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The equation $$ \sin(\theta) = \sin(\phi) $$ has the set of solutions $$ \big\{\, (\theta, \phi) \in \mathbb{R}^{2} : \theta - \phi \equiv 0 \,\,\mathrm{(mod}\,\,2\pi) \hspace{10pt}\text{or}\hspace{10pt} \theta + \phi \equiv \pi \,\,\mathrm{(mod}\,\,2\pi) \,\big\},\tag{$1$} $$ and similarly the equation $$ \cos(\theta) = \cos(\phi) $$ has the set of solutions $$ \big\{\, (\theta, \phi) \in \mathbb{R}^{2} : \theta \pm \phi \equiv 0 \,\,\mathrm{(mod}\,\,2\pi) \,\big\}.\tag{$2$} $$

The solution set $(1)$ can be written equivalently as $$ \theta = n\pi + (−1)^{n}\phi $$ for some integer $n \in \mathbb{Z}$.

I am similarly trying to find the solutions to the (separate) equations $$ \sin(\theta) = -\sin(\phi) \quad\text{and}\quad \cos(\theta) = -\cos(\phi) $$ in terms of $\theta$ and $\phi$.

Workings

Since $\sin$ is an odd function, $$ \sin(\theta) = -\sin(\phi) = \sin(-\phi) $$ has the set of solutions $$ \big\{\, (\theta, \phi) \in \mathbb{R}^{2} : \theta + \phi \equiv 0 \,\,\mathrm{(mod}\,\,2\pi) \hspace{10pt}\text{or}\hspace{10pt} \theta - \phi \equiv \pi \,\,\mathrm{(mod}\,\,2\pi) \,\big\}\tag{3} $$ by substituting $-\phi$ in place of $\phi$ in $(1)$.

By drawing key points on the cosine graph, the set of solutions to the cosine equation is $$ \big\{\, (\theta, \phi) \in \mathbb{R}^{2} : \theta \pm \phi \equiv \pi \,\,\mathrm{(mod}\,\,2\pi) \,\big\}.\tag{4} $$ I would like some clarification that this is correct. I have checked them numerically for a range of values, and it is sensible that these should be the answers.

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  • $\begingroup$ Seems correct to me. $\endgroup$ – SinTan1729 Jul 8 '18 at 16:01
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    $\begingroup$ Looks fine now. $\endgroup$ – David K Jul 8 '18 at 16:53
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Your equations say that $$e^{i\theta}=\cos\theta+i\sin\theta=-(\cos\phi+i\sin\phi)=-e^{i\phi}=e^{i(\phi+\pi)}\ .$$ This is true iff $$\theta-\phi=(2n+1)\pi,\qquad n\in{\mathbb Z}\ .$$

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