# 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...

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

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

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

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

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

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

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

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

I cannot do anything better than it.

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

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.

• Thank you very much for your edit on my questions. May 2 at 22:46

Hint: $$\cos t = \dfrac{e^{it}+e^{-it}}{2}, \sin t = \dfrac{e^{it}-e^{-it}}{2i}, t = \theta, \phi$$.

• Why do you think it makes things simpler ?
– user65203
Apr 30 at 9:53
• I know but why do you think it makes things simpler ? Did you try ?
– user65203
Apr 30 at 10:07
• 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.
– Ella
Apr 30 at 10:14
• This should be a comment, not an answer.
– user65203
Apr 30 at 10:33