Clearly not Understanding Photon Motion When the motion of a photon is described, it's always shown as a sine wave. So, the question I have is this: A sine wave describes a change in direction as well as a change in acceleration. If a photon's straight line movement as it travels from point A to point B takes place in the form of a sine wave, then what force (if any) is applied to it to alter its course into the sine wave shape? In other words, what causes it to accelerate along the slopes of the sine wave and decrease in velocity (and reverse it's direction of movement) at the peaks?
 A: You are ignoring that the photon is a quantum mechanical elementary point particle  and in the particle table it has: zero mass , spin + or - to its direction of motion and energy=momentum in units where  c=1. Those are the only measureable quantities.
The sinusoidal description is the quantum mechanical wave function of the photon., a solution of  quantized Maxwell's equations. The complex conjugate square of the wavefunction gives the probability of the photon being at (x,y,z,t)
 
its Heisenberg uncertainty envelope.
The E and B fields are the E and B fields that will emerge from  a large number of photons in confluence, building the classical electromagnetic field , which does vary in space and time and can be measured.
Maybe this can help:

On the left is the build up of a polarized classical electromagnetic wave, the red arrow depicting the electric field maximum. The middle describes the photon, which individually has only forward and backward spin, nevertheless in confluence builds up the classical wave.
A: A photon is (classically) described by an electric- and magnetic field. You are confusing waves in the electric field with the photon waving in space. Let's ignore the magnetic field for a second to simplify things. The electric field is a vector field which simply means that every point in space is assigned a different vector:
$$\vec E(x,y,z)=\pmatrix{E_x(x,y,z)\\E_y(x,y,z)\\E_z(x,y,z)}.$$
As you can you see there are three functions, each depending on three coordinates. Since humans can only think in three dimensions this is hard to display naturally. To visualize this people often choose the following type of diagram, which is what likely confused you in the first place.

They considered a 1-dimensional slice of the entire space, here along the z-axis so $x=y=0$. This gives us a new function $\vec E(z)$. This still allows $\vec E$ to point in any direction it wants. We actually know that the electric field never points in the direction of travel, so $E_z=0$. If we know arbitrarily pick $\vec E$ to point in the y-direction we get the same situation as in the picture:
$$\vec E(z)=\pmatrix{0\\E_y(z)\\0}$$
Now we plug in a travelling sine-wave solution for $E_y$ to get exactly the solution in the picture:
$$E_y(z, t)=\sin{\left(\frac{2\pi}\lambda(z-ct))\right)}.$$
Again, you can assign a field value to every point in space. Not just on the z-axis. A well-known 3D solution is the plane wave. You get this by taking the above solution and extending it such that for every value of z the field is constant along x and y:
$$\vec E(x,y,z)=\pmatrix{0\\\sin{\left(\frac{2\pi}\lambda(z-ct))\right)}\\0}.$$
The photon is moving straight in the z-direction but the electric field is 'waving' in the y-direction.
Source of picture: http://physicsbuzz.physicscentral.com/2015/12/seeing-photons-in-new-light.html
