Simplifying $\cos^2\theta\sin\phi+\sin^2\theta\cos\phi$ or $\sin^2\theta\sin\phi-\cos^2\theta\cos\phi$ Can anyone see a way to simplify one of these expressions? It looks so neat, there has got to be a way! :D
$$\cos^2\theta\sin\phi+\sin^2\theta\cos\phi \tag1$$
or
$$\sin^2\theta\sin\phi-\cos^2\theta\cos\phi \tag2$$
I've tried identities like the following, but am only making whole thing more complicated...
$$\sin2x=2\sin x\cos x \qquad \sin^2x=\frac12(1-\cos2x)$$
 A: $\cos^2(\theta)\sin(\phi)+\sin^2(\theta)\cos(\phi)=$
$=\left(1-\sin^2(\theta)\right)\sin(\phi)+\sin^2(\theta)\cos(\phi)=$
$=\sin(\phi)+\sin^2(\theta)\big(\cos(\phi)-\sin(\phi)\big)=$
$=\sin(\phi)+\sin^2(\theta)\big(\cos(\phi)+\cos(\phi+\frac{\pi}2)\big)=$
$=\sin(\phi)+2\sin^2(\theta)\big(\cos(\phi+\frac{\pi}4)\cos(\frac{\pi}4)\big)=$
$=\sin(\phi)+\sqrt2\sin^2(\theta)\cos\left(\phi+\frac{\pi}4\right)\;.$
I cannot do anything better than it.
A: The expressions are pretty simple. You could write
$$\begin{align}
f :=\; \cos^2\theta\sin\phi+\sin^2\theta\cos\phi &= \phantom{-}\frac1{\sqrt{2}} \left(\sin\left(\frac\pi4+\phi\right)- \cos2\theta\sin\left(\frac\pi4-\phi\right)\right) \\[6pt]
g:=\; \sin^2\theta\sin\phi-\cos^2\theta\cos\phi &= -\frac1{\sqrt{2}} \left(\sin\left(\frac\pi4-\phi\right)+\cos2\theta\sin\left(\frac\pi4+\phi\right)\right)
\end{align}$$ but those are not what I'd call "simpler". (That said, if your context lends some significance to the quantity $\phi\pm\pi/4$, then there could be some benefit here.)
Depending upon your needs, it could be useful to write them as
$$\begin{align}
f &= \phantom{-}\cos^2\theta\cos\phi(\tan^2\theta+\tan\phi)\\[4pt]
g &= -\cos^2\theta\cos\phi(1-\tan^2\theta\tan\phi)
\end{align}$$
Those may-or-may-not seem better, but note that the ratio $f/g$ looks an awful lot like (the negative of) the angle-addition formula for tangent ... if only we had $\tan^2\theta=\tan\psi$ for some $\psi$. But in that case, we'd have
$$\begin{align}
\cos^2\theta &= \frac{1}{1+\tan^2\theta} = \frac{1}{1+\tan\psi}=\frac{\cos\psi}{\sin\psi+\cos\psi}=\frac{\cos\psi}{\sqrt{2}\sin(\psi+\frac\pi4)} \\[6pt]
\sin^2\theta &= \frac{\sin\psi}{\sqrt{2}\sin(\psi+\frac\pi4)}
\end{align}$$
whereupon, we'd obtain
$$f=\frac{\sin(\psi+\phi)}{\sqrt2\sin(\psi+\frac\pi4)} \qquad\qquad
g = -\frac{\cos(\psi+\phi)}{\sqrt2\sin(\psi+\frac\pi4)}$$
Those are also not necessarily "simpler" than the original forms, and the $\tan^2\theta\to\tan\psi$ re-parameterization may not be appropriate for your particular needs, but they seem pretty neat to me. :)
We couuld even take it further, defining $\psi':=\psi+\pi/4$ and $\phi':=\phi-\pi/4$:
$$f = \frac{\sin(\psi'+\phi')}{\sqrt2\sin\psi'} \qquad\qquad g = -\frac{\cos(\psi'+\phi')}{\sqrt2\sin\psi'}$$
but there's a danger of veering too far outside the context of your investigation.
A: Hint: $\cos t = \dfrac{e^{it}+e^{-it}}{2}, \sin t = \dfrac{e^{it}-e^{-it}}{2i}, t = \theta, \phi$.
