Can anyone see a way to simplify one of these expressions? It looks so neat, there has got to be a way! 😀
$$\cos^2\theta\sin\phi+\sin^2\theta\cos\phi \tag1$$
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…
$$\sin2x=2\sin x\cos x \qquad \sin^2x=\frac12(1-\cos2x)$$
Best Answer
$\cos^2(\theta)\sin(\phi)+\sin^2(\theta)\cos(\phi)=$
$=\left(1-\sin^2(\theta)\right)\sin(\phi)+\sin^2(\theta)\cos(\phi)=$
$=\sin(\phi)+\sin^2(\theta)\big(\cos(\phi)-\sin(\phi)\big)=$
$=\sin(\phi)+\sin^2(\theta)\big(\cos(\phi)+\cos(\phi+\frac{\pi}2)\big)=$
$=\sin(\phi)+2\sin^2(\theta)\big(\cos(\phi+\frac{\pi}4)\cos(\frac{\pi}4)\big)=$
$=\sin(\phi)+\sqrt2\sin^2(\theta)\cos\left(\phi+\frac{\pi}4\right)\;.$
I cannot do anything better than it.