi}{2a}\int^{a}_{-a}sin\frac{n\pi}{a}xdx

The Attempt at a Solution

I really don't understands why this comes about and can make no sense from the trig tables, I'm sure the answer is trivial, yet it still alludes me, if anyone could shed any light on this I would be very grateful

by the way this is my first post here so forgive me if the latex is formatted incorrectly! :/

 

[Ritorufon]

whoops the last equation should be

<p_x> = B^2 (+{i} {\hbar} ) \frac{n\pi}{4a} \int^{a}_{-a} sin \frac{n\pi}{a} x dx

 

[Sethric]

It's just the trig formula for double angle sine:

sin(2 \theta) = 2 cos(\theta) sin(\theta)

 

[AlexChandler]

Yes it comes from the angle addition formula

sin (2 \alpha ) = sin( \alpha + \alpha ) = sin( \alpha ) cos( \alpha ) + cos( \alpha ) sin ( \alpha ) = 2 sin( \alpha ) cos( \alpha )

and the angle addition formula comes from some geometry. you can see it here if you want.

http://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities

 

[Ritorufon]

I would understand if the equation was multiplied by 2 but its multipied by 1/2 as the fraction in the end equation is
\frac{n\pi}{4a},
and the integral inside is multiplied by 1/2 so the identity doesn't apply right?

or am i missing something stupidly obvious?

 

[Sethric]

What happens when you take the identity and divide it by 2?

 

[AlexChandler]

I would understand if the equation was multiplied by 2 but its multipied by 1/2 as the fraction in the end equation is
\frac{n\pi}{4a},
and the integral inside is multiplied by 1/2 so the identity doesn't apply right?

or am i missing something stupidly obvious?

It may be more helpful to see the formula as:

\frac{1}{2} sin( \alpha ) = sin( \frac{1}{2} \alpha ) cos( \frac{1}{2} \alpha )

 

[Ritorufon]

ah! stupidly obvious like I thought! thanks for your help and patience guys! that's been annoying me for weeks!