# Independent solutions to $\sin(\theta) = -\sin(\phi)$ and $\cos(\theta) = -\cos(\phi)$

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.

• Seems correct to me. – SinTan1729 Jul 8 '18 at 16:01
• Looks fine now. – David K Jul 8 '18 at 16:53

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 $$\theta-\phi=(2n+1)\pi,\qquad n\in{\mathbb Z}\ .$$