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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...

$$\sin2x=2\sin x\cos x \qquad \sin^2x=\frac12(1-\cos2x)$$

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    $\begingroup$ These are already simple. Seems hard to do anything better. $\endgroup$
    – user65203
    Apr 30 at 9:48
  • $\begingroup$ It would look even neater if you put a backslash before "sin" and "cos". Like this: $\sin(\phi)$ instead of $sin(\phi)$. $\endgroup$ Apr 30 at 10:25
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$\cos^2(\theta)\sin(\phi)+\sin^2(\theta)\cos(\phi)=$

$=\left(1-\sin^2(\theta)\right)\sin(\phi)+\sin^2(\theta)\cos(\phi)=$

$=\sin(\phi)+\sin^2(\theta)\big(\cos(\phi)-\sin(\phi)\big)=$

$=\sin(\phi)+\sin^2(\theta)\big(\cos(\phi)+\cos(\phi+\frac{\pi}2)\big)=$

$=\sin(\phi)+2\sin^2(\theta)\big(\cos(\phi+\frac{\pi}4)\cos(\frac{\pi}4)\big)=$

$=\sin(\phi)+\sqrt2\sin^2(\theta)\cos\left(\phi+\frac{\pi}4\right)\;.$

I cannot do anything better than it.

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    $\begingroup$ @ella: This result has lost any symmetry. It is ugly. $\endgroup$
    – user65203
    Apr 30 at 11:35
  • $\begingroup$ You are absolutely right, my result is not symmetric, but can you simplify that expression without losing symmetry? $\endgroup$
    – Angelo
    Apr 30 at 11:39
  • $\begingroup$ @a: I already said no, and I don't see any simplification here. $\endgroup$
    – user65203
    Apr 30 at 12:26
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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.

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  • $\begingroup$ Thank you very much for your edit on my questions. $\endgroup$
    – Sebastiano
    May 2 at 22:46
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Hint: $\cos t = \dfrac{e^{it}+e^{-it}}{2}, \sin t = \dfrac{e^{it}-e^{-it}}{2i}, t = \theta, \phi$.

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    $\begingroup$ Why do you think it makes things simpler ? $\endgroup$
    – user65203
    Apr 30 at 9:53
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    $\begingroup$ I know but why do you think it makes things simpler ? Did you try ? $\endgroup$
    – user65203
    Apr 30 at 10:07
  • $\begingroup$ thank you for your answer. I don't think translating sin^2 with Euler is useful here. ill have to square my imaginary e-functions too. i believe it will get messy. $\endgroup$
    – Ella
    Apr 30 at 10:14
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    $\begingroup$ This should be a comment, not an answer. $\endgroup$
    – user65203
    Apr 30 at 10:33

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