# Find $\cos(x+y)$ if $\sin(x)+\sin(y)= a$ and $\cos(x)+\cos(y)= b$

Find $\cos(x+y)$ if $\sin(x)+\sin(y)= a$ and $\cos(x)+\cos(y)= b$.

-
Jay, you should know that on this site it is considered impolite to issue commands to other users. It would help if you asked kindly, and also explained what you have tried so far, or what your thoughts are about the problem. –  Zev Chonoles Jul 12 '11 at 18:01
I've also edited your question so that the equations are in $\LaTeX$. –  Zev Chonoles Jul 12 '11 at 18:02
These questions sometimes have many different solutions. Sometimes you just have to play around with things. With sines and cosines, squaring things might help. Then, maybe take advantage of the fact that $\sin^2 u + \cos^2 u = 1$. –  idmercer Jul 12 '11 at 18:06
What have you tried? Have you expanded the left hand side to see if that gives any ideas? –  Ross Millikan Jul 12 '11 at 18:16
It could be a typo, the obvious thing to ask about is $\cos(x-y)$. –  André Nicolas Jul 12 '11 at 18:39

Square and add the two to get $$2 + 2 \cos(x-y) = a^2 + b^2$$ $$2 \cos(x-y) = a^2 + b^2 - 2$$ Square and subtract the two to get $$\cos(2x) + 2 \cos(x+y) + \cos(2y) = b^2 - a^2$$

Now, $$\cos(2x) + \cos(2y) = 2 \cos(x+y) \cos(x-y) = \cos(x+y) (a^2+b^2-2)$$

Hence, we get $$\cos(x+y) (a^2 + b^2) = b^2 - a^2$$ Hence, $$\cos(x+y) = \frac{b^2-a^2}{b^2+a^2}$$

-
+1 I used the same logic, but first I did a change of variables with $x=u+v$ and $y=u-v$ and proceeded the same way. I think yours is simpler. –  ja72 Jul 12 '11 at 19:58

The following identities should help you: $$\tan \bigg(\frac{{x + y}}{2}\bigg) = \frac{{\sin x + \sin y}}{{\cos x + \cos y}}$$ and $$\tan \bigg(\frac{{x + y}}{2}\bigg) = \pm \sqrt {\frac{{1 - \cos (x + y)}}{{1 + \cos (x + y)}}} .$$

-
For the first identity, see en.wikipedia.org/wiki/…. –  Shai Covo Jul 12 '11 at 18:28
For the second identity, see en.wikipedia.org/wiki/…. –  Shai Covo Jul 12 '11 at 18:29

If you know the identities $$\displaylines{\sin x+\sin y=2\sin{x+y\over2}\cos{x-y\over2}\cr\cos x+\cos y=2\cos{x+y\over2}\cos{x-y\over2}\cr\tan2A={2\tan A\over1-\tan^2A}\cr}$$ then dividing the first one by the second one you get $$\tan{x+y\over2}={a\over b}$$ whence the third one gives $$\tan(x+y)={2(a/b)\over1-(a/b)^2}$$ Now looking at a right triangle shows that if $\tan B=u/v$ then $\cos B=v/\sqrt{u^2+v^2}$, so with a little algebra you get $$\cos(x+y)={1-(a/b)^2\over1+(a/b)^2}={b^2-a^2\over b^2+a^2}$$

If you don't know those first two identities, they come from the identities for $\sin(r\pm s)$ and $\cos(r\pm s)$.

-

An approach using complex numbers/geometry:

If $c = b + ia$, and if $z = \cos x + i \sin x$ and $z_1 = c - z = \cos y + i \sin y$, you are looking at the real part of $w = \cos(x+y) + i \sin (x+y) = zz_1 = z(c-z)$ with the restriction that $|z| = 1$ and $|z_1| = |c - z| = 1$.

Now the points satisfying $|z| = 1$ and $|c-z| = 1$ are given by the intersection of two unit circles: one centered at the origin and the other at $c$.

Thus $w = z(c-z)$ is a constant dependent only on $c$: it is the product of the two intersection points and can be computed easily as follows:

Each point of intersection can be written as $\dfrac{c}{2} \pm d$, where $d$ is perpendicular to $c$ (Why?) i.e. $d = kic$ for some real constant $k$. (Why?) (It might help to draw a figure here).

Thus the product of points of intersection is $w = (\dfrac{c}{2} + d)(\dfrac{c}{2} - d) =\dfrac{c^2}{4} + k^2 c^2$.

This implies that $w$ is a multiple of $c^2$ and the argument of $w$ is twice that of $c$. Since $|w| = 1$, $w$ can be computed easily without having to worry about $k$.

We get $w = \cos 2 \alpha + i \sin 2 \alpha$ where $\alpha = \tan^{-1}(\frac{a}{b})$.

PS: We find that $|c|^2(1/4 + k^2) = 1$ and hence $d = (\sqrt{\frac{1}{|c|} - \frac{1}{4}})\ i c$ and so we can easily solve the given system of equations: i.e. find $\cos x, \cos y, \sin x, \sin y$

-