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

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    It could be a typo, the obvious thing to ask about is $\cos(x-y)$.2011-07-12

4 Answers 4

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

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    +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.2011-07-12
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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)}}} . $

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    For the second identity, see http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Double-.2C_triple-.2C_and_half-angle_formulae.2011-07-12
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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 $\cos(r\pm s)$.

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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 $\cos x, \cos y, \sin x, \sin y$