How to solve trigonometric equations in 2 variables?

Find out both $$x$$ and $$y$$, $$\cos(x)=-\cos(x+y)$$ I come up with this equation when I was finding out maxima and minima of a two variable function $$f(x,y)=\sin x+\sin y+\sin (x+y);$$ however I get a solution by hit and trial approach, that is $$x=y={\pi\over 3}$$ will satisfy this equation, but how to solve it, as I have many other problems of the same kind and this hit and trial is time consuming.

• Welcome to Math. SE. Next time, please use formatting Sep 23, 2015 at 6:43
• There are other solutions, such as $x=\pi, y=0$, and indeed for every $x$ there are values of $y$ which make $\cos(x)=-\cos(x+y)$. For example $y=\pi$ works for all $x$, but it is not the only answer. Sep 23, 2015 at 6:46
• @Shailesh how to do this formatting?
– Onix
Sep 24, 2015 at 14:48
• Do a help on MathJax. Also click edit on your question and you still start the learning process Sep 24, 2015 at 14:50

I would try $$x=a+b$$, $$x+y=a-b$$. Then \begin{align}\cos x&=\cos(a+b)=\cos a\cos b-\sin a\sin b\\ &=-\cos(x+y)=-\cos(a-b)=-\cos a\cos b-\sin a\sin b\end{align} So that brings us to $$2\cos a\cos b=0$$ so either $$\cos a = 0$$ in which case $$a=\left(n+\frac12\right)\pi$$, $$y=(a-b)-(a+b)=-2b$$, so $$x=\left(n+\frac12\right)\pi-\frac12y$$ or $$\cos b=0$$ so $$b=\left(n+\frac12\right)\pi$$, $$y=-(2n+1)\pi$$, and $$x=a-(2n+1)\pi$$.

Summarizing, we could have $$y=\text{anything}$$ but then $$x=\left(n+\frac12\right)\pi-\frac12y$$ or $$x=\text{anything}$$ but $$y=(2m+1)\pi$$.

EDIT: I guess another way of expressing the solution would be to say that $$x=\text{anything}$$ and $$y=(2n+1)\pi$$ or $$y=(2n+1)\pi-2x$$.

To find two variables you need two equations.

If you had tried $y = 0$ you would have get $x = \pi/2, 3\pi/2...$

On the top of your solution, finding $x$ and $y$ for: $$Cos(x)=-Cos(x+y)$$

I. If $y = 0$ then any x$\in$ $\Re$ is a solution.

II. By the trigonometric identity $Cos(x)=-Cos(x+\pi)$ we get that if $y = \pi$, x can be again any real number.

To solve these type of problems in general, I would look through the identities first and see if any of them are similar to my problem. Identities