Compound angle formulae in 3D

Question

We know that for $\phi=a-b$ $$cos(\phi)=cos(a)\cdot cos(b)+sin(a) \cdot sin(b)$$ $$sin(\phi)=sin(a)\cdot cos(b)-cos(a)\cdot sin(b)$$ the formula for $cos(\phi)$ also have a similar version in 3D geometry $$ cos(\phi)=cos(a_x)\cdot cos(b_x) +cos(a_y)\cdot cos(b_y)+ cos(a_z)\cdot cos(b_z) $$ where $a_x, b_x, a_y,b_y,a_z,b_z$ are the angles two lines A and B will make the axes x,y and z respectively. My question is, is there a similarly simple enough formula for $sin(\phi)$?
Here is what I tried: $$ 1-cos^2(\phi)=1-(cos(a_x)\cdot cos(b_x) +cos(a_y)\cdot cos(b_y)+ cos(a_z)\cdot cos(b_z))^2 $$ $$ sin^2(\phi)=1-(1-sin^2(a_x))\cdot cos^2(b_x) -(1-sin^2(a_y))\cdot cos^2(b_y)- (1-sin^2(a_z))\cdot cos^2(b_z)-2(cos(a_x)\cdot cos(a_y)\cdot cos(b_x)\cdot cos(b_y)+cos(a_y)\cdot cos(b_y)\cdot cos(b_z)\cdot cos(a_z)+cos(a_x)\cdot cos(b_x)\cdot cos(b_z)\cdot cos(a_z))) $$ $$ sin^2(\phi)=sin^2(a_x)\cdot cos^2(b_x) +sin^2(a_y)\cdot cos^2(b_y)+sin^2(a_z)\cdot cos^2(b_z)-2(cos(a_x)\cdot cos(a_y)\cdot cos(b_x)\cdot cos(b_y)+cos(a_y)\cdot cos(b_y)\cdot cos(b_z)\cdot cos(a_z)+cos(a_x)\cdot cos(b_x)\cdot cos(b_z)\cdot cos(a_z))) $$ I couldn't proceed beyond this point, I then tried using vector cross product for two unit vectors as, I know $cos(\phi)$ can be derived using dot product, but since it gives a vector as an output I do not know if that is the correct approach, also upon taking absolute value of the output vector the formula was not a simple one.